given teach trig ratio as a fraction in simplest form:sin(N)

The measure of the length and width of the rectangle is given as 135 feet and 92 feet respectively.

From the information given, we have the following proofs;

Length, l = 2x + 30 - 49

Length = 2x - 19

The formula for perimeter of a rectangle is given as;

Perimeter = 2( length + width)

Substitute the expressions into the formula

Perimeter =  2 ( x+ 15 + 2x - 19 )

Perimeter = 2 (3x - 4)

Perimeter = 6x - 8

We have that the area is  162 more than 27 times the perimeter, which is Area = 27 (perimeter )+ 163

Area = 27(6x-8) + 162

Expand the bracket

Area= 162x - 216 + 162

Area = 162x - 54

But we know that

Area = length × width

Substitute the expressions

Area = (x+15)(2x-19)

Area =  2x² - 19x +30x - 285

Area = 2x² + 11x - 285

Equate the two formulas for area

162x - 54 = 2x² +11x - 285

Collect like terms

2x² + 11x - 285 - 162x + 54 = 0

2x²  - 151x - 231 = 0

Solve the quadratic equation

(2x + 3)(x-77) = 0

Let's solve for x

x - 77 = 0

x = 77

The expression for the width;

Width = x+15

Width = 77 + 15

Width = 92 feet

The expression for the length

Length = 2(x+15) - 49

Length = 2 ( 77 + 15) - 49

Length = 154 + 30 - 49

Length = 184 - 49

Length = 135 feet

Thus, the measure of the length and width of the rectangle is given as 135 feet and 92 feet respectively.

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[tex] \underline{ \orange{\huge \boxed{ \frak{Answer : }}}}[/tex]

Let ,

[tex] \sf \large \color{purple} y = 2 {x}^{3} - 3 {x}^{2} - 12x + 10 \: --( \: 1 \: )[/tex]

[tex] \: \: \: [/tex]

Now , Diff wrt ' x ' , we get :

[tex] \sf \: \frac{dy}{dx} = \frac{d}{dx} (2 {x}^{3} - 3 {x}^{2} - 12x + 10) \\ \sf \: \sf \: \frac{dy}{dx} = \frac{d}{dx} \: 2(3 {x}^{2} ) - \frac{d}{dx} 3 {x}^{2} - \frac{d}{dx} 12x + \frac{d}{dx} 10 \\ \sf \: \frac{dy}{dx} =2(3 {x}^{2} ) - 3(2x) - 12(1) + 0 \\ \sf \: \frac{dy}{dx} =6 {x}^{2} - 6x - 12 + 0 \\ \: \sf \red{\frac{dy}{dx} = 6 {x}^{2} 6x - 12 -- (2)}[/tex]

[tex] \: \: \: [/tex]

For maxima or minima \frac{dy}{dx} = 0

[tex] \: \: \: [/tex]

[tex] \sf \: 6 {x}^{2} - 6x - 12 = 0[/tex]

[tex] \: \: \: [/tex]

Divided by 6 on both side , we get.

[tex] \: \: \: [/tex]

[tex] \sf \: {x}^{2} - x - 2 = 0 \\ \sf \: {x}^{2} - 2x + x - 2 = 0 \\ \sf \: x(x - 2) + 1(x - 2) = 0 \\ \sf \: (x - 2)(x + 1) = 0 \\ \sf \: x - 2 = 0 \: \: \bold or \: \: x + 1 = 0 \\ \sf \fbox{x = 2 \: } \: \bold or \: \fbox{ x = - 1}[/tex]

[tex] \: \: \: [/tex]

Again Diff wrt ‘ x ’ , we get.

[tex] \sf \: \frac{d}{dx} =(\frac{dy}{dx} ) = 6\frac{d}{dx} - 6\frac{d}{dx}x - \frac{d}{dx}12 \\ \sf \: \frac{ {d}^{2}y }{ {dx}^{2} } = 6(2x) - 6(1) - 0 \\ \sf \: \sf \bold{ \frac{ {d}^{2}y }{ {dx}^{2} } =12x - 6}[/tex]

[tex] \: \: \: [/tex]

At x = 2

[tex] \: \: \: [/tex]

[tex]\sf \: \frac{ {d}^{2}y }{ {dx}^{2} } =12(2) - 6 \\ \: \: \: \sf \: = 24 - 6 \\ \: \: \: \: \sf \red{ = 18 > 0}[/tex]

At x = -1

[tex] \: \: \: [/tex]

[tex]\sf \: \frac{ {d}^{2}y }{ {dx}^{2} } =12( - 1) - 6 \\ \: \: \: \sf \: = - 12 - 6 \\ \: \: \: \: \sf \red{ = - 18 < 0 }[/tex]

[tex] \: \: \: [/tex]

x = 2 gives minima value of function.

[tex] \: \: \: [/tex]

x = -1 gives maxima value of function.

[tex] \: \: \: [/tex]

Now, put x = 2 in eqⁿ ( 1 )

[tex] \: \: \: [/tex]

[tex] \sf \: y \: minima \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2( {2})^{3} - 3 ({2})^{2} - 12(2) + 10 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: \: \: \: = 2(8) - 3(4) - 24 + 10 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: \: \: \: \: = 16 - 12 - 24 + 10 \\\sf \: \: \: \: \: \: \: \: \: \: = - 20 + 10 \\\sf \color{red}{\boxed{ = - 10}}[/tex]

[tex] \: \: \: [/tex]

The Point of minima is ( 2 , -10 ).

[tex] \: \: \: [/tex]

Now , put x = -1 in eqⁿ ( 1 )

[tex]\sf \: y \: maxima \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2( { - 1})^{3} - 3 ({ - 1})^{2} - 12( - 1) + 10 \\\sf \color{red}{\boxed{ = 17}}[/tex]

[tex] \: \: \: [/tex]

The point of maxima value is ( -1 , 17 ).

[tex] \: \: \: [/tex]

[tex] \: \: [/tex]

Hope Helps! :)

Answer:

5

Step-by-step explanation:

(2x - 6) + 1 + (x - 4) = 18

3x - 9 = 18

3x = 27

x = 9

RS = x - 4 = 9 - 4 = 5

Answer:

B. 5

Step-by-step explanation:

Pls mark brainiest if helpful!!

PQ + QR + RS = 18

This statement is equivalent to the following:

(2x - 6) + 1 + (x - 4) = 18.

This is because we substitute the line segments with their values.

Now that we have an equation to work with, we will solve it. We start by removing the parenthesis as they are not necessary here.

2x - 6 + 1 + x - 4 = 18

Next, we will make it, so x is alone on the left. We will do this by adding 6 and 4 to both sides

2x - 6 (+6) + 1 + x - 4 (+4) = 18 (+6) (+4)

After doing this we are left with the following equation:

2x + 1 + x = 28

Now we will subtract 1 from both sides to make x alone of the left side of the equation.

2x + 1 (-1) + x = 28 (-1)

After we solve this, we get the following:

2x + x = 27

We can add 2x and x and get 3x because 2 + 1 = 3.

3x = 27

We now simplify this equation by dividing both sides by 3.

[tex]\frac{3x}{3}[/tex] = [tex]\frac{27}{3}[/tex]

We again simplify the equation to the following:

x = 9

BUT THATS NOT THE ANSWER!

Now that we know what x is, we can solve for the value of RS.

We input 9 for x in the equation x - 4.

When we do this, we get 9 - 4 which is 5.

And this is our final answer!!


I hope I could help! Have an amazing day and good luck on your homework!

"You can't predict the future, but you can create it"
            -Juliana Palazzolo, 2022

Pls mark brainiest if helpful!!


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Answer:

30

Step-by-step explanation:

Since ABC is isosceles, BC = CA.

So, AB = 9(3)+3 = 30.

Monthly interest payment is $100.

According to the statement

balance in account = $100

APR = 12%

we use the formula Balance formula to solve this problem.

So, BALANCE * [APR / 12 month] to find monthly interest payment

Here balance is the starting amount in the bank account

And APR is the type of interest rate applied on the amount

And 12 month is the time period for which APR is applicable.

So, substitute the values in it then

Monthly interest payment = 100 * [12/12]

After solving the equation become

Monthly interest payment = 100*1

Monthly interest payment = 100$

So, Monthly interest payment is $100.

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The area of the paper that remains is 391.04 cm²

From the question, we are to calculate the area of the remaining part of the paper

Area of the paper that remains = Area of rectangle - 2×Area of semicircle

Area of the paper that remains = (l×w) - 2(πr²/2)

Area of the paper that remains = (l×w) - (πr²)

Where l is the length of the paper

w is the width if the paper

and r is the radius of the semicircle

From the given information,

l = 37 cm

w = 16 cm

r = w/2 = 16/2 = 8 cm

Putting the parameters into the equation,

Area of the paper that remains = (37×16) - (3.14(8)²)

Area of the paper that remains = 592 - (3.14×64)

Area of the paper that remains = 592 - 200.96

Area of the paper that remains = 391.04 cm²

Hence, the area of the paper that remains is 391.04 cm²

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The percentage error of the given estimation is 17.7%.

The percentage error of any estimation is calculated as:

Percentage error = {(|Actual Value - Estimated Value|)/Actual Value}*100%.

In the question,

The estimated value is given to be 40 marbles.

The actual value is given to be 34 marbles.

We are asked to find the percentage error of this estimation.

We know that:

Percentage error = {(|Actual Value - Estimated Value|)/Actual Value}*100%.

Substituting the values, we get:

Percentage error = {(|34 - 40|)/34}*100%,

or, Percentage error = 6/34*100%,

or, Percentage error = 17.6470 % = 17.7% (Rounding to the nearest tenth, that is, up to one decimal place).

Thus, the percentage error of the given estimation is 17.7%.

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A paragraph proof D. contains a set of sentences explaining the steps needed to reach a conclusion.

It should be noted that a paragraph proof simply means a way of presenting a mathematical proof.

In this case, it contains a set of sentences explaining the steps needed to reach a conclusion.

In conclusion, the correct option is D.

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Answer:

D: contains a set of sentences explaining the steps needed to reach a conclusion.  

The graph of a function f exists the set of all points in the plane of the form (x, f(x)).

The graph of a function f exists the set of all points in the plane of the form (x, f(x)). We could even describe the graph of f to be the graph of the equation y = f(x). So, the graph of a function exists as a special case of the graph of an equation.

Given: [tex]h(t)=0.6\cdot 3.2^t[/tex]

Substitute the values in the function, we get

t = -2, h(t) = 0.05859375

t = -1, h(t) = 0.1875

t = 0, h(t) = 0.6

t = 1, h(t) = 1.92

t = 2, h(t) = 6.144

t = 3, h(t) = 19.6608

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Yes, because both figures are rectangles and all rectangles are similar. Option B is correct option.

According to the statement

We have given that the two rectangles EFGH and ABCD and we have show that the these rectangles Are dilation at origin or not.

So, According to the given diagram

Yes, the rectangle EFGH is a result of dilation of rectangle ABCD with a center of dilation of rectangle at the origin.

Also The scale factor of the dilation is greater than one as the image is bigger than the pre-image i.e. there is a stretch.

The scale factor could be calculated by the ratio of the sides of the image to the pre-image rectangle.

According to the diagram In Rectangle ABCD:

Its vertices have coordinates A(-3,3), B(3,3), C(3,0) and D(-3,0).

Now consider rectangle EFGH:  

Its vertices have coordinates E(-4,4), F(4,4), G(4,0) and H(-4,0).

Hence the scale factor is becomes from these values is:

EF/AB = EH/AD = FG/BC = HG/DC = 4/3.

Hence the scale factor becomes 4/3.

Also

∠A=∠B=∠C=∠D=∠E=∠F=∠G=∠H=90°

( When a shape is dilated the two shapes are similar.And similar shapes have equal interior angles , corresponding sides are proportional ).

So, Yes, because both figures are rectangles and all rectangles are similar. Option B is correct option.

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First, find the supplement of the angle that is 105 degrees.

105 + 75 = 180

Now, we can find complete the lower left triangle's angles.

39 + 75 + ? = 180

? = 66 degrees

The 66 degree angle of the lower left triangle and angle x are vertical angles. Vertical angles are congruent.

Therefore, the measure of angle x is 66 degrees.

Hope this helps!

Answer:

- 11 < x < 8

Step-by-step explanation:

inequalities of the type | x | < a have solutions of the form

- a < x < a

then

| 2x + 3 | < 19

- 19 < 2x + 3 < 19 ( subtract 3 from each interval )

- 22 < 2x < 16 ( divide each interval by 2 )

- 11 < x < 8

The exponential function that models this situation is:

[tex]D(t) = 10000(1.045)^t[/tex]

Using the function, in 2050, she will have $37,453.

An increasing exponential function is modeled by:

[tex]A(t) = A(0)(1 + r)^t[/tex]

In which:

For this problem, the parameters are:

A(0) = 10000, r = 0.045.

Hence the equation is:

[tex]D(t) = 10000(1.045)^t[/tex]

2050 is 30 years after 2020, hence the amount is:

[tex]D(30) = 10000(1.045)^{30} = 37453[/tex]

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Answer:

First function:

Zeros = 6 and -6. Y-intercept = (0,72). X-intercepts = (6,0) and (-6,0).

Second function:

Zeros = 3 and -3. Y-intercept = (0, -27). X-intercepts: (3, 0) and (-3, 0).

Step-by-step explanation:

The zeros are x-intercept numbers.

The y-intercept can be found when the function is in standard form. Plug in 0 for x, then solve.

Hope this helps!

Answer:

1. x₁ = 6, x₂ = -6

2. x₁ = 3, x₂ = -3

Step-by-step explanation:

Given functions:

[tex]1)\ f(x) = -2(x-6)(x+6)[/tex]

[tex]2)\ f(x)=(x-3)(3x+9)[/tex]

..................................................................................................................................................

Zero Product Property: If m • n = 0, then m = 0 or n = 0.

Standard Form of a Quadratic: ax² + bx + c = 0.

..................................................................................................................................................

1. f(x) = -2(x - 6)(x + 6)

Step 1: Set the function to zero.

[tex]\implies 0 = -2(x-6)(x+6)[/tex]

Step 2: Divide both sides of equation by [tex]-2[/tex].

[tex]\implies \dfrac{0}{-2} = \dfrac{-2(x-6)(x+6)}{-2}[/tex]

[tex]\implies 0=(x-6)(x+6)[/tex]

Step 3: Apply the Zero Product Property.

[tex]x_1 \implies x-6=0[/tex]

[tex]x_2 \implies x+6=0[/tex]

Step 4: Solve for x in both equations.

[tex]x-6+6=0+6 \implies \boxed{x_1 = 6}[/tex]

[tex]x+6-6=0-6 \implies \boxed{x_1 = -6}[/tex]

The zeros (x-intercepts) of this function are: [tex]x_1=6,\ x_2=-6[/tex].

.................................................................................................................................................

2. f(x) = (x - 3)(3x + 9)

Step 1: Set the function to zero.

[tex]\implies 0 = (x - 3)(3x + 9)[/tex]

Step 2: Apply the Zero Product Property.

[tex]x_1 \implies x-3=0[/tex]

[tex]x_2 \implies 3x + 9=0[/tex]

Step 3: Solve for x in both equations.

[tex]x-3+3=0+3 \implies \boxed{x_1 = 3}[/tex]

[tex]3x=-9 \implies \dfrac{3x}{3}=\dfrac{-9}{3} \implies \boxed{x_1 = -3}[/tex]

The zeros (x-intercepts) of this function are: [tex]x_1=3,\ x_2=-3[/tex].

..................................................................................................................................................

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Answer: 32

Step-by-step explanation:

The line is straight meaning 180 angle. When finding B you subtract 148 by 180 which would be 32.

Answer:

D.

[tex] \frac{1}{16?} [/tex]

[tex]\huge\text{Hey there!}[/tex]

[tex]\huge\textbf{Equation: }[/tex]

[tex]\mathbf{2^{-4}}[/tex]

[tex]\huge\textbf{Simplify it: }[/tex]

[tex]\mathbf{2^{-4}}[/tex]

[tex]\mathbf{\approx \dfrac{1}{2^4}}[/tex]

[tex]\mathbf{= \dfrac{1}{2\times2\times2\times2}}[/tex]

[tex]\mathbf{= \dfrac{1}{4\times4}}[/tex]

[tex]\mathbf{= \dfrac{1}{16}}[/tex]

[tex]\huge\textbf{Therefore, your answer should be: }[/tex]

[tex]\huge\boxed{\frak{Option\ D. \ \dfrac{1}{16}}}\huge\checkmark[/tex]

[tex]\huge\text{Good luck on your assignment \& enjoy your day!}[/tex]

Answer:

  35:9

Step-by-step explanation:

The ratios can be combined by making the common element be represented by the same number.

The common element in the two ratios is the number of girls. In the first ratio, girls are represented by 3 ratio units. In the second ratio, they are represented by 7 ratio units. Multiplying the first ratio by 7 and the second ratio by 3 will make the number of girls the same in both.

  boys : girls = 2 : 3 = 14 : 21

  girls : teachers = 7 : 3 = 21 : 9

Then the extended ratio can be written ...

  boys : girls : teachers = 14 : 21 : 9

The number of students is the total of the numbers of boys and girls, so we have ...

  students : teachers = (14 +21) : 9

  students : teachers = 35 : 9

The Outcome of event A and B is simply the intersection of both sets written as A ∩ B is; C: {seagull, crow}

From the image, we are given the animals namely; Pig, Penguin, Seagull, Tiger, Crow, Bat, Mosquito.

Now, these animals are classified as either a bird or can fly.

Animals that are birds are; Penguin, Seagull, Crow

Animals that can fly are; Seagull, Crow, Bat, Mosquito.

Now, we are told that;

Event A is the animal is a bird. Thus, the set notation that represents this event A is written as;

A = {Penguin, Seagull, Crow}

Event B is that the animal can fly and again the set notation that represents event B is written as;

B = {Seagull, Crow, Bat, Mosquito}

Now, Outcome of event A and B is simply the intersection of both sets. Thus; A ∩ B = {seagull, crow}

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Answer is 5/4The slope is the number next to the “x” in the y intercept formula

y= m (slope) x + b

Answer: m= 5/4

Step-by-step explanation:

La ecuación general de la recta es y=mx+by=mx+b, donde m es la pendiente y b es la intersección en y

y=mx+by=mx+b

The scale factor that can be applied to Cone 1 to make Cone 2 is 0.8

This are constants that is used to enlarge of diminish a given figure of sides of a figure

We can determine the scale factor by finding the ratio of the similar sides of two figures, From the given cones, the ratio of their radius can determine the scale factor that is applied to Cone 1 to make Cone 2

From the given figure;

scale factor = radius of cone 2/radius of cone 1

Given the following parameters

radius of cone 2 = 2 ft

radius of cone 1 = 2.5ft

Substitute

scale factor =  2/2.5

scale factor = 0.8

Hence the scale factor that can be applied to Cone 1 to make Cone 2 is 0.8

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The equation of the line of best fit is y = 1.7x  - 58

We start by drawing the line of best fit (see attachment)

From the attached graph, we have the following points

(x, y) = (70, 75) and (61, 60)

The slope (m) is:

m = (y2 - y1)/(x2 - x1)

This gives

m = (60 - 75)/(61 - 70)

Evaluate

m = 1.7

The line of best fit is then calculated as:

y = m(x - x1) + y1

This gives

y = 1.7(x - 70) + 61

This gives

y = 1.7x - 119 + 61

Evaluate

y = 1.7x  - 58

Hence, the equation of the line of best fit is y = 1.7x  - 58

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Answer:

d

Step-by-step explanation:

the equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k ) are the coordinates of the centre and r is the radius

(x - 5 )² + (y + 1)² = 4 ← is in standard form

with centre (5, - 1 ) and r = [tex]\sqrt{4}[/tex] = 2

Answer: 3.217 & 12.434

Step-by-step explanation:

If we use w to represent the width, the length will be 6 more than 2 times w.

Hence, the length is [tex]2w+6[/tex].

The area of a rectangle would be its length times its width, so let's make an equation to represent it's area.

[tex]A=w(2w+6)[/tex]

We can also substitute 40 in for A as it's given in the question.

[tex]40 = w(2w+6)[/tex]

Distributing w by multiplying it by both terms in the parentheses, we get

[tex]40 = 2w^2+6w[/tex]

We can make the equation simpler by dividing both sides by 2.

[tex]20 = w^2+3w[/tex]

Subtracting both sides by 20 will make the left-hand side 0.

[tex]0=w^2+3w-20[/tex]

Now that we have put this quadratic equation into standard form (ax²+bx+c), we can find its solutions using the quadratic formula.

For reference, the quadratic formula is

[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]

In this case, a is 1, b is 3, and c is -20.

Substituting, we get

[tex]w=\frac{-3\pm\sqrt{3^2-4(1)(-20)}}{2(1)}[/tex]

[tex]w= \frac{-3\pm\sqrt{9+80}}{2}[/tex]

[tex]w=\frac{-3+\sqrt{89}}{2}\hspace{0.1cm}or\hspace{0.1cm}\frac{-3-\sqrt{89}}{2}[/tex]

Since the second solution results in a negative number, it cannot be the length of w.

[tex]w=\frac{-3+\sqrt{89}}{2}\approx3.217[/tex]

The width/breadth of the rectangle is 3.217 cm.

To calculate the length, let's substitute the width into the expression for the length:

[tex]l=2(3.217)+6[/tex]

[tex]l=12.434[/tex]

The length of this rectangle is 12.434 cm.

The other factor of the polynomial is a. 3x2-4x+11

Given polynomial is 3x3 + 20x2 - 21x + 88 and the factor is (x+8)

We need to find another factor using long division method

So,

We will divide the polynomial by the factor to find the another factor

Therefore,

[tex]\sqrt[x+8]{3x^3+20x^2-21x+88}[/tex]

Now calculating

First multiplying [tex]3x^2[/tex] with (x+8) so , [tex]3x^2[/tex] will be in the quotient

We get [tex]3x^3+24x^2[/tex]

simplifying the calculation for [tex]3x^3+20x^2[/tex] and [tex]3x^3+24x^2[/tex]

We get the remainder is [tex]-4x^2-21x+88[/tex]

Second we will multiply -4x with (x+8) Where -4x will be in the quotient

We get [tex]-4x^2-32x[/tex] and then we will simplify the equation

We get 11x +88 as a remainder

The quotient we get is [tex]3x^2-4x[/tex]

Third we will multiply +11 with (x+8) Where +11 will be in the quotient

we get 11x+88

Simplifying the equation we get the remainder 0

So the quotient we get is  (3x2 - 4x+11)

Hence the another factor of the polynomial is  (3x2 - 4x+11)

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abc =32 ft³  is the maximum volume of an open rectangular box (with no top face) if its surface area is 1 square foot.

A rectangle with all equal sides, commonly known as a square.  Multiplying the length by the length is the. Using L as the length of each side, solve for L X L = L2,

The maximum volume of an open rectangular box (with no top face) if its surface area is 1 square foot This Lagrange multiplier optimization is standard. If the box has a base of a, a height of c, and an area constraint of ab+2ac+2bc−48=0  we wish to optimize V= abc.

L(a,b,c,λ)= abc−λ(ab+2ac+2bc−48)

The four partial derivatives are zero at an ideal position, so:

δLδa=bc−λ(b+2c)=0

δLδb=ac−λ(a+2c)=0

δLδc=ab−λ(2a+2b)=0

Plus the restriction. The first two enlighten:

λ=bcb+2c=aca+2c

Consequently,  b(a+2c)=a(b+2c)  implies to b=a. The third partial, where b=a, now informs us that   a2=4aλ  and so  λ=a/4  nd by using this information in the second partial, we obtain  4c=a+2c  which informs us that c=a/2 .

Now that we've inserted these b and c expressions into the constraint, we get [tex]3a^2=1[/tex]  which means that a=4 feet, b=4 feet, and c=2 feet.

The maximum volume is therefore, abc=32 ft³

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Answer:

13 + [tex]\sqrt{89}[/tex]

or

22.43398... (depends on how you round it)

Step-by-step explanation:

1. The question stated that the radius of the circle O is 5, so the length of AO and CO is 5.

2. Since line AB and line CB are both tangent to the circle, they have the same length. CB is 8, so AB will also be 8.

--> Both triangle AOB and COB share one side, and the other side (radius) has the same length, so the third side must be the same length

3. Tangent means having a 90-degree angle with the radius. We know that the triangle AOB is a right triangle since the angle OAB is 90 degrees.

We can use the Pythagorean theorem to find side OB. OB^2 = AO^2 + AB^2.

--> OB^2 = 25 + 64

--> OB^2 = 89

--> OB = [tex]\sqrt{89}[/tex]

Now that we know the lengths of all three sides, we can add them up.

--> 5 + 8 + [tex]\sqrt{89}[/tex]

--> 13 + [tex]\sqrt{89}[/tex]

or

--> 22.43398113....

The total possible number of outcomes for the season record is [tex]2^{33}[/tex].

An event that may be given a mathematical probability is known as an outcome in mathematics (an outcome of an experiment). The amount of potential outcomes for a particular experiment determines the probability assigned to each result.The result of an experiment is a number of outcomes.

According to the question,

The basketball team at a college will play 33 games next winter. There are only two possible outcomes for each game: a win or a loss.

So, the total number of possible outcomes for the season record=[tex]2^{33}[/tex].

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The exponent Marina should use for b is: D. 7.

When dividing, you subtract the exponents. I.e. [tex]a^m \div a^n = a^{m - n}[/tex].

Using the rule, we would also subtract the exponents of b in the simplified expression. Thus:

[tex]b^4 \div b^{-3}[/tex]

Subtract the exponents

4 -(-3) = 4 + 3 = 7.

Therefore, the Maria should use the exponent b = 7.

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Answer:  D

Step-by-step explanation:

Answer:

21:14:8

Step-by-step explanation:

X:Y=3:2

Y:Z=7:4

X:Y=21:14

Y:Z=14:8

X:Y:Z=21:14:8

Question 4

1) [tex]\angle B[/tex] and [tex]\angle E[/tex] are right angles, [tex]\overline{AB} \cong \overline{DE}[/tex], [tex]\angle A \cong \angle D[/tex] (given)

2) [tex]\triangle CBA[/tex] and [tex]\triangle FED[/tex] are right triangles (a triangle with a right angle is a right triangle)

3) [tex]\triangle CBA \cong \triangle FED[/tex] (LA)

Question 5

1) [tex]\overline{XY} \perp\overline{WZ}[/tex], [tex]\overline{UV} \perp \overline{WZ}[/tex], [tex]\overline{VW} \cong \overline{YZ}[/tex], and [tex]\angle Z \cong \angle W[/tex] (given)

2) [tex]\angle XYZ[/tex] and [tex]\angle UVW[/tex] are right angles (perpendicular lines form right angles)

3) [tex]\triangle XYZ[/tex] and [tex]\triangle UVW[/tex] are right triangles (a triangle with a right angle is a right triangle)

4) [tex]\triangle XYZ \cong \triangle UVW[/tex] (LA)

5) [tex]\overline{UW} \cong \overline{XZ}[/tex] (CPCTC)

Question 6

1) [tex]\overline{PQ} \perp \overline{QT}[/tex], [tex]\overline{ST} \perp \overline{QT}[/tex], [tex]\overline{PQ} \perp \overline{ST}[/tex] (given)

2) [tex]\angle PQR[/tex] and [tex]\angle RTS[/tex] are right angles (perpendicular lines form right angles)

3) [tex]\triangle PQR[/tex] and [tex]\triangle STR[/tex] are right triangles (a triangle with a right angle is a right triangle)

4) [tex]\angle PRQ \cong \angle SRT[/tex] (vertical angles are congruent)

5) [tex]\triangle PQR \cong \triangle STR[/tex] (LA)

6) [tex]\overline{QR} \cong \overline{TR}[/tex] (CPCTC)