A rock climber climbs at a rate of 720 feet per hour. Write and solve an equation to find the number of minutes
m it takes for the rock climber to climb 288 feet.
An equation that represents this situation is

The set of inequalities y ≤ x² - 3 and y > -x² + 2 do not have a solution

From the graph, we have:

f(x) = x² - 3

g(x) = -x² + 2

Next, we change the equations to inequalities as follows:

y ≤ x² - 3

y > -x² + 2

To modify the graph, we then perform the following transformations:

After the modifications in (a), we have:

y ≤ x² - 3 and y > -x² + 2

Substitute y > -x² + 2 in y ≤ x² - 3

-x² - 2 ≤ x² - 3

This gives

2x² ≤ - 1

Divide by 2

x² ≤ - 0.5

The square root of numbers less than 0 is a complex number

Hence, the set of inequalities do not have a solution

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Find y intercept

Point(0,-7)

Find vertex

x coordinate

y coordinate

Vertex at (-2,-27)

Slope

Answer:

10

Step-by-step explanation:

The x-coordinate of the vertex of a quadratic equation in the form

[tex]f(x)=ax^2+bx+c\quad \textsf{is} \quad -\dfrac{b}{2a}[/tex]

Given function:

[tex]f(x)=5x^2+20x-7[/tex]

[tex]\implies a=5, \quad b=20, \quad c=-7[/tex]

x-coordinate of the vertex

[tex]\implies -\dfrac{b}{2a}=-\dfrac{20}{2(5)}=-2[/tex]

To find the y-coordinate of the vertex, substitute the found value of x into the function:

[tex]\begin{aligned}\implies f(-2) & =5(-2)^2+20(-2)-7\\& = 5(4)-40-7\\& = 20-47\\& = -27\end{aligned}[/tex]

Therefore, the coordinates of the vertex are (-2, -27).

The y-intercept is when the curve crosses the y-axis, so when x = 0.

To find the y-coordinate of the y-intercept, substitute x = 0 into the function:

[tex]\begin{aligned}\implies f(0) & =5(0)^2+20(0)-7\\& = 0 + 0-7\\& = -7\end{aligned}[/tex]

Therefore, the coordinates of the y-intercept are (0, -7).

To find the slope of the line passing through the vertex and the y-intercept, simply substitute the found points into the slope formula:

[tex]\implies \sf slope=\dfrac{change\:in\:y}{change\:in\:x}=\dfrac{-27-(-7)}{-2-0}=\dfrac{-20}{-2}=10[/tex]

Therefore, the slope of the line passing through the vertex and the y-intercept of the given quadratic function is 10.

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Answers:

i) [tex]\sf (x + 2)(x+ 3)[/tex]

ii) [tex]\sf \left(3x+1\right)\left(3x+2\right)\left(9x^2+9x-16\right)[/tex]

Factorize expression's:

i.

[tex]\sf (x + 1)^2 + 3(x + 1) + 2[/tex]

apply perfect square and distributive method

[tex]\sf (x^2 + 2(x)(1) + 1^2) + 3x + 3 + 2[/tex]

expand

[tex]\sf x^2 + 2x + 1 + 3x + 3 + 2[/tex]

collect like terms

[tex]\sf x^2 + 2x + 3x + 3 + 2 + 1[/tex]

add/subtract like terms

[tex]\sf x^2 + 5x + 6[/tex]

breakdown

[tex]\sf x^2 + 3x + 2x+ 6[/tex]

factor common term

[tex]\sf x(x + 3) + 2(x+ 3)[/tex]

collect into groups

[tex]\sf (x + 2)(x+ 3)[/tex]

ii.

[tex]\sf (9x^2 + 9x - 4)(9x^2 + 9x - 10) - 72[/tex]

breakdown

[tex]\sf (9x^2 + 12x - 3x - 4)(9x^2 + 15x - 6x - 10) - 72[/tex]

factor common term

[tex]\sf (3x(3x + 4) -1( 3x + 4)) ( (3x(3x + 5)- 2(3x +5) )- 72[/tex]

collect like terms

[tex]\sf (3x -1)( 3x + 4) (3x- 2)(3x +5) - 72[/tex]

expand

[tex]\sf 81x^4+162x^3-45x^2-126x-32[/tex]

factor

[tex]\sf \left(3x+1\right)\left(3x+2\right)\left(9x^2+9x-16\right)[/tex]

Answer:

[tex]\textsf{1.} \quad (x+3)(x+2)[/tex]

[tex]\textsf{2.} \quad (3x+1)(3x+2)(9x^2+9x-16)[/tex]

Step-by-step explanation:

Question 1

[tex]\textsf{Given expression}: \quad(x+1)^2+3(x+1)+2[/tex]

[tex]\textsf{Let }u=(x+1) \implies u^2+3u+2[/tex]

[tex]\textsf{To factor }\:\:u^2+3u+2:[/tex]

Rewrite the middle term as u + 2u:

[tex]\implies u^2+u+2u+2[/tex]

Factorize the first two terms and the last two terms separately:

[tex]\implies u(u+1)+2(u+1)[/tex]

Factor out the common term (u+1):

[tex]\implies (u+2)(u+1)[/tex]

Replace [tex]u[/tex] with [tex](x+1)[/tex] :

[tex]\implies (x+1+2)(x+1+1)[/tex]

Simplify:

[tex]\implies (x+3)(x+2)[/tex]

Question 2

[tex]\textsf{Given expression}: \quad (9x^2+9x-4)(9x^2+9x-10)-72[/tex]

Expand:

[tex]\implies 9x^2(9x^2+9x-10)+9x(9x^2+9x-10)-4(9x^2+9x-10)-72[/tex]

[tex]\implies 81x^4+81x^3-90x^2+81x^3+81x^2-90x-36x^2-36x+40-72[/tex]

Collect like terms:

[tex]\implies 81x^4+81x^3+81x^3-90x^2+81x^2-36x^2-90x-36x+40-72[/tex]

Combine like terms:

[tex]\implies 81x^4+162x^3-45x^2-126x-32[/tex]

Use the Factor Theorem:

If f(x) is a polynomial, and f(a) = 0, then (x – a)  is a factor of f(x).

[tex]\begin{aligned}\implies f \left(-\dfrac{1}{3}\right) & =81\left(-\dfrac{1}{3}\right)^4+162\left(-\dfrac{1}{3}\right)^3-45\left(-\dfrac{1}{3}\right)^2-126\left(-\dfrac{1}{3}\right)-32\\ & = 1-6-5+42-32\\ & = 0\end{alilgned}[/tex]

Therefore (3x + 1) is a factor.

[tex]\begin{aligned}\implies f \left(-\dfrac{2}{3}\right) & =81\left(-\dfrac{2}{3}\right)^4+162\left(-\dfrac{2}{3}\right)^3-45\left(-\dfrac{2}{3}\right)^2-126\left(-\dfrac{2}{3}\right)-32\\ & = 16-48-20+84-32\\ & = 0\end{alilgned}[/tex]

Therefore (3x + 2) is a factor.

Therefore:

[tex]\implies f(x)=(3x+1)(3x+2)(ax^2+bx+c)[/tex]

Compare the coefficient of x⁴ and the constant to find a and c:

[tex]\implies 3 \cdot 3 \cdot a=81 \implies a=9[/tex]

[tex]\implies 2c=-32 \implies c=-16[/tex]

Therefore:

[tex]\implies f(x)=(3x+1)(3x+2)(9x^2+bx-16)[/tex]

Expand:

[tex]\implies f(x)=81x^4+(81+9b)x^3-(126-9b)x^2-(144-2b)x-32[/tex]

Compare the coefficient of x³ to find b:

[tex]\implies 81+9b=162 \implies b=9[/tex]

Therefore, the fully factorized expression is:

[tex]\implies (3x+1)(3x+2)(9x^2+9x-16)[/tex]

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Answer:

Step-by-step explanation:

To find the inverse of a function [tex]f(x)[/tex], we let [tex]f(y)=x[/tex] and then solve again for y.

For example, using [tex]g(x)=-0.5x[/tex], we will let [tex]g(y)=x[/tex], giving [tex]x=-0.5y[/tex]. This means that [tex]y=-2x[/tex], and thus [tex]g^{-1}(x)=-2x[/tex].

Using similar logic, we see that:

Answer:

90 is the right answer

Step-by-step explanation:

because it give the value pf o is tan o

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

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)

Step-by-step explanation:

[tex]f(x) = ln( \frac{1}{x} ) [/tex]

To find the derivative, notice we have a function 1/x inside of another function, ln(x)

We use what we call the chain rule,

It states that

derivative of a '

[tex] \frac{d}{dx} f(g(x)) = f '(g(x)) \times \: g '(x)[/tex]

Here f is ln(x)

f is 1/x

So first, we know that

[tex] \frac{d}{dx} ( ln(x) = \frac{1}{x} [/tex]

so

[tex]f'(g(x)) = \frac{1}{ \frac{1}{x} } [/tex]

We know that

[tex]g'(x) = - \frac{1}{ {x}^{2} } [/tex]

So we have

[tex]x \times - \frac{1}{ {x}^{2} } = \frac{ - 1}{x} [/tex]

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

Using the binomial distribution, it is found that the standard deviation of the distribution is of 0.67.

It is the probability of exactly x successes on n repeated trials, with p probability of a success on each trial.

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

The parameters are:

n = 5, p = 0.1.

Hence the standard deviation is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{5 \times 0.1 \times 0.9} = 0.67[/tex]

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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:

4x^3 + 7x^2 - 3x + 1

Step-by-step explanation:

BTW your question, there is a sign misssing

Option B. The situation that would be more likely to show a constant rate of change would be B. The distance traveled by a truck driving at a constant speed compared with time.

This is the factor that is used to determine if there is a decrease or an increase in a given factor.

A good way to d this is by the use of time and speed to compare the distance that is being traveled.

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Complete question

Which situation is most likely to show a constant rate of change?

A. The weight of a child compared with its age in months

B. The distance traveled by a truck driving at a constant speed compared with time

C. The cost of a new car compared with the number of tires it has

D. The outside temperature compared with the time of day

Answer: c

Step-by-step explanation:

The equation of the line is

[tex]y+2=\frac{1}{3}(x-1)\\\\y+2=\frac{1}{3}x-\frac{1}{3}\\\\y=\frac{1}{3}x-\frac{7}{3}[/tex]

To determine which point lies on the line, we can substitute in the x coordinate and see if the equation gives the same y-coordinate.

Therefore, the answer is c.

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

Answer:

False

Step-by-step explanation:

They both have 1 solution

15x + 4 = 3(6x +5)

15x +4 = 18x + 15  Multiple everything in the parentheses by 3.

4 = 3x + 15  Subtract 15 x from both sides

-11 = 3x   Subtract 15 from both sides

-11/3 = x  Divide both sides by 3

There is only one solution for this equation.

2x -8 = 8x + 36

-8 = 6x + 36  Subtract 2x from both sides

--44 = 6x  Subtract 36 from both sides

-44/6 = x Divide both sides by 6.

There is only one solution for this equation.

Answer:

false

Step-by-step explanation:

They both have solutions

Answer:

Zero real number solutions

Step-by-step explanation:

Doesn't intercept the x-axis -> No real solutions

1. 1/2

2.2/3

3. 1

4. 1/3

5. 2/7

6. 1/4

7. 4/3

8. 3/2

To find 2x of the vale of x, we have to multiply the value of 'x' by 2

1. x = 1/4

[tex]2x = 2 * \frac{1}{4}[/tex] ⇒ [tex]\frac{2}{4}[/tex] ⇒[tex]\frac{1}{2}[/tex]

2. x = 1/3

[tex]2x = 2 * \frac{1}{3}[/tex] ⇒[tex]\frac{2}{3}[/tex]

3. x = 1/2

[tex]2x = 2* \frac{1}{2}[/tex] ⇒ [tex]\frac{2}{2}[/tex] ⇒ [tex]1[/tex]

4. x= 1/6

[tex]2x = 2* \frac{1}{6}[/tex] ⇒ [tex]\frac{2}{6}[/tex] ⇒ [tex]\frac{1}{3}[/tex]

5. x = 1/ 7

[tex]2x = 2 * \frac{1}{7}[/tex] ⇒ [tex]\frac{2}{7}[/tex]

6. x = 1/8

[tex]2x = 2 * \frac{1}{8}[/tex] ⇒ [tex]\frac{2}{8}[/tex] ⇒ [tex]\frac{1}{4}[/tex]

7. x = 2/3

[tex]2x = 2 * \frac{2}{3}[/tex] ⇒ [tex]\frac{4}{3}[/tex]

8. x = 3/4

[tex]2x = 2 * \frac{3}{4}[/tex] ⇒ [tex]\frac{6}{4}[/tex] ⇒ [tex]\frac{3}{2}[/tex]

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Answer:

ans is 42635 when expressed in standard form

Step-by-step explanation:

Answer:

-1/2

Step-by-step explanation:

cos(x+pi) = -cos(x), so the answer is -1/2.

The measure of angle D in the convex pentagon ABCDE is 132°

An equation is an expression that shows the relationship between two or more numbers and variables.

Let x represent the measure of angle D, hence:

angle A = x - 40.

∠A + ∠B + ∠C + ∠D + ∠E = 540° (sum of angle in a pentagon)

3(x - 40) + 2x = 540

x = 132°

The measure of angle D in the convex pentagon ABCDE is 132°

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The volume of the prism would be enlarged by a factor of 8

The scale factor is given as:

k = 2

Let the initial volume be v and the final volume be V.

The relationship between both volumes is

V = k^3 * v

This gives

V = 2^3 * v

Evaluate

V = 8v

Hence, the volume of the prism would be enlarged by a factor of 8

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Answer: 2400

Step-by-step explanation:

3000 x 20/100 = 600

Discount = 600

Sale price of the article: 3000-600 = 2400

Answer:

5445

Step-by-step explanation:

We can use distributive property.  a*(b - c) = (a*b) - (a *c)

  99 = 100 - 1

55 * 99 = 55 * (100 - 1)

            = 55*100 - 55*1

            = 5500 - 55

            = 5445

Answer:

4 : 9

Step-by-step explanation:

ratio of corresponding sides and ratio of perimeter are equal

then ratio of perimeter = 4 : 9

Based on the number of shares that Ralph Warren purchased, the total cost of the stock was $469.63. The amount he received from sales was $447.13. The capital loss was $22.50.

The cost of the stock was:

= (27 x 16³/₈) + 27.50

= $469.63

The amount received from sales:

= (27 x 17⁵/₈) - 28.75

= $447.13

This capital gain (loss):

= 447.13 - 469.63

= -$22.50

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The plane is parallel to the base of the cone describes the relationship between the plane and the cone

A circle is a spherical form without boundaries or edges or we can say that A circle is a closed, curved object with two dimensions in geometry.

The produced cross-section is a circle when a plane slices a cone such that the plane is parallel to the base of the cone.

Ellipse is the shape that is produced when a plane cuts a cone at an angle such that it misses the base or the vertex. (As shown in figure)

Hence A circle is formed when a plane slices through a cone and the plane is parallel to the base of the cone

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