Can someone pls help answer this asap!! I’d really appreciate it!

Step-by-step explanation:

please refer to the sketch

The decimal value is exact without any rounding.

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Explanation:

If Jackson saves 22%, then he pays the remaining 100% - 22% = 78%, which leads to the decimal value 0.78

An increase of 6% leads to 100%+6% = 106% and that converts to 1.06

Multiply those decimal values: 0.78*1.06 = 0.8268; this trick is useful to chain together multiple discounts and/or tax increases.

Therefore, the final cost expression is 0.8268y where the variable y is the original price before the discount and taxes.

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Extra info:

The expression 0.8268y tells us that Jackson is paying 82.68% of the original price. Note how 100% - 82.68% = 17.32%

This means Jackson is getting a real discount of 17.32% after the taxes have been considered.

Answer:

Step-by-step explanation:

Midpoint formula is x3 = (x1+x2)/2, y3=(y1+y2)/2

so x3=(9-5)/2=2, y3=(-5/2-3/2)/2=-2.

the coordinates is (2, -2).

Answer:

1 unit to the left, 3 units down

Step-by-step explanation:

Take any point: I will take D. In order to get to D', you need to travel 1 unit along the x-axis towards the left, then 3 until along the y-axis downwards. The same applies for any other point.

25 kilograms of fertilizer would be needed to fertilize the lawn.

In essence, percentages are fractions with 100 as the denominator. We place the percent symbol (%) next to the number to indicate that the number is a percentage.

given, A certain type of fertilizer contains 8 percent of nitrogen. If 2 kilograms of nitrogen are needed to fertilize a lawn.

Since fertilizer contain 8% nitrogen in it

thus for 80 grams of nitrogen = 1 kg of fertilizer

for 1 gram of nitrogen = 1/80 kg fertilizer

for 2000 gram of nitrogen = (1/80)* 2000 kg fertilizer

for 2 kilograms of nitrogen = 25 kg fertilizer

therefore, To fertilize the yard, 25 kg of fertilizer would be required.

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The solution to the equation log₇343 = 3 is 7³ = 343

An equation is used to show the relationship between numbers and variables.

The logarithm can be defined as the exponent that raises a base to get a particular number. The law of logarithm states:

logₓm = n is the same as xⁿ = m

Given the equation:

log₇343 = 3

Applying the law of logarithm gives:

7³ = 343

The solution to the equation is 7³ = 343

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The unit rate is 18 commercials per hour.

A division is a process of splitting a specific amount into equal parts.

On Saturday morning, Wendy watched one and a half hours of her favorite shows on television.

During the shows there were 27 commercials.

We need to find the unit rate of commercials per hour.

To find this we have to divide 27 by number of hours

27/[tex]1\frac{1}{2}[/tex]

Convert [tex]1\frac{1}{2}[/tex] to improper fraction

27/(3/2)

When whole number is divided by fraction, the denominator is multiplied with whole number

27×2/3

54/3

18

Hence, the unit rate is 18 commercials per hour.

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Answer:18 commercials per hour

Step-by-step explanation:

4.8

Divide 250 by 2 which is 125 then do 600 divided by 125. Then your answer should be 4.8.

Answer:

A 1 in 4 chance or a probability of 0.25

Step-by-step explanation:

There is only 3 months that start with J and 12 months in a year so simplify 3/12 you get 1/4 or 0.25

Answer:

Step-by-step explanation:

If m∠ABJ = 28 and ∠ABC ≅ ∠DBJ, then we can set up the following equation:

m∠ABC = m∠DBJ

Since angles ∠ABC and ∠DBJ are congruent, they have the same measure, so the equation above is true.

Since we are trying to find m∠JBC, we can use the fact that the measure of an angle is equal to the sum of the measures of the other two angles in the triangle to set up another equation:

m∠JBC = m∠ABC + m∠ABJ

Substituting the value for m∠ABC from the first equation into the second equation, we get:

m∠JBC = m∠DBJ + m∠ABJ

Substituting the given value of m∠ABJ into this equation, we get:

m∠JBC = m∠DBJ + 28

Since m∠ABJ = m∠DBJ, we can substitute m∠ABJ in for m∠DBJ:

m∠JBC = m∠ABJ + 28

Substituting the given value of m∠ABJ into this equation, we get:

m∠JBC = 28 + 28

Solving this equation, we find that m∠JBC = 56.

Therefore, the measure of angle ∠JBC is 56 degrees.

From the exponential function, the average rate of change from year 1 to year 4 is 62.43, the average rate of change from year 5 to year 8 is 67.57 and the rate of change per year is 65

An exponential function is a type of function that involves a variable in an exponent. It is of the form f(x) = a⋅b^x where a and b are constants and b > 0, b ≠ 1. The input to the function, x, is the exponent that the base, b, is raised to. The output, f(x), is the result of raising the base to the exponent.

In this problem, we have to calculate the rate of increase per year.

Data;

rate = 2%

P = 3000

The exponential growth can be calculated by;

A = P(1 + r)ⁿ

In year 1

A = 3000(1 + 0.02)¹

A = 3060

In year 4

A = 3000(1 + 0.02)⁴

A = 3247.3

a)

The average rate of change from year 1 to year 4 will be

A = [3000(1 + 0.02)⁴ - 3000(1 + 0.02)¹] / 4 - 1

A = 62.43

The average rate of change from year 1 to year 4 is 62.43

b)

The average rate of change from year 5 to year 8 is calculated as

A = [3000(1 + 0.02)⁸ - 3000(1 + 0.02)⁵] / 8 - 5

A = 67.57

C)

The rate of change increase can be calculated as;

A = [3000(1 + 0.02)⁸ - 3000(1 + 0.02)¹] / 8 - 1

A =65

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1a) An equation that gives the width `y` of the book, if it has `x`-hundred pages printed on bond paper is; y = ¹/₂ + ¹/₄x

1b) Ledger paper has a thickness of 2/5 in. per `100` pages, the equation is; y = ¹/₂ + ²/₅x

2) The equation will be; y = ²/₃ +¹/₄x

1) We are told that the front cover and the back cover have a thickness of 1/4 inches and they also have a thickness of 1/4 inch. Thus;

The addition of both gives to the book thickness of;

1/4 + 1/4 = 1/2 inch, despite how many pages the book  has.

We are told that for every hundred pages, a thickness of 1/4 inch is added to the book, then the equation is:

y = ¹/₂ + ¹/₄x

where;

y is the width of the book in inches.

x is every hundred pages printed on bond paper.

1b) If ledger paper has a thickness of 2/5 in. per 100 pages, then the equation is now;

y = ¹/₂ + ²/₅x

2) If both front and back covers now, have a thickness of 1/3 inches instead of 1/4 inches each, then the equation will be;

y = 2(1/3) +¹/₄x

y = ²/₃ +¹/₄x

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By adding the weight of the triangle, the weight of the circle heavier than that of the square.

The linear equation that results from examining the weights will contain the following parameters.

Let t stand for the triangle.

Let c stand for the circle.

Let's use the shape square.

left side of the equation = 6s + 3t

right side of the equation = 3s + 3c

equating both sides

6s + 3t = 3s + 3c

6s - 3s + 3t = 3s + 3c

3s + 3t = 3c

3(s + t) = 3c

dividing all through by 3

s + t = c

10 + t = c

Because of the resulting equation, we can conclude that the the circle is weightier than the square by the weight of the triangle

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Answer:

3.2 mph

Step-by-step explanation:

If it takes her 3 hours to walk 9.6 miles, we need to divide 9.6 by 3.

To do so, we can remove the decimal point in 9.6, to give us 96.

Then, we can divide 96 by 3 to get 32.

Add the decimal back to 32.

Doing so would give us 3.2, and our answer.

We can check this answer by multiplying 3.2 by 3, which gives us 9.6, telling us our answer is correct.

Answer: I don't remember how to solve, but, I figured out the lengths of each side in the shaded region:

11, 11, 7, 7, 4, 4

Step-by-step explanation:

Hope this helps!! :D

Answer:

157.08 is the circumference. Hope it helps you

Step-by-step explanation:

The linear equation in this case is . The [tex]$3 x^2+4 x+4=0$[/tex] answer is [tex]x=-\frac{2}{3}+i \frac{2 \sqrt{2}}{3}, x=-\frac{2}{3}-i \frac{2 \sqrt{2}}{3}[/tex] and [tex]$3 y^2+2 x+4=0: \quad$[/tex]: quadratic [tex](h, k)=(-2,0), p=-\frac {1} {6}[/tex] , is the solution.

[tex]$3 x^2+4 x+4=0 \quad: \quad x=-\frac{2}{3}+i \frac{2 \sqrt{2}}{3}, x=-\frac{2}{3}-i \frac{2 \sqrt{2}}{3}$[/tex]

[tex]$3 x^2+4 x+4=0$[/tex]

Apply the quadratic formula to the problem. [tex]$x_{1,2}=\frac{-4 \pm \sqrt{4^2-4 \cdot 3 \cdot 4}}{2 \cdot 3}$[/tex]

Simplify [tex]$\sqrt{4^2-4 \cdot 3 \cdot 4}: \quad 4 \sqrt{2} i$\\[/tex]

[tex]$x_{1,2}=\frac{-4 \pm 4 \sqrt{2} i}{2 \cdot 3}$[/tex]

Distinguish the answers

[tex]x_1=\frac{-4+4 \sqrt{2} i}{2 \cdot 3}, x_2=\frac{-4-4 \sqrt{2} i}{2 \cdot 3}\\x=\frac{-4+4 \sqrt{2} i}{2 \cdot 3}: \quad-\frac{2}{3}+i \frac{2 \sqrt{2}}{3}\\x=\frac{-4-4 \sqrt{2} i}{2 \cdot 3}: \quad-\frac{2}{3}-i \frac{2 \sqrt{2}}{3}[/tex]

The following are the quadratic equation's solutions:

[tex]x=-\frac{2}{3}+i \frac{2 \sqrt{2}}{3}, x=-\frac{2}{3}-i \frac{2 \sqrt{2}}{3}[/tex]

2) [tex]$3 y^2+2 x+4=0: \quad$[/tex] A parabola with a focal length of and a vertex at (h,k)=(-2, 0) [tex]$|p|=\frac{1}{6}$[/tex]

[tex]$3 y^2+2 x+4=0$[/tex]

[tex]$4 p(x-h)=(y-k)^2$[/tex] is the common formula for a right-to-left parabola with a vertex at (h, k) and a focal length of |p|.

Rewrite [tex]$3 y^2+2 x+4=0$[/tex]  using the proper form: [tex]$4\left(-\frac{1}{6}\right)(x-(-2))=(y-0)^2$[/tex]

The following are characteristics of parabolas:

[tex](h, k)=(-2,0), p=-\frac {1} {6}[/tex]

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If the HCF is 5 and the product of the two numbers is 70, then the value of LCM is 14.

The smallest number that can be divided by both numbers is known as the least common multiple, or LCM, of two numbers. The product of the two numbers is the LCM if their greatest common divisor is 1, in which case their product is 1.

First, multiply both numbers by their prime factors to express them. Then, multiply the common factor with the highest power by the remaining factors to obtain the LCM. Take the common factors with the lowest power and multiply them to determine HCF.

To find the LCM of two numbers, you can use the formula: LCM = (a * b) / HCF.

In this case, we know that the HCF is 5, and the product of the two numbers is 70. Therefore, we can substitute these values into the formula:

LCM = (70) / 5 = 14.

So the LCM of the two numbers is 14.

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The sale price of the item is given by the equation A = $ 26.40

Equations are mathematical statements with two algebraic expressions flanking the equals (=) sign on either side.

It demonstrates the equality of the relationship between the expressions printed on the left and right sides.

Coefficients, variables, operators, constants, terms, expressions, and the equal to sign are some of the components of an equation. The "=" sign and terms on both sides must always be present when writing an equation.

Given data ,

Let the sale price of the item be represented as A

Now , the equation will be

The original price of the item = $ 33

An item on sale costs 80% of the original price

So , the sale price of the item A = 80 % x original price of the item

Substituting the values in the equation , we get

The sale price of the item A = ( 80 / 100 ) x 33

The sale price of the item A = 0.80 x 33

The sale price of the item A = $ 26.40

Therefore , the value of A is $ 26.40

Hence , the sale price is $ 26.40

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Answer:

x = 55

Step-by-step explanation:

Since the sides of the shape are parallel, AngleC and AngleD will add up to 180° (see "parallel lines cut by a transversal" or maybe you have learned a big pile of parallelogram theorems)

Use this relationship to set up an equation.

2x + 20 + 50 = 180

combine like terms

2x + 70 = 180

subtract 70

2x = 110

divide by 2

x = 110/2

x = 55

x is 55

Answer:

The value of x is 55°

Step-by-step explanation:

Since we know that opposite angles of a parallelogram are equal

The sum of any two adjacent angles is supplementary. This means, ∠A + ∠B = 180°, ∠B + ∠C = 180°, ∠C + ∠D = 180°, and ∠D + ∠A = 180.

So,

∠C + ∠D = 180°

(2x + 20)° + 50° = 180°

2x + 70° = 180°

2x + 70° - 70° = 180° - 70°

2x = 110°

2x/2 = 110°

x = 55°

Thus, The value of x is 55°

Answer:

11

Step-by-step explanation:

29-20= 91

2+9=11

11+9=

5a) The total cost of buying the car, including the harmonized sales tax (HST) of 13%, is $36,160.

5b) Under the lease, the customer saves $14,760.

5c) The lessee after returning the car has the following lease options:

6a) The after-tax cost of the car that Jordan wants to buy is $32,199.35.

6b) The part of the Honda Civic that needs to be financed after the down payment is $26,199.35.

6c) The monthly payment will be $567.26.

6d) After four years, the total amount Jordan had paid for the vehicle is $33,228.48.

A lease is a financing option for the use of capital assets by which the lessee utilizes an asset for a stated period while making periodic payments to the lessor.

A car purchase can also be financed through loans, like home mortgages.

Value of the car = $32,000

Harmonized Sales Tax (HST) = 13%

Cost of the car with HST = $36,160 ($32,000 x 1.13)

Down payment = $1,000

Financing part = $35,160 ($36,160 - $1,000)

Installment payments = 48

Periodic payments = $425

Total lease cost = $21,400 ($1,000 + 48 x $425)

Savings from leasing = $14,760 ($36,160 - $21,400)

The value of the Honda Civic = $28,495

Harmonized Sales Tax (HST) = 13%

After-tax cost = $32,199.35 ($28,495 x 1.13)

Annual interest rate = 1.9% compounded monthly

Down payment = $6,000

Financing part = $26,199.35 ($32,199.35 -$6,000)

Financing period = 4 years

N (# of periods) = 48 months (4 years x 12)

I/Y (Interest per year) = 1.9%

PV (Present Value) = $26,199.35

FV (Future Value) = $0

P/Y (# of periods per year) = 12

C/Y (# of times interest compound per year) = 12

Results:

Monthly Payments (PMT) = $567.26

Sum of all periodic payments = $27,228.48

Total Interest $1,029.13

Total amount paid by Jordan after 4 years = $33,228.48 ($27,228.48 + $6,000)

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In the right triangle in the figure the value of x is solved to be 4.5

Using SOH CAH TOA for the right triangle

Sin = opposite / hypotenuse - SOH

Cos = adjacent / hypotenuse - CAH

Tan = opposite / adjacent - TOA

The figures show that

hypotenuse = 6

adjacent = x

given angle = 41 degrees

The value of x will be calculated using cos, CAH let the angle be y

cos y = adjacent / hypotenuse

cos 41 = x / 6

x = 6 * cos 41

x = 4.528

x = 4.5

Hence the value of x is 4.5

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The value of the function f(x) at x = - 2, - 1, 0, 1 are - 4, - 1, 2, 5.

A function can be defined as the outputs for a given set of inputs.

The inputs of a function are known as the independent variable and the outputs of a function are known as the dependent variable.

Given, A function f(x) = 3x + 2.

Now,

At x = - 2, f(x) = 3(-2) + 2 ⇒ f(x) = - 4.

At x = - 1, f(x) = 3(-1) + 2 ⇒ f(x) = - 1.

At x = 0, f(x) = 3(0) + 2 ⇒ f(x) = 2.

At x = 1, f(x) = 3(1) + 2 ⇒ f(x) = 5.

This can also be written in the form of ordered pairs

(- 2, - 4), (- 1, - 1), (0, 2), (1, 5).

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A) The linear equation is:

y = (3/4)*x + 3

B) The sets with only positive elements are:

A general linear equation can be written as:

y = a*x + b

Where a is the slope and b is the y-intercept.

Here, we do know that the y-intercept of the linear equation is (0, 3), then we know that b = 3.

Then the line is something like:

y = a*x + 3

We also know that the line passes through (-4, 0), then we can replace these values in the equation above to get:

0 = a*-4 + 3

4a = 3

a = 3/4

Then the linear equation is:

y = (3/4)*x + 3

The sets where all the numbers are positive are:

{x| x ∈ N}

Where N is the set of the natural numbers, these are all the integers equal to or larger than 1, so all are positives.

{y| 0 < y < 1}

There we can see that y must be larger than zero, so all the elements are positive.

{y| 3 ≤ y ≤22}

Here the elements are between 3 and 22, so all of these are positve.

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Answer:

Step-by-step explanation:

i have no idea

Answer:

Step-by-step explanation:

To prove that a quadrilateral ABCD is a trapezium when the diagonals intersect at a point O such that AO/BO=CO/DO, we can use the fact that opposite sides of a trapezium are parallel.

First, let's draw a diagram to represent the given information:

[asy]

pair A, B, C, D, O;

A = (0,0);

B = (2,0);

C = (3,2);

D = (1,2);

O = intersection point(A-C, B--D);

draw(A--B--C--D--cycle);

draw(A-C);

draw(B--D);

label("A", A, W);

label("B", B, E);

label("C", C, E);

label("D", D, W);

label("O", O, N);

label("$\theta_1$", (B+O)/2, NE);

label("$\theta_2$", (O+D)/2, NE);

label("$\theta_3$", (O+C)/2, NW);

label("$\theta_4$", (A+O)/2, NW);

[/asy]

Since the diagonals intersect at point O, the angles formed by the diagonals at O are supplementary (that is, they add up to 180 degrees). In other words, we have:

θ1 + θ2 = 180 degrees

θ3 + θ4 = 180 degrees

Since AO/BO=CO/DO, the triangles AOB and COD are similar. This means that the ratios of their sides are equal. We can write this as:

AO/BO = CO/DO

We can also write this as:

AO/DO = CO/BO

Then, since the angles formed by the diagonals at O are supplementary, we have:

θ1 + θ2 = 180 degrees

θ1 + θ4 = 180 degrees

Subtracting these two equations gives us:

θ2 - θ4 = 0

Since the angles are equal, they are also opposite angles of the quadrilateral ABCD. Therefore, the opposite sides of AB and CD are parallel, which means that ABCD is a trapezium.