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3 years ago

7

How can you use equations with the variable on both sides to solve real-world problems?

1 answer:

dsp733 years ago

5 0

I could join a game club for $15 and rent games for $2 each, or I could pay $3 each video rental, which is the better deal? Another example could be

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Please help as soon as possible

Leokris [45]

Lamar used 80% of his data, so he used a greater percentage.

5 0

3 years ago

Read 2 more answers

PLS HELP Choose the solution to this inequality. 43<−83y y < −12 y > 12 y < 43 y > 4

OlgaM077 [116]

Given:

The inequality is:

$\dfrac{4}{3}$

To find:

The solution of the given inequality.

Solution:

We have,

$\dfrac{4}{3}$

Multiply both sides by 3.

$4$

Divide both sides by -8 and change the sign of inequality.

$\dfrac{4}{-8}>\dfrac{-8y}{-8}$

$-\dfrac{1}{2}>y$

It can be written as:

$y$

Therefore, the correct option is A.

8 0

3 years ago

gizmo_the_mogwai [7]

Answer:

the selling price is $28.50

i m confused on part b if you explain it i could help

Step-by-step explanation:

6 0

3 years ago

Someone help me please

garri49 [273]

Answer:

42 seats

Step-by-step explanation:

This represents an arithmetic progression with the first term of a = 15 and common difference of d = 3.

<u>The tenth row is the 10th term:</u>

aₙ = a + (n - 1)d

<u>The row 10 has:</u>

a₁₀ = 15 + 9*3 = 15 + 27 = 42 seats

7 0

2 years ago

Read 2 more answers

A figure is composed of a rectangle and a right triangle. What is the area of the figure?

strojnjashka [21]

QUESTION 1

The figure is a tra-pezoid.

The area of a trapezoid is given by the formula;

$A=\frac{1}{2}(Sum\:of\:parallel\:sides)\times h$

We substitute the values to obtain;

$Area=\frac{1}{2}(13.2cm+8.4cm)\times 6cm$

$Area=\frac{1}{2}(21.6cm)\times 6cm$

$Area=64.8cm^2$

QUESTION 2

The approximate length of the ribbon is equal to the circumference of the circular table cloth.

We can find this by using the formula for calculating the circumference of a circle.

$C=2\pi r$

where $r=3.5ft$ is the radius of the circular table cloth.

We substitute this value and $\pi=3.14$ into the formula to get;

$C=2(3.14)(3.5)ft$

$C=21.98ft$

The approximate length of the ribbon is 22.0ft to the nearest tenth.

QUESTION 3

Type of quadrilateral:Rectangle

Explanation: A rectangle has 4 angles that are right angles.

The two pairs of opposite sides of a rectangle are also congruent.

QUESTION 4.

The circumference of a semi circle is calculated using the formula;

$C=\pi r$

where $r=12.5cm$ is the radius of the semicircle.

$C=3.14\times 12.5cm$

$C=39.25cm$

QUESTION 5

The area of the entire figure is the area of the semicircle plus the area of the isosceles triangle.

$Area=\frac{1}{2}\pi r^2+\frac{1}{2}bh$

$Area=\frac{1}{2}\times3.14\times12.5^2+\frac{1}{2}\times25\times 24$

$Area=345.3125+300$

$Area=645.3125cm^2$

$Area\approx645cm^2$

QUESTION 6

The area of the parallelogram is given by the formula;

$A=bh$

From the diagram, $b=y\:in.,h=3in.$ .

Given, A=23.7 inches squared.

We substitute the values to obtain;

$23.7=3y$

$y=\frac{23.7}{3}$

$y=7.9in.$

QUESTION 7

The area of a rectangle is given by the formula;

$Area=l\times b$

The area of the bigger rectangle $=9\times15=135in^2$

The area of the smaller rectangle $=8\times 13=104in^2$

The area of the shaded region $=135-104=31in^2$

5 0

3 years ago