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2 years ago

What is the area of one of the small trapezoids? show your work and explain your reasoning.

1 answer:

8 0

The area of a trapezoid is found using the formula, A = ½ (a + b) h, where 'a' and 'b' are the bases (parallel sides) and 'h' is the height (the perpendicular distance between the bases) of the trapezoid.

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sarah can complete a project in 90 minutes and her sister betty can complete it in 120 minutes if they both work on the project

alukav5142 [94]

Answer:

It will take them approximately 51.43 minutes to complete the project together

Step-by-step explanation:

This is what is called a "shared job" problem.

The best way to work on them is to start by finding the "portion" of the job done by each of the people in the unit of time.

So, for example, Sarah completes the project in 90 minutes, so in the unit of time (that is 1 minute) she completed 1/90 of the total project

Betty completes the project in 120 minutes, so in the unit of time (1 minute) she completes 1/120 of the total project.

We don't know how long it would take for them to complete the project when working together, so we call that time "x" (our unknown).

Now, when they work together completing the entire job in x minutes, in the unit of time they would have done 1/x of the total project.

In the unite of time, the fraction of the job done together (1/x) should equal the fraction of the job done by Sarah (1/90) plus the fraction of the job done by Betty. This in mathematical form becomes:

$\frac{1}{x} =\frac{1}{90} +\frac{1}{120}\\\frac{1}{x} =\frac{4}{360} +\frac{3}{360}\\\frac{1}{x} =\frac{7}{360} \\x=\frac{360}{7} \\x=51.43\,\,min$

So it will take them approximately 51.43 minutes to complete the project together.

8 0

3 years ago

A square with side length 4 cm represents 20% of a larger figure what is the area of the larger figure?

victus00 [196]

First we are going to find divided by 100 and multiply 20 is equal to four.

We are gonna use x to represent this number.

x/100 * 20 = 4

Solve for x

Therefore x = 20cm

Now that we found x the side of the larger square we are gonna find the area which is side^2.

20 * 20

Therefore the area of the larger square is 400cm^2

8 0

3 years ago

Is -3+3-3+3-3+3 a arithmetic?

bezimeni [28]

Answer:

$- 3 + 3 = 0 - 3 = - 3 \\ - 3 + 3 = 0 - 3 = - 3 \\ - 3 + 3 = 0$

3 0

2 years ago

Read 2 more answers

Solve for the x <br> (Round to the nearest hundred)

stiks02 [169]

Answer:

17.97=x

Step-by-step explanation:

tan theta = opposite side/ adjacent side

tan 38 = x /23

Multiply each side by 23

23 * tan 38 = x/23 * 23

23 tan 38 = x

17.96956941 =x

To the nearest hundredth

17.97 =x

8 0

2 years ago

Read 2 more answers

Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10

zvonat [6]

The approximate difference in the ages of the two cars, which depreciate to 60% of their respective original values, is 1.7 years.

<h3>What is depreciation?</h3>

Depreciation is to decrease in the value of a product in a period of time. This can be given as,

$FV=P\left(1-\dfrac{r}{100}\right)^n$

Here, (<em>P</em>) is the price of the product, (<em>r</em>) is the rate of annual depreciation and (<em>n</em>) is the number of years.

Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10%.

Suppose the original price of the first car is x dollars. Thus, the depreciation price of the car is 0.6x. Let the number of year is $n_1$ . Thus, by the above formula for the first car,

$0.6x=x\left(1-\dfrac{10}{100}\right)^{n_1}\\0.6=(1-0.1)^{n_1}\\0.6=(0.9)^{n_1}$

Take log both the sides as,

$\log 0.6=\log (0.9)^{n_1}\\\log 0.6={n_1}\log (0.9)\\n_1=\dfrac{\log 0.6}{\log 0.9}\\n_1\approx4.85$

Now, the second car depreciates at an annual rate of 15%. Suppose the original price of the second car is y dollars.

Thus, the depreciation price of the car is 0.6y. Let the number of year is $n_2$ . Thus, by the above formula for the second car,

$0.6y=y\left(1-\dfrac{15}{100}\right)^{n_2}\\0.6=(1-0.15)^{n_2}\\0.6=(0.85)^{n_2}$

Take log both the sides as,

$\log 0.6=\log (0.85)^{n_2}\\\log 0.6={n_2}\log (0.85)\\n_2=\dfrac{\log 0.6}{\log 0.85}\\n_2\approx3.14$

The difference in the ages of the two cars is,

$d=4.85-3.14\\d=1.71\rm years$

Thus, the approximate difference in the ages of the two cars, which depreciate to 60% of their respective original values, is 1.7 years.

Learn more about the depreciation here;

brainly.com/question/25297296

4 0

2 years ago