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Setler [38]
3 years ago
13

Two sides of a parallelogram are 65 feet and 87 feet. The measure of the angle between these sides is 52°. Find the area of the

parallelogram to the nearest square foot.
Mathematics
1 answer:
telo118 [61]3 years ago
3 0

Answer:

Area of the parallelogram is approximately 4,456.2 square feet

Step-by-step explanation:

The lengths of the sides of the parallelogram are 65 feet and 87 feet

The measure of the angle between the sides, Y = 52°

The formula for finding the area of a parallelogram is given as follows;

Area of a parallelogram = A·B·sin(Y)

Where;

'A' and 'B' are the length of the parallel sides and 'Y' is the measure of the angle between them

Let 'A' and 'B' represent the 65 feet and the 87 feet respectively, we get;

Area of the parallelogram = 65 ft. × 87 ft. × sin(52°) ≈ 4,456.2 ft.²

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Answer: Eve is 3

Melissa is 29. (3 + 26 = 29)

Paul is 33 ( 29 + 4 = 33)

3+ 29 + 33 = 65

Step-by-step explanation:

7 0
3 years ago
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Softa [21]

Answer:

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Step-by-step explanation:

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3 0
3 years ago
How do you write a an equation to represent a real world situation
Sauron [17]
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8 0
3 years ago
The product of a number, x, and 3 is six more than the product of the number and 2.
Pavel [41]

Answer:

  • 3x = 6 + 2x

Step-by-step explanation:

The word "product" defines multiplication between two numbers. Those numbers are x and 3. The word "is" defines the "equal sign". The word "more" defines addition. Now, using these definitions, let's create an equation.

  • => (3 × x) = 6 + (2 × x)
  • => 3x = 6 + 2x

Hence, Option C is correct.

7 0
2 years ago
Use integration by parts to find the integrals in Exercise.<br> x^3 ln x dx.
34kurt

Answer:

\frac{\text{ln}(x)x^4}{4}-\frac{x^4}{16}+C.

Step-by-step explanation:

We have been given an indefinite integral \int \:x^3\:ln\:x\:dx. We are asked to find the value of the integral using integration by parts.

\int\: u\text{dv}=uv-\int\: v\text{du}

Let u=\text{ln}(x), v'=x^3.

Now, we will find du and v as shown below:

\frac{du}{dx}=\frac{d}{dx}(\text{ln}(x))

\frac{du}{dx}=\frac{1}{x}

du=\frac{1}{x}dx

v=\frac{x^{3+1}}{3+1}=\frac{x^{4}}{4}

Upon substituting our values in integration by parts formula, we will get:

\int \:x^3\:\text{ln}\:(x)\:dx=\text{ln}(x)*\frac{x^4}{4}-\int\: \frac{x^4}{4}*\frac{1}{x}dx

\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\int\: \frac{x^3}{4}dx

\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\frac{1}{4}\int\: x^3dx

\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\frac{1}{4}*\frac{x^{3+1}}{3+1}+C

\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\frac{1}{4}*\frac{x^4}{4}+C

\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\frac{x^4}{16}+C

Therefore, our required integral would be \frac{\text{ln}(x)x^4}{4}-\frac{x^4}{16}+C.

5 0
3 years ago
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