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Andru [333]
1 year ago
5

Can you please help me find the area? Thank you. :)))

Mathematics
1 answer:
Phoenix [80]1 year ago
8 0

The figure shown in the picture is a rectangular shape that is missing a triangular piece. To determine the area of the figure you have to determine the area of the rectangle and the area of the triangular piece, then you have to subtract the area of the triangle from the area of the rectangle.

The rectangular shape has a width of 12 inches and a length of 20 inches. The area of the rectangle is equal to the multiplication of the width (w) and the length (l), following the formula:

A=w\cdot l

For our rectangle w=12 in and l=20 in, the area is:

\begin{gathered} A_{\text{rectangle}}=12\cdot20 \\ A_{\text{rectangle}}=240in^2 \end{gathered}

The triangular piece has a height of 6in and its base has a length unknown. Before calculating the area of the triangle, you have to determine the length of the base, which I marked with an "x" in the sketch above.

The length of the rectangle is 20 inches, the triangular piece divides this length into three segments, two of which measure 8 inches and the third one is of unknown length.

You can determine the value of x as follows:

\begin{gathered} 20=8+8+x \\ 20=16+x \\ 20-16=x \\ 4=x \end{gathered}

x=4 in → this means that the base of the triangle is 4in long.

The area of the triangle is equal to half the product of the base by the height, following the formula:

A=\frac{b\cdot h}{2}

For our triangle, the base is b=4in and the height is h=6in, then the area is:

\begin{gathered} A_{\text{triangle}}=\frac{4\cdot6}{2} \\ A_{\text{triangle}}=\frac{24}{2} \\ A_{\text{triangle}}=12in^2 \end{gathered}

Finally, to determine the area of the shape you have to subtract the area of the triangle from the area of the rectangle:

\begin{gathered} A_{\text{total}}=A_{\text{rectangle}}-A_{\text{triangle}} \\ A_{\text{total}}=240-12 \\ A_{\text{total}}=228in^2 \end{gathered}

The area of the figure is 228in²

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Answer:

x=5

Step-by-step explanation:

77=4x+57

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4x=20

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x=5

3 0
3 years ago
Can I get help with finding the Fourier cosine series of F(x) = x - x^2
trapecia [35]
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You have

a_0=\displaystyle\frac1L\int_{-L}^Lf(x)\,\mathrm dx
a_0=\dfrac1L\left(\dfrac{x^2}2-\dfrac{x^3}3\right)\bigg|_{x=-L}^{x=L}
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Two successive rounds of integration by parts (I leave the details to you) gives an antiderivative of

\displaystyle\int(x-x^2)\cos nx\,\mathrm dx=\frac{(1-2x)\cos nx}{n^2}-\dfrac{(2+n^2x-n^2x^2)\sin nx}{n^3}

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f_C(x)=-\dfrac{L^2}3+\displaystyle\sum_{n\ge1}\left(-\dfrac{4L\cos nL}{n^2L}+\dfrac{(4-2n^2L^2)\sin nL}{n^3L}\right)\cos nx
4 0
3 years ago
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Likurg_2 [28]

Answer:

True!

Step-by-step explanation:

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3 0
3 years ago
Read 2 more answers
Find all the critical points of the function <img src="https://tex.z-dn.net/?f=%20f%28x%29%3D%20%28x%2B1%29%2F%28x-3%29%20" id="
yulyashka [42]

The function

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... y' = -1/x²

which has no zeros. It is undefined at x=0, the only critical point. The derivative is negative for all values of x, so the function is decreasing everywhere in its domain.

Your function

... y = (x+1)/(x-3)

can be written as

... y = 1 +4/(x-3)

which is a version of y = 1/x that has been vertically scaled by a factor of 4, then shifted 1 unit up and 3 units to the right. Shifting the function to the right means x=3 is excluded from the domain (and the interval on which the function is decreasing).

The critical point is x=3.

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5 0
2 years ago
What is mental math?
ryzh [129]
Math you do in your head
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that is mental math
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it's helpful if at the supermarket or at a resaurant

mental math is math done in the mind
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3 years ago
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