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frosja888 [35]
3 years ago
8

What is the approximate area of the shaded portion of the diagram? (Use 3.14 as an estimate for pi.)

Mathematics
1 answer:
Nadusha1986 [10]3 years ago
6 0

Answer:

218

Step-by-step explanation:

first find the area of the circle-

A=\pi r^{2} \\A=\pi 10^{2} \\A=100\pi \\A=314

to find the area of the triangle, we need the base and the height. The base we know is 16, since we don't know the height yet, use the Pythagorean theorem to find the height. One leg is 16 the hypotenuse is 20 (2* the radius).

a^{2} +b^{2} =c^{2} \\a^{2} +16^{2} =20^{2} \\a^{2} +256=400\\a^{2} =144\\a=12

Now I know the height is 12. Find the area of the triangle

A=\frac{1}{2} bh\\A=\frac{1}{2} (16)(20)\\A=96

subtract the area of the triangle from the area of the circle

314-96=218

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Answer:

+0.1

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Note that -4.5 + 4.4 = -0.1.

Adding 0.1 to -0.1 results in 0.

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3 years ago
-3x^2 plus 24x = 27 <br> complete the square
Rudik [331]

You can divide the whole expression by -3:

-3x^2+24x=27 \iff x^2-8x = -9

Recalling that

(x-a)^2=x^2-2ax+a^2

Let's compare the middle term:

-8x = -2ax \iff a=4

So, we want to complete

(x-4)^2 = x^2-8x+16

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x^2-8x=-9 \iff x^2-8x+16=7 \iff (x-4)^2=7

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3 years ago
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Step-by-step explanation:

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In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini
svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2

The augmented matrix is

\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]

The reduction of this matrix to row-echelon form is outlined below.

R_2\rightarrow R_2-R_1

\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]

R_3\rightarrow R_3-2R_1

\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]

R_3\rightarrow R_3-R_2

\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]

The last row determines, if there are solutions or not. To be consistent, we must have k such that

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\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

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This gives the infinite many solution.

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3 years ago
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If we simplify like terms on left and right sides we gwt

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Its B
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