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2 years ago

YOU GET CORRECT YOU GET BRAINLIEST!!!!!!!

Mathematics

1 answer:

Virty [35]2 years ago

5 0

The answer is 6 you just divide

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Use a protractor to find the measure of each angle in the circle

zzz [600]

You need to show us the circle in order to let us answer the question.

7 0

2 years ago

Serika has hired Tom to build her a dollhouse. The dollhouse will be a scale model of her home, and the scale will be 2 inches :

sveticcg [70]

Answer:

96 square inches

Step-by-step explanation:

The scale is 2 inches : 3 feet

Thus for 18 feet in Serika's real house, the model is 18 * (2"/3') = 12 inches

For 12 feet in Serika's real house, the model is 12 * (2"/3') = 8 inches

Area of carpeting needed in Serika's dollhouse = 12" * 8" = 96 square inches

7 0

1 year ago

What is x, the length of the diagonal, to the nearest whole number

lions [1.4K]

brainly being slow today so i did my work on this attachment

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│Hope this helped _____________________│

│~Xxxtentaction _______________________│

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6 0

3 years ago

Read 2 more answers

Consider the following points on cartesian plane. A (1;2) BC3; 1) and CG-3; 4) prove that Points A, B and C are collinear.

Arlecino [84]

Step-by-step explanation:

Collinear points have the same gradient

find the gradient(slope) between AB first

$= \frac{1 - 2}{3 - 1} \\ = \frac{ - 1}{2}$

Find the one for BC

$= \frac{4 - 1}{ - 3 - 3} \\ = \frac{3}{ - 6} \\ = \frac{ - 1}{2}$

Find the gradient between AC now

$= \frac{4 - 2}{ - 3 - 1} \\ = \frac{2}{ - 4} \\ = \frac{ - 1}{2}$

YOUR LAST STATEMENT NOW WILL BE

POINTS A,B AND C ARE COLLINEAR SINCE THE GRADIENTS BETWEEN THEM IS THE SAME!!!

5 0

1 year ago

Find a compact form for generating functions of the sequence 1, 8,27,... , k^3

pantera1 [17]

This sequence has generating function

$F(x)=\displaystyle\sum_{k\ge0}k^3x^k$

(if we include $k=0$ for a moment)

Recall that for $|x|$ , we have

$\displaystyle\frac1{1-x}=\sum_{k\ge0}x^k$

Take the derivative to get

$\displaystyle\frac1{(1-x)^2}=\sum_{k\ge0}kx^{k-1}=\frac1x\sum_{k\ge0}kx^k$

$\implies\dfrac x{(1-x)^2}=\displaystyle\sum_{k\ge0}kx^k$

Take the derivative again:

$\displaystyle\frac{(1-x)^2+2x(1-x)}{(1-x)^4}=\sum_{k\ge0}k^2x^{k-1}=\frac1x\sum_{k\ge0}k^2x^k$

$\implies\displaystyle\frac{x+x^2}{(1-x)^3}=\sum_{k\ge0}k^2x^k$

Take the derivative one more time:

$\displaystyle\frac{(1+2x)(1-x)^3+3(x+x^2)(1-x)^2}{(1-x)^6}=\sum_{k\ge0}k^3x^{k-1}=\frac1x\sum_{k\ge0}k^3x^k$

$\implies\displaystyle\frac{x+4x^3+x^3}{(1-x)^4}=\sum_{k\ge0}k^3x^k$

so we have

$\boxed{F(x)=\dfrac{x+4x^3+x^3}{(1-x)^4}}$

5 0

3 years ago