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3 years ago

13

Help me

1 answer:

Rashid [163]3 years ago

4 0

The answer to your question would be B because you divide 10 by 3.5 to get 3, then you would divide 12 by 3 to get 4. Multiply them together to get 14

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Vincent works 8 hours a day for 5 days. How much does he earn at the rate of $.97 per hour?

Tresset [83]

Answer:

$38.8

Step-by-step explanation:

First, find how much money he gets paid in a day. Multiply 8 x .97, which equals $7.76. Then, multiply $7.76 x 5 to see how much he gets paid in 5 days. $7.76 x 5= $38.8

Hope this helps!

6 0

3 years ago

Read 2 more answers

Please answer this correctly

TiliK225 [7]

Answer:

22 meters

Step-by-step explanation:

Area of yellow rectangle: 6x4=24

Area of gray rectangle: 8x=24

x=3

Thus, the perimeter is 3+3+8+8, which is 22.

5 0

3 years ago

Read 2 more answers

Convert from general form to standard form x^2 + 4y^2 + 6x-40y+9 =0

Burka [1]

$x^2 + 4y^2 + 6x-40y+9 =0\\x^2 + 6x+9+4y^2-40y=0\\x^2+6x+9+4y^2-40y+100-100=0\\(x+3)^2+(2y-10)^2=100$

P.S. Hello from Russia

7 0

3 years ago

If m \angle∠ 13 = 59°, find the measure of \angle∠ 8.

Sophie [7]

Answer:

59

Step-by-step explanation:

alternate interior angles are congruent

6 0

2 years 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