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vlabodo [156]
3 years ago
7

Jane wants to share her pizza with two friends. She wants to make sure they all get the same amount. How should she cut the pizz

a?
Mathematics
2 answers:
Alexxandr [17]3 years ago
8 0

Answer:

<em>Hello there the Correct answer is Jane should cut 2 equal pieces, as a diameter of a pizza.</em>

Hope it helps!

From ItsNobody

Anna11 [10]3 years ago
3 0

Answer:

Into 1/3 pieces

Step-by-step explanation:

1/3 + 1/3 + 1/3 = 1 the whole pizza pie, yum

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Can someone help me find the diameter? <br><br> also what is a diameter even?
lukranit [14]

Answer:

24

Step-by-step explanation:

area of circle = πr²

πr² = 144π

Cancel out π

r² = 144

r = √144

r = 12

diameter = 2 x radius

diameter = 2 x 12

diameter = 24m

* diameter is the "width of a circle", the distance from the widest points of the circle

3 0
2 years ago
A ball is thrown from an initial height of 3 feet with an initial upward velocity of 37 fts. The ball's height h (in feet) after
tatyana61 [14]

Answer:

t=(37+root89)/32,(37-root89)/32

Step-by-step explanation:

Solve the equation:

23=3+37t-16t^2

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2 years ago
Algebra 1 - which equation represents exponential growth?
Nataly_w [17]

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d. y =5(1.06)^x

Step-by-step explanation:

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Each statement describes a transformation of the graph of y= x. Which statement correctly describes the graph of yu x- 8?
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3 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
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