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dimulka [17.4K]
3 years ago
8

15y^2-15y+2y+7-2y^2+4

Mathematics
1 answer:
scoray [572]3 years ago
8 0

Answer:

13 y^2 - 13 y + 11

Step-by-step explanation:

Simplify the following:

15 y^2 - 2 y^2 + 2 y - 15 y + 4 + 7

Grouping like terms, 15 y^2 - 2 y^2 + 2 y - 15 y + 4 + 7 = (15 y^2 - 2 y^2) + (-15 y + 2 y) + (7 + 4):

(15 y^2 - 2 y^2) + (-15 y + 2 y) + (7 + 4)

15 y^2 - 2 y^2 = 13 y^2:

13 y^2 + (-15 y + 2 y) + (7 + 4)

2 y - 15 y = -13 y:

13 y^2 + -13 y + (7 + 4)

7 + 4 = 11:

Answer: 13 y^2 - 13 y + 11

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a circle with a radius of 1 cm sits inside a 3cm x 3cm rectangle. What is the area of the shaded region (the rectangle is shaded
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Answer:

The area of the shaded region is 5.86\ cm^{2}

Step-by-step explanation:

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If x+y = 1, and x^2 + y^2 = -1, what is x^7 + y^13?<br> 1) 0<br> 2) 1<br> 3) -2<br> 4) 2<br> 5) -1
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Solve for <em>x</em> and <em>y</em> :

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2<em>x</em> ² - 2<em>x</em> + 2 = 0

<em>x</em> ² - <em>x</em> + 1 = 0

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<em>x</em> - 1/2 = ±√(-3/4)

<em>x</em> - 1/2 = ±√3/2 <em>i</em>

<em>x</em> = 1/2 ± √3/2 <em>i</em>   →   <em>x</em> = exp(± <em>iπ</em>/3)

<em>y</em> = 1 - (1/2 ± √3/2 <em>i</em> )   →   <em>y</em> = -1/2 ± √3/2 <em>i</em>   →   <em>y</em> = exp(± 2<em>iπ</em>/3)

Then

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… = exp(± <em>iπ</em>/3) + exp(± 2<em>iπ</em>/3)

since 7<em>π</em>/3 is equivalent to <em>π</em>/3, and 26<em>π</em>/3 is equivalent to 2<em>π</em>/3 (both modulo 2<em>π</em>).

In either case, we get

<em>x </em>⁷ + <em>y </em>¹³ = <em>x</em> + <em>y</em> = 1

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AfilCa [17]

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Since, the graph of the function passes through a point (-2, 2),

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2 = k

Therefore, the function is h(x)=2(x + 3)^{\frac{1}{2} }.

7 0
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