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ivann1987 [24]
3 years ago
11

What expression is equivalent to 5y^-3?

Mathematics
1 answer:
sineoko [7]3 years ago
7 0

Answer:

\frac{5}{y^{3}}

Step-by-step explanation:

apply exponent rule:

a^{-b}=\frac{1}{a^b}

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Please help me! And get marked as brainlist!
Luda [366]

Answer:

Step-by-step explanation:

a) area of the rectangle = length × breadth

Therefore area should be = 12cm × 6cm

Which should give you the answer 72centimetre squared.

b) area of triangle A = 1/2 × b × h

So that is 1/2 × 3 × 6

Which is 18 centimetres × 1/2

Which is 9 centimetres squared

c) area of triangle B = 1/2 × b × h

So 1/2×3×6=18

Now 1/2×18 should give you 9 centimetres squared

d) therefore the area of the parallelogram = 72+9+9=90 centimetres squared

Therefore the area of the parallelogram is 90 centimetres squared

8 0
3 years ago
Find the missing side. Round it to the nearest tenth.​
Alenkasestr [34]

Answer:

19.1

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
What is the volume of the rectangular prism in cubic centimeters?
son4ous [18]
The answer is 348 so D
6 0
3 years ago
-9 + ( -16) =
Advocard [28]

Answer:

-9 + ( -16) = -25

( -4) – (-6) + (-5) – (8) =-11

( 3) – (-9) + (-3) – (-2) =11

- ( 5) – (-7) + (-8) =-6

- ( -8) – (-9) + (-4) – (-1) =14

( -5) – (-11) + (10) – (8) =8

( -1) – (-2) + (-3) – (-6) =4

( -4) – ( 4) – (6) + (-5) – (-8) =-11

(-6) + (-5) – (-8) =-3

Step-by-step explanation:

Hope it helps

6 0
3 years ago
Read 2 more answers
Expand:<br><img src="https://tex.z-dn.net/?f=f%28z%29%20%3D%20%20%5Cfrac%7B1%7D%7Bz%28z%20-%202%29%7D%20" id="TexFormula1" title
Monica [59]

Expand f(z) into partial fractions:

\dfrac1{z(z-2)} = \dfrac12 \left(\dfrac1{z-2} - \dfrac1z\right)

Recall that for |z| < 1, we have the power series

\displaystyle \frac1{1-z} = \sum_{n=0}^\infty z^n

Then for |z| > 2, or |1/(z/2)| = |2/z| < 1, we have

\displaystyle \frac1{z-2} = \frac1z \frac1{1 - \frac2z} = \frac1z \sum_{n=0}^\infty \left(\frac 2z\right)^n = \sum_{n=0}^\infty \frac{2^n}{z^{n+1}}

So the series expansion of f(z) for |z| > 2 is

\displaystyle f(z) = \frac12 \left(\sum_{n=0}^\infty \frac{2^n}{z^{n+1}} - \frac1z\right)

\displaystyle f(z) = \frac12 \sum_{n=1}^\infty \frac{2^n}{z^{n+1}}

\displaystyle f(z) = \sum_{n=1}^\infty \frac{2^{n-1}}{z^{n+1}}

\displaystyle \boxed{f(z) = \frac14 \sum_{n=2}^\infty \frac{2^n}{z^n} = \frac1{z^2} + \frac2{z^3} + \frac4{z^4} + \cdots}

6 0
2 years ago
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