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Anna35 [415]
3 years ago
8

7 + a/3 = 5 HELP ASAPPP :)

Mathematics
1 answer:
serg [7]3 years ago
4 0
A = -6 , that is the answer
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Find the value of x.
lidiya [134]

Answer:

x = 19

Step-by-step explanation:

180-123 = 57

x = 57/3

x = 19

5 0
3 years ago
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Solve and show steps F(-5)=-2x+1 / 3
Blizzard [7]

Answer:

11/3

Step-by-step explanation:

-2×-5 +1 /3

=10+1/3

5 0
3 years ago
Which inequailities has no solution? I need this to be answer ASAP!
Nadusha1986 [10]
Let's simplify them top to bottom.

6(x + 2) < x - 3
6x + 12 < x - 3
5x + 12 < -3
5x < -15
x < -3
This one checks out.


3 + 4x <= 2(1 + 2x)
4x + 3 <= 4x + 2

This is a problem. In no world is 4x + 3 ever less than or equal to 4x + 2. This one has no solution.


3 0
3 years ago
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
to make 2 batches of nut bars, Jayda needs to use 4 eggs. how many eggs are used in each batch of nut bars?
maxonik [38]
2 eggs are needed because if you need 4 to make 2 batches, you have to divide 4 by 2. 
5 0
3 years ago
Read 2 more answers
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