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4 years ago

Solve the equation 7m-5-13m=25 by combining like terms. Show all work please need asap

2 answers:

7 0

You must combine like terms.
First add 7m and -13m which makes -6m. That makes the equation
-6m - 5 =25.
Then add 5 on both sides, which makes -6m = 30.
Finally divide -6 on both sides and the answer will be
m= -5

Strike441 [17]4 years ago

4 0

You subtract 7m and 13m which would give you -6m after you have -6m - 5= 25 then you add 5 which gives you 30 you then you divide that by -6 which gives you m=-5

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6x+3y=24 y=-2x+8 plz help

Whitepunk [10]

Solve for the first variable in one of the equations, then substitute the result into the other equation.

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(x,−2x+8)(x,-2x+8)

Equation Form:x=xy=−2x+8

8 0

3 years ago

Read 2 more answers

This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive an

GaryK [48]

Answer:

The series representation is $\sum_{n=0}^{\infty}(-1)^n\frac{x^n}{6^{n+1}}$ and the interval of convergence is (-6,6)

Step-by-step explanation:

We want to find a series, such that f(x) = \sum_{n=0}{\infty}a_n(x-a)^{n}[/tex], were a is the value that we are using to center the series expansion. In our case, a=0.

We will use the geometric series formula as follows. For |r|<1 then

$\frac{1}{1-r}=\sum_{n=0}^{\infty} r^n$

In our case, with some algebreaic manipulation we have that

$f(x) = \frac{1}{6+x} = \frac{1}{6}\frac{1}{1-(\frac{-x}{6})}$

Taking $r = \frac{-x}{6}$ we get that

$f(x) = \frac{1}{6}\sum_{n=0}^{\infty} (\frac{-x}{6})^n = \sum_{n=0}^{\infty}(-1)^n\frac{x^n}{6^{n+1}}$

This representation is valid (that means that the series converges to the value of f(x)) only for |r|<1. That is

$\left|\frac{-x}{6}\right|= \left|\frac{x}{6}\right|$

which implies that |x|<6. So the interval of convergence is (-6,6).

4 0

4 years ago

6 (2p + 3 ) = -37 +7p

Diano4ka-milaya [45]

12p+18= -37+7p

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3 years ago

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Sonja [21]

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4 0

3 years ago

(5t^3)^-4 1. 625/t^12 2. 20/t^7 3. 1/625t^12 4. 1/20t^7

rusak2 [61]

Answer:

$\large\boxed{3.\ \dfrac{1}{625t^{12}}}$

Step-by-step explanation:

$\text{Use}\\\\(ab)^n=a^nb^n\\\\(a^n)^m=a^{nm}\\\\a^{-n}=\dfrac{1}{a^n}\\---------------\\\\(5t^3)^{-4}=\dfrac{1}{(5t^3)^4}=\dfrac{1}{5^4(t^3)^4}=\dfrac{1}{625t^{(3)(4)}}=\dfrac{1}{625t^{12}}$

4 0

3 years ago