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enyata [817]
3 years ago
13

Rewrite the following expression x^9/7

Mathematics
1 answer:
Alexxx [7]3 years ago
4 0

Answer:

\sqrt[7]{x^9}

Step-by-step explanation:

its kinda difficult for me to explain so i guess just use this equation next time:

x^{m/n} = \sqrt[n]{x^m}

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Solve this parallelogram problem. Find x and y. Show your work
krok68 [10]

Answer:

2x+9 = 27 => 2x =18 => x =9

5y = 125 => y = 25

8 0
3 years ago
A flagpole that was originally 24 feet tall has cracked 9 feet from the ground and has fallen as if hinged. Find
Dimas [21]
The top of the flag pole touched the ground 12 feet from the base.

Since it cracked 9 feet from the base, 9 is the vertical leg of the right triangle formed.  The rest of the flagpole, 24-9=15, is the hypotenuse, as it is slanted down to the ground.  The missing side is the horizontal leg, which we will call b:

9²+b²=15²
81+b²=225

Subtract 81 from both sides:
81+b²-81=225-81
b²=144

Take the square root of both sides:
√b²=√144
b=12
8 0
3 years ago
What is the slope of -1,7 and 3,10
Diano4ka-milaya [45]

Answer:

m=3/4

Step-by-step explanation:

slope=\frac{y_{2}-y_{1} }{x_{2}-x_{1}} \\

\frac{10-(7)}{3-(-1)} \\\frac{10-7}{3+1}\\\frac{3}{4} \\ =3/4\\

Therefore the slope is 3/4

3 0
3 years ago
I NEED HELP ASAP PLS ANSWER
viktelen [127]
I gotchu!! It’s 40:))
3 0
3 years ago
Find a power series representation for the function. (give your power series representation centered at x = 0. ) f(x) = ln(5 − x
klio [65]

Recall that for |x|, we have the convergent geometric series

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

Now, for \left|\frac x5\right| < 1, we have

\dfrac1{5 - x} = \dfrac15 \cdot \dfrac1{1 - \frac x5} = \dfrac15 \displaystyle \sum_{n=0}^\infty \left(\frac x5\right)^n = \sum_{n=0}^\infty \frac{x^n}{5^{n+1}}

Integrating both sides gives

\displaystyle \int \frac{dx}{5-x} = C + \int \sum_{n=0}^\infty \frac{x^n}{5^{n+1}} \, dx

\displaystyle -\ln(5-x) = C + \sum_{n=0}^\infty \frac{x^{n+1}}{5^{n+1}(n+1)}

If we let x=0, the sum on the right side drops out and we're left with C=-\ln(5).

It follows that

\displaystyle \ln(5-x) = \ln(5) - \sum_{n=0}^\infty \frac{x^{n+1}}{5^{n+1}(n+1)}

or

\displaystyle \ln(5-x) = \boxed{\ln(5) - \sum_{n=1}^\infty \frac{x^n}{5^n n}}

3 0
2 years ago
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