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timurjin [86]
2 years ago
9

PLEASE HELP I WILL GIVE BRANLIEST IF YOUR ANSWER IS CORRECT!

Mathematics
2 answers:
Tamiku [17]2 years ago
5 0
1:4 2:8 3:12 4:16 because every one is adding 4 hope this helps. ;)
Vilka [71]2 years ago
3 0

Answer:

1:4

2:8

3:12

4:16

Step-by-step explanation:

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For what value of x does 32X – 9^3x-4?
Lubov Fominskaja [6]

Answer: X = 4/61

Step-by-step explanation

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3 years ago
Eremy works two jobs. He makes $8/hr working at a deli and $12/hr working for a landscaper. Let x represent the number of hours
bezimeni [28]

Answer:

A.) combinations of the hours he can work at the two jobs and earn at least $300

Step-by-step explanation:

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Read 2 more answers
4. Using the geometric sum formulas, evaluate each of the following sums and express your answer in Cartesian form.
nikitadnepr [17]

Answer:

\sum_{n=0}^9cos(\frac{\pi n}{2})=1

\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=0

\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})=\frac{1}{2}

Step-by-step explanation:

\sum_{n=0}^9cos(\frac{\pi n}{2})=\frac{1}{2}(\sum_{n=0}^9 (e^{\frac{i\pi n}{2}}+ e^{\frac{i\pi n}{2}}))

=\frac{1}{2}(\frac{1-e^{\frac{10i\pi}{2}}}{1-e^{\frac{i\pi}{2}}}+\frac{1-e^{-\frac{10i\pi}{2}}}{1-e^{-\frac{i\pi}{2}}})

=\frac{1}{2}(\frac{1+1}{1-i}+\frac{1+1}{1+i})=1

2nd

\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=\frac{1-e^{\frac{i2\pi N}{N}}}{1-e^{\frac{i2\pi}{N}}}

=\frac{1-1}{1-e^{\frac{i2\pi}{N}}}=0

3th

\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})==\frac{1}{2}(\sum_{n=0}^\infty ((\frac{e^{\frac{i\pi n}{2}}}{2})^n+ (\frac{e^{-\frac{i\pi n}{2}}}{2})^n))

=\frac{1}{2}(\frac{1-0}{1-i}+\frac{1-0}{1+i})=\frac{1}{2}

What we use?

We use that

e^{i\pi n}=cos(\pi n)+i sin(\pi n)

and

\sum_{n=0}^k r^k=\frac{1-r^{k+1}}{1-r}

6 0
3 years ago
What is the result of isolating y2 in the equation below? 9x2 + 7y2 = 42
ElenaW [278]
First, move 9x^2 to the right:

7y^2 = 42 - 9x^2

Then, divide 7 on both sides:

y^2 = 6 - 9/7 * x^2
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Find the y intercept of a line with a slope of -4 and point (2,1)
Oxana [17]

Is this a rise over run question

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