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3 years ago

Help me with this please

Mathematics

1 answer:

lara [203]3 years ago

3 0

a)Image of 9=-9 (on the x line)

image of (9,3) across y line is(-9,3)

across x line is (9,-3)

b) (0,0)

(1,1)

(4,2)

(9,3)

(16,4)

(25,5)

(36,6)

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Solve the following equation 6x-1=3 (2x+4)-1

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$6x - 1 = 3(2x + 4) - 1$
Distribute
$6x - 1 = 6x + 12 - 1$
Combine like terms
$6x - 1 = 6x + 11$
Add 1 and subtract 6x from each side
0 = 12
The equation has no real solution

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Tonya plans to join a fitness club. She researched the costs to join different clubs in her area and found that there is a linea

Ne4ueva [31]

Answer: 1 club - Club 2

Step-by-step explanation:

You can find the monthly rates by deducting the cost at 12 months from the cost at 24 months and dividing it by 12.

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= 18 * 6 = 15 * 6 = 21 * 6

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Same as total cost at Joining fee of $30; No joining fee as

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2 years ago

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If f (n)(0) = (n + 1)! for n = 0, 1, 2, , find the taylor series at a=0 for f.

Pie

Given that $f^{(n)}(0)=(n+1)!$ , we have for $f(x)$ the Taylor series expansion about 0 as

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

Replace $n+1$ with $n$ , so that the series is equivalent to

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and notice that

$\displaystyle\frac{\mathrm d}{\mathrm dx}\sum_{n=0}^\infty x^n=\sum_{n=1}^\infty nx^{n-1}$

Recall that for $|x|$ , we have

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

which means

$f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}=\frac{\mathrm d}{\mathrm dx}\frac1{1-x}$
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2 years ago

On-the-Go Phone Company has two monthly plans for their customers. The EZ Pay Plan costs $0.15

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Answer:

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