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3 years ago

What is wrong with this solution

Mathematics

1 answer:

zzz [600]3 years ago

7 0

Answer:

A. -5m/5 does not equal m

Step-by-step explanation:

-5m/5 = -m, not m.

In step 2, both sides should be divided by -5, not by 5.

Answer: A. -5m/5 does not equal m

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10(1+3x)=-20 and -5x-8(1+7x)=-8 <br> With steps

SOVA2 [1]

(1)

10(1+3x)=-20

Cross multiply.

10+30x=-20

Isolate x on one side. So you would subtract 10 on each side. and one side will cross each other out. leaving you with,

30x=-20-10

Subtract 10 from -20.

30x=-30

Divide each side by 30 to get x.

30x÷30=-30÷30

Therefore,

x=-1

(2)

-5x-8(1+7x)=-8

Cross multiply -8(1+7x)

-5x-8-56x=-8

Isolate all x's on one side. So you would add 8 on each side. and one side will cross each other out. leaving you with,

-5x-56x=-8+8

Add 8 to -8.

-5x-56x=0

Subtract -56x from -5x

-61x=0

Divide each side by -61

-61x÷-61=0÷-61

Therefore,

x=0

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3 years ago

Find the surface area of the prism. Write your answer as a fraction or mixed number

Serjik [45]

Answer:

2 and 2/3

Step-by-step explanation:

(2/3)x(2/3)x6=

2 and 2/3

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2 years ago

Bob paid 17.50 for a box of cards and .35 each for 18 stamps. What was the total cost in dollars and cents of the box of cards a

alexgriva [62]

$23.80 for a box of cards and 18 stamps

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2 years ago

Use the power series for 1 1−x to find a power series representation of f(x) = ln(1−x). What is the radius of convergence? (Note

Viktor [21]

a. Recall that

$\displaystyle\int\frac{\mathrm dx}{1-x}=-\ln|1-x|+C$

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

By integrating both sides, we get

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

If $x=0$ , then

$\displaystyle-\ln1=C+\sum_{n=0}^\infty\frac{0^{n+1}}{n+1}\implies 0=C+0\implies C=0$

so that

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

We can shift the index to simplify the sum slightly.

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

b. The power series for $x\ln(1-x)$ can be obtained simply by multiplying both sides of the series above by $x$ .

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

c. We have

$\ln2=-\dfrac\ln12=-\ln\left(1-\dfrac12\right)$

$\displaystyle\implies\ln2=\sum_{n=1}^\infty\frac1{n2^n}$

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drek231 [11]

Parallelograms. Rectangles. Trapezoids.

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