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2 years ago

12

Write a equation that represents this comparison sentence 32 is 8 times as many as 4

1 answer:

inna [77]2 years ago

7 0

32 = 8 x 4 I think that is the answer

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A couple plans to have 3 children, what is the probability that atleast two will be boys?

Olin [163]

1/4 or 25 percent because there is a 1/2 chance to get a boy or girl so the probability that will be boys us 1/2 x 1/2= 1/4

6 0

2 years ago

Read 2 more answers

. (01.02 HC) Javier has a job installing windows. He earns a rate of $75 per day plus $8 per window installed. He also receives

Anettt [7]

Part A: Coefficient is either the 75 or 8 since it's linked with the Variable(letter) d or w. The only constant is 25
Part B: 75(5)+8(48)+25=$784 Just plug in for the variables.
Part C: Only the constant since it is a set amount and it is not changed based on days or windows installed.

6 0

3 years ago

In this figure AB||CD and m<1=120°<br>What is m<5?

mars1129 [50]

Answer:

120 degrees

Step-by-step explanation:

Angle 1 is congruent to angle 5 by the corresponding angles theorem.

3 0

1 year ago

8 cubed in standard form and exponential form

masya89 [10]

Answer:

so 8 cubed (8 × 8 × 8) in standard form is: 512

so 8 cubed (8 × 8 × 8) in exponential form is: 8³

Step-by-step explanation:

As we know that

The cube of any given number is that number times itself.

For example, 8 cubed, denoted by 8³ is equal to 8 × 8 × 8.

An exponent basically indicates the number of times we must multiply the base number. For example, to compute 8³, we multiply 3 three times (8×8×8).

so 8 cubed (8 × 8 × 8) in standard form is: 512

so 8 cubed (8 × 8 × 8) in exponential form is: 8³

8 0

3 years ago

SOLUTION We observe that f '(x) = -1 / (1 + x2) and find the required series by integrating the power series for -1 / (1 + x2).

Ann [662]

Answer:

Required series is:

$\int{\frac{-1}{1+x^{2}} \, dx =-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....$

Step-by-step explanation:

Given that

$f'(x) = -\frac{1}{1 + x^{2}}$ ---(1)

We know that:

$\frac{d}{dx}(tan^{-1}x)=\frac{1}{1+x^{2}}$ ---(2)

Comparing (1) and (2)

$f'(x)=-(tan^{-1}x)$ ---- (3)

Using power series expansion for $tan^{-1}x$

$f'(x)=-tan^{-1}x=-\int {\frac{1}{1+x^{2}} \, dx$

$= -\int{ \sum\limits^{ \infty}_{n=0} (-1)^{n}x^{2n}} \, dx$

$= -\sum{ \int\limits^{ \infty}_{n=0} (-1)^{n}x^{2n}} \, dx$

$=-[c+\sum\limits^{ \infty}_{n=0} (-1)^{n}\frac{x^{2n+1}}{2n+1}]$

$=C+\sum\limits^{ \infty}_{n=0} (-1)^{n+1}\frac{x^{2n+1}}{2n+1}$

$=C-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....$

as

$tan^{-1}(0)=0 \implies C=0$

Hence,

$\int{\frac{-1}{1+x^{2}} \, dx =-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....$

7 0

3 years ago