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4 years ago

2.80=9(70)/I. I need a answer for i

Mathematics

1 answer:

Westkost [7]4 years ago

8 0

I would be 225 since 9(70) is 630 and since you are solving for I, you multiply both sides by i and divide by 2.80

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Pls pls pls answer this fast !!

scoundrel [369]

Answer:

negatives on the right and positives on the right if you can do tht

Step-by-step explanation:

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3 years ago

Simplify (6x2 + 11x − 3) + (2x2 − 17x − 4).

d1i1m1o1n [39]

Answer:

B. 8x^2-6x-7

Step-by-step explanation:

All you have to do is combine the like terms.

Like terms are the terms that have the same variable and same exponent.

The like terms in this equation are 6x^2 and 2x^2, 11x and -17x, and -3 and -4.

When you add 6x^2 and 2x^2, you get 8x^2

When you add 11x and -17x, you get -6x

When you add -3 and -4, you get -7.

Putting these all in order, your answer is

8x^2 - 6x - 7

8 0

3 years ago

Need help with these 2 questions

posledela

Answer:

I worked really hard on this problem, I had to manually edit with my editing tool for 15 minutes lol.

4 0

3 years ago

Determine the domain and range of (g ○ f)(x) if f of x is equal to 4 over the quantity x squared minus 4 end quantity and g(x) =

lara [203]

The domain and range of a function are the possible x and y values of the function.

The domain and the range of the function is: (a) $\mathbf{D:\{x \in R|x \ne -2,2\}}$ and $\mathbf{R:\{(-\infty,1) \cup (2,\infty)\}}$

The functions are given as:

$\mathbf{f(x) = \frac{4}{x^2 - 4}}$

$\mathbf{g(x) = x + 2}$

(g o f)(x) is calculated as:

$\mathbf{(g\ o\ f)(x) = g(f(x))}$

So, we have:

$\mathbf{(g\ o\ f)(x) = \frac{4}{x^2 - 4} + 2}$

Take LCM

$\mathbf{(g\ o\ f)(x) = \frac{4 + 2x^2 - 8}{x^2 - 4} }$

$\mathbf{(g\ o\ f)(x) = \frac{2x^2 - 4}{x^2 - 4} }$

Represent the denominator as follows, to calculate the domain

$\mathbf{x^2 - 4 \ne 0 }$

Add 4 to both sides

$\mathbf{x^2 \ne 4 }$

Take square roots of bot sides

$\mathbf{x \ne \±2 }$

Hence, the domain of the function is:

$\mathbf{D:\{x \in R|x \ne -2,2\}}$

On the graph of $\mathbf{(g\ o\ f)(x) = \frac{2x^2 - 4}{x^2 - 4} }$ (see attachment), the function does not have a value from y = 1 to 2.

Hence, the range is:

$\mathbf{R:\{(-\infty,1) \cup (2,\infty)\}}$

Read more about domain and range at:

brainly.com/question/1632425

5 0

2 years ago

Read 2 more answers

What is this expression in scientific notation

goblinko [34]

76 x 10^5 or 7.6 x 10^6

5 0

3 years ago