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3 years ago

Given: f(x) = 3 - X; g(x) = -2x Find g[f(x)]

Mathematics

1 answer:

OLEGan [10]3 years ago

5 0

Answer:

g[f(x)] = -2(3 - x) = -6 + 2x

Step-by-step explanation:

When we substitute the function f(x) for x in g(x) = -2x, we get the composition g[f(x)]:

g[f(x)] = -2(3 - x) = -6 + 2x

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Help! I don't know how to do this! I will mark Brainliest!

gogolik [260]

Answer:

- 3y² - 7y + 6

Step-by-step explanation:

Given

(- 3y + 2)(y + 3)

Each term in the second factor is multiplied by each term in the first factor, that is

- 3y (y + 3) + 2(y + 3) ← distribute both parenthesis

= - 3y² - 9y + 2y + 6 ← collect like terms

= - 3y² - 7y + 6

4 0

3 years ago

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Consider the average daily winter temperature and your heating bill- is there likely to be a correlation? If so does the correla

creativ13 [48]

The answer to this is that t<span>here is a negative correlation and a causal correlation. The higher the average daily winter temperature the lower your heating bill.</span>

7 0

3 years ago

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103,727,495 in a word form

3241004551 [841]

103,727,495 in word form is: one hundred three million, seven hundred twenty-seven thousand, four hundred ninety-five.

3 0

3 years ago

A rectangular swimming pool is twice as long as it is wide. A small concrete walkway surrounds the pool. The walkway is a 2 feet

Ksivusya [100]

Answer:

The width and the length of the pool are 12 ft and 24 ft respectively.

Step-by-step explanation:

The length (L) of the rectangular swimming pool is twice its wide (W):

$L_{1} = 2W_{1}$

Also, the area of the walkway of 2 feet wide is 448:

$W_{2} = 2 ft$

$A_{T} = W_{2}*L_{2} = 448 ft^{2}$

Where 1 is for the swimming pool (lower rectangle) and 2 is for the walkway more the pool (bigger rectangle).

The total area is related to the pool area and the walkway area as follows:

$A_{T} = A_{1} + A_{w}$ (1)

The area of the pool is given by:

$A_{1} = L_{1}*W_{1}$

$A_{1} = (2W_{1})*W_{1} = 2W_{1}^{2}$ (2)

And the area of the walkway is:

$A_{w} = 2(L_{2}*2 + W_{1}*2) = 4L_{2} + 4W_{1}$ (3)

Where the length of the bigger rectangle is related to the lower rectangle as follows:

$L_{2} = 4 + L_{1} = 4 + 2W_{1}$ (4)

By entering equations (4), (3), and (2) into equation (1) we have:

$A_{T} = A_{1} + A_{w}$

$A_{T} = 2W_{1}^{2} + 4L_{2} + 4W_{1}$

$448 = 2W_{1}^{2} + 4(4 + 2W_{1}) + 4W_{1}$

$224 = W_{1}^{2} + 8 + 4W_{1} + 2W_{1}$

$224 = W_{1}^{2} + 8 + 6W_{1}$

By solving the above quadratic equation we have:

W₁ = 12 ft

Hence, the width of the pool is 12 feet, and the length is:

$L_{1} = 2W_{1} = 2*12 ft = 24 ft$

Therefore, the width and the length of the pool are 12 ft and 24 ft respectively.

I hope it helps you!

8 0

3 years ago

Reduce fraction: a^3+a^2b/5a times 25/3b+3a

Margaret [11]

ANSWER

$\frac{5a}{3}$

EXPLANATION

The given fractions are:

$\frac{{a}^{3} + {a}^{2} b}{5a} \times \frac{25}{3b + 3a}$

We factor to obtain:

$\frac{{a}^{2}(a + b)}{5a} \times \frac{25}{3(a + b)}$

We cancel the common factors to get:

$\frac{{a}(1)}{1} \times \frac{5}{3(1)}$

We multiply the numerators and also multiply the denominators to get:

$\frac{5a}{3}$

Therefore the two fractions simplifies to $\frac{5a}{3}$

3 0

3 years ago