All categories

3 years ago

Consider f(x) = 1.8x – 10 and g(x) = −4.

Mathematics

2 answers:

erastova [34]3 years ago

6 0

Answer:

Step-by-step explanation:

The input value at which f(x) = g(x) can be found by setting the right sides of each of those equal to each other and solving for x.

1.8x - 10 = -4 and

1.8x = 6 so

x = 3 1/3

The bold equation is the answer to the first part, and the second part is that x = 3 1/3

Naily [24]3 years ago

4 0

Answer:

Step-by-step explanation:

yw egen

You might be interested in

PLEASE HELP!!! Julianne opens a dance studio. Her start-up costs for the building, advertising, and supplies total $52,000. Each

givi [52]

Alright, lets gets started.

Jullianne's start-up costs for the building, advertising, and supplies total $52,000.

she spends $680 on operating costs on daily basis.

Means Suppose the days d, the operating cost will be= $680 d$

This cost will be added in her start up cost.

The earning for d days will be = $960 d$

So, part A :

$5200 + 680 d < 960 d$

Now suppose after d days, she will start making profit means, after d days, her cost will be equal to her earning.

So, part B:

$5200 + 680 d = 960 d$

Subtracting 680 d from both sides

$5200 + 680 d - 680 d = 960 d - 680 d$

$5200 = 280 d$

Dividing 280 in both sides

$d = \frac{5200}{280}=18.6$

Means after 19 days, she will start making profit. : Answer

Hope it will help :)

7 0

3 years ago

Allison has 1/2 cup of yogurt for making fruit parfait

ivolga24 [154]

Allison can make four parfaits because you need 1/8 cup of yogurt and Allison has 1/2 cup of yogurt. So the half of the cup would give her 4 * 1/8 to equal 1/2.

4 0

3 years ago

Find the derivative of the following functions: a. f(x) = (x^3 + 5)^1/4 - 15e^x^3 b. f(x) = (x - 3)^2 (x - 5)/(x - 4)^2(x^2 + 3)

Darya [45]

Answer:

Step-by-step explanation:

Given function is

(a)F(x)= $\left ( x^{3}+5\right )^{0.25}-15e^{x^{3}}$

$F^{'}\left ( x\right )$ = $0.25\left ( x^{3}+5\right )^{-0.75}\frac{\mathrm{d} x^{3}}{\mathrm{d} x}-15e^{x^{3}}\frac{\mathrm{d} x^{3}}{\mathrm{d} x}$

$F^{'}\left ( x\right )=0.25\left ( x^{3}+5\right )^{-0.75}\times 3x^{2}-15e^{x^{3}}\times \left ( 3x^{2}\right )$

(b)F(x)= $\frac{\left ( x-3\right )^2\left ( x-5\right )}{\left ( x-4\right )^2\left ( x^{2}+3\right )^5}$

$F^{'}\left ( x\right )$ = $\frac{\left [ 2\left ( x-3\right )\right \left ( x-5\right )+\left ( x-3\right )^2]\left [ \left ( x-4\right )^2\left ( x^2+3\right )^5\right ]-\left [ 2\left ( x-4\right )^{3}\left ( x^2+3\right )^5+5\left ( x^2+3\right )^4\left ( 2x\right )\left ( x-4\right )^2\right ]\left [ \left ( x-3\right )^2\left ( x-5\right )\right ]}{\left [\left ( x-4\right )^2\left ( x^2+3\right )^5\right ]^2}$