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3 years ago

6

ASAP! GIVING BRAINLIEST! Please read the question THEN answer CORRECTLY! NO guessing. I say no guessing because people usually g

uess on my questions.

1 answer:

Goshia [24]3 years ago

8 0

Answer:

A

Step-by-step explanation:

(f-g)(x) means that the 2 functions are being subtracted.

$3^{x}$ +10x -(5x-3) = $3^{x}$ +10x-5x+3

simplify!

$3^{x}$ +5x+3

the answer is A

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Lostsunrise [7]

Answer:

A. x=0

Step-by-step explanation:

x+20+10x=20+9x

cancel equal terms

x+10x+9x

add x to 10x

11x=9x

move variable to left

11x-9x=0

subtract 11x to 9x

2x=0

divide Both sides by 2

x=0

4 0

3 years ago

Question is in screenshot

yan [13]

Answer:

100/10 by 2/1 10.$01

20/10 by 2/1

6 0

3 years ago

Two urns each contain green balls and blue balls. urn i contains 4 greenballs and 6 blue balls, and urn ii contains 6 green ball

Lemur [1.5K]

1st urn....4 green balls, 6 blue balls.....10 total balls....probability of drawing a blue ball in urn 1 is : 6/10 reduces to 3/5

2nd urn...6 green balls, 2 blue balls....8 total balls...probability of drawing a blue ball in urn 2 is : 2/8 reduces to 1/4

probability both balls are blue : 3/5 * 1/4 = 3/20 <==

5 0

3 years ago

Find the product of 5/6 and 42

Keith_Richards [23]

Answer: Its 35 hope this helped

3 0

3 years ago

Use the denition of the derivative to find f 0 (3), where f (x) = 3x+5 / 2x−1

Crazy boy [7]

Answer:

$f'(3)= -\frac{13}{25}$

Step-by-step explanation:

We are asked to find $f'(3)$ of function $f(x)=\frac{3x+5}{2x-1}$ using definition of derivatives.

Limit definition of derivatives:

$f'(x)= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$

Let us find $f(3+h)$ and $f(3)$ .

$f(3+h)=\frac{3(3+h)+5}{2(3+h)-1}$

$f(3+h)=\frac{9+3h+5}{6+2h-1}\\\\f(3+h)=\frac{3h+14}{2h+5}$

$f(3)=\frac{3(3)+5}{2(3)-1}$

$f(3)=\frac{9+5}{6-1}\\\\f(3)=\frac{14}{5}$

Substituting these values in limit definition of derivatives, we will get:

$f'(3)= \lim_{h \to 0} \frac{f(3+h)-f(3)}{h}$

$f'(3)= \lim_{h \to 0} \frac{\frac{3h+14}{2h+5}-\frac{14}{5}}{h}$

Make a common denominator:

$f'(3)= \lim_{h \to 0} \frac{\frac{(3h+14)*5}{(2h+5)*5}-\frac{14(2h+5)}{5(2h+5)}}{h}$

$f'(3)= \lim_{h \to 0} \frac{\frac{5(3h+14)-14(2h+5)}{5(2h+5)}}{h}$

$f'(3)= \lim_{h \to 0} \frac{5(3h+14)-14(2h+5)}{5h(2h+5)}$

$f'(3)= \lim_{h \to 0} \frac{15h+70-28h-70}{5h(2h+5)}$

$f'(3)= \lim_{h \to 0} \frac{-13h}{5h(2h+5)}$

Cancel out h:

$f'(3)= \lim_{h \to 0} \frac{-13}{5(2h+5)}$

$f'(3)= \frac{-13}{5(2(0)+5)}$

$f'(3)= \frac{-13}{5(5)}$

$f'(3)= -\frac{13}{25}$

Therefore, $f'(3)= -\frac{13}{25}$ .

8 0

3 years ago