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3 years ago

11

Dont not take answer from Zayar223, or Zepav03

2 answers:

Agata [3.3K]3 years ago

7 0

Answer:

thanks

Step-by-step explanation:

Aleksandr-060686 [28]3 years ago

3 0

Wym??????? who are they

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The nakagin capsule tower has 140 modules and is 14 stories high if the modules were divided evenly among the number of stories

ArbitrLikvidat [17]

To determine the number of modules that is in each story, we simply divide the number of modules by the number of stories.
= 140 / 14
Simplifying,
= 10 modules / story
Therefore, by even distribution we note that there are 10 modules per story.

8 0

3 years ago

PLEASE PLEASE HELP ME IM HORRIBLE AT MATH

belka [17]

Step-by-step explanation:

can u show the list of options?

3 0

4 years ago

21 A stick 7 inches long is broken into two pieces, so that one piece is twice as long as the other one. How long are the two pi

Hunter-Best [27]

Answer:

2.33 inches and 4.66 inches

Step-by-step explanation:

2x+x=7 or 3x=7

divide by three and x =2.33

that's for the first one and the second one is twice the size so it'll be 4.66 inches

3 0

3 years ago

Read 2 more answers

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

The sum of their age 97. The difference of their ages is 19. Find their ages?

arsen [322]

Im not sure but I can at least try so I got one (29) and one (48)because I divided 97 by 2 since the sum is addition then I got 48 a subtracted 19 because subtraction is "difference"so I dont know if I'm wrong but at least I helped

5 0

4 years ago