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3 years ago

14

HELP PLEASE FOR BRAINLEIST

1 answer:

strojnjashka [21]3 years ago

5 0

Answer:

First one is c because 3 1/2 - 1/2=3

Second one is also c because 0.9x2=1.8

And third one c is because 2 2/3:3/1=1 1/8

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Reptile [31]

Answer:

B I'm pretty sure

Step-by-step explanation:

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3 years ago

If a relationship can be written as s=−5+2t, how does s change if t is increasing

Alenkinab [10]

S would also increase if t is being increased

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3 years ago

Read 2 more answers

The sum of two terms of gp is 6 and that of first four terms is 15/2.Find the sum of first six terms.

Gnoma [55]

Given:

The sum of two terms of GP is 6 and that of first four terms is $\dfrac{15}{2}.$

To find:

The sum of first six terms.

Solution:

We have,

$S_2=6$

$S_4=\dfrac{15}{2}$

Sum of first n terms of a GP is

$S_n=\dfrac{a(1-r^n)}{1-r}$ ...(i)

Putting n=2, we get

$S_2=\dfrac{a(1-r^2)}{1-r}$

$6=\dfrac{a(1-r)(1+r)}{1-r}$

$6=a(1+r)$ ...(ii)

Putting n=4, we get

$S_4=\dfrac{a(1-r^4)}{1-r}$

$\dfrac{15}{2}=\dfrac{a(1-r^2)(1+r^2)}{1-r}$

$\dfrac{15}{2}=\dfrac{a(1+r)(1-r)(1+r^2)}{1-r}$

$\dfrac{15}{2}=6(1+r^2)$ (Using (ii))

Divide both sides by 6.

$\dfrac{15}{12}=(1+r^2)$

$\dfrac{5}{4}-1=r^2$

$\dfrac{5-4}{4}=r^2$

$\dfrac{1}{4}=r^2$

Taking square root on both sides, we get

$\pm \sqrt{\dfrac{1}{4}}=r$

$\pm \dfrac{1}{2}=r$

$\pm 0.5=r$

Case 1: If r is positive, then using (ii) we get

$6=a(1+0.5)$

$6=a(1.5)$

$\dfrac{6}{1.5}=a$

$4=a$

The sum of first 6 terms is

$S_6=\dfrac{4(1-(0.5)^6)}{(1-0.5)}$

$S_6=\dfrac{4(1-0.015625)}{0.5}$

$S_6=8(0.984375)$

$S_6=7.875$

Case 2: If r is negative, then using (ii) we get

$6=a(1-0.5)$

$6=a(0.5)$

$\dfrac{6}{0.5}=a$

$12=a$

The sum of first 6 terms is

$S_6=\dfrac{12(1-(-0.5)^6)}{(1+0.5)}$

$S_6=\dfrac{12(1-0.015625)}{1.5}$

$S_6=8(0.984375)$

$S_6=7.875$

Therefore, the sum of the first six terms is 7.875.

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3 years ago

What is the slope between<br> the points (0,-3) and (-6,7)?

IceJOKER [234]

Answer:

-5/3

Step-by-step explanation:

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3 years ago

Show work. This is a BC calculus problem.

stepan [7]

Answer:

D.

Step-by-step explanation:

Remember that the limit definition of a derivative at a point is:

$\displaystyle{\frac{d}{dx}[f(a)]= \lim_{x \to a}\frac{f(x)-f(a)}{x-a}}$

Hence, if we let f(x) be ln(x+1) and a be 1, this will yield:

$\displaystyle{\frac{d}{dx}[f(1)]= \lim_{x \to 1}\frac{\ln(x+1)-\ln(2)}{x-1}}$

Hence, the limit is equivalent to the derivative of f(x) at x=1, or f’(1).

The answer will thus be D.

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2 years ago