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Veseljchak [2.6K]

3 years ago

7

I need help with this

1 answer:

bonufazy [111]3 years ago

5 0

It should be 1, 5, and 6

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Solve for x.<br> −7≥13−5x

Sergio039 [100]

Answer:

$=>x\geq 4$

Step-by-step explanation:

Given inequality is $−7\geq13−5x$

Simplifying:

$5x\geq13+7$

$=>5x\geq20$

$=>x\geq \frac{20}{5}$

$=>x\geq 4$

Hence the solution for the given inequality is represented by $=>x\geq 4$

This means that the original inequality will be true only for those values of x which are greater than or equal to 4.

4 0

3 years ago

Someone please help!

svetoff [14.1K]

Answer:

a =6

b=4

c=2

step-by-step explanation:

$\sqrt{x^{12} y^{9}z^{5} }$ = $\sqrt{x^{12} y^{8}y z^{4} z}$ = $\sqrt{(x^{6}) ^{2} (y^{4}) ^{2} (z^{2})^{2} yz }$ = $x^{6}y^{4} z^{2} \sqrt{yz}$

7 0

2 years ago

PLEASE HELP ME!!<br> Find the length of CU.

Viktor [21]

Answer:

Did you ever find the answer? I have the same question!

Step-by-step explanation:

3 0

3 years ago

Use the Distributive Property of Multiplication over Addition to find each missing number.

Svet_ta [14]

Answer:

(3*9)+(3*6)

Step-by-step explanation:

7 0

2 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.

5 0

3 years ago