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3 years ago

10

6(3n+1)-3(-1+2n)=n+2

1 answer:

kap26 [50]3 years ago

7 0

First, you want to distribute the 6 and the -3. This turns you equation into 18n+6+3-6n=n+2. Next, put all the n's on one side and numbers on the other. 18n-6n-n=2-6-3. Then, combine. 11n=-7 -> n=-7/11. Hope this helps!

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Which statement is true about the end behavior of the graphed function?

Art [367]

Answer:

The correct answer is "As the x-values go to positive infinity the function's value go to positive infinity".

Step-by-step explanation:

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Consider the 6 x 9 inch rectangular cake that is 2 inches deep. If you cover the cake with frosting excluding the bottom, what i

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3 years ago

Read 2 more answers

the slope of a line passing through h (-2, 5) is -3/4. Which ordered pair represents a point on this line?

rosijanka [135]

Answer:

A

Step-by-step explanation:

Calculate the slope of the given points with the point (- 2, 5 )

If the slope is - $\frac{3}{4}$ then the point is on the line

Calculate slope m using the slope formula

m = $\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

with (x₁, y₁ ) = (- 2, 5) and (x₂, y₂ ) = (6, - 1)

m = $\frac{-1-5}{6+2}$ = $\frac{-6}{8}$ = - $\frac{3}{4}$ ← point (6, - 1) is on the line

Repeat with (x₁, y₁ ) = (- 2, 5 ) and (x₂, y₂ ) = (2, 8)

m = $\frac{8-5}{2+2}$ = $\frac{3}{4}$ ← point (2, 8) is not on the line

Repeat with (x₁, y₁ ) = (- 2, 5) and (x₂, y₂ ) = (- 5, 1)

m = $\frac{1-5}{-5+2}$ = $\frac{-4}{-3}$ = $\frac{4}{3}$ ← point (- 5, 1) is not on the line

Repeat with (x₁, y₁ ) = (- 2, 5) and (x₂, y₂ ) = (1, 1)

m = $\frac{1-5}{1+2}$ = - $\frac{4}{3}$ ← point (1, 1) is not on the line

Thus the point on the line is (6, - 1 ) → A

4 0

3 years ago

The first term of a geometric sequence is 32, and the 5th term of the sequence is 818 .

sammy [17]

Answer:

$32,24,18,\frac{27}{2} ,\frac{81}{8}$

Step-by-step explanation:

Let x, y , and z be the numbers.

Then the geometric sequence is $32,x,y,z,\frac{81}{8}$

Recall that term of a geometric sequence are generally in the form:

$a,ar,ar^2,ar^3,ar^4$

This implies that:

a=32 and $ar^4=\frac{81}{8}$

Substitute a=32 and solve for r.

$32r^3=\frac{81}{8}$

$r^4=\frac{81}{256}$

Take the fourth root to get:

$r=\sqrt[4]{\frac{81}{256} }$

$r=\frac{3}{4}$

Therefore $x=32*\frac{3}{4} =24$

$y=24*\frac{3}{4} =18$

$z=18*\frac{3}{4} =\frac{27}{2}$

8 0

3 years ago