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3 years ago

6

You have $12,000 to invest and want to keep your money invested for 8 years. You are considering the following investment option

s. Choose the investment option that will earn you the most money.

1 answer:

kozerog [31]3 years ago

7 0

Ngzktzkg mf kg247;77. Kyzjfsktxhgtvt

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Question 1(Multiple Choice Worth 4 points)

Airida [17]

I think you mean x^2 - 7x + 8
This is a basic factoring problem:
What two numbers multiply to make 8 and add to make -7?
There are no two numbers that work!
Therefore, it is prime

4 0

3 years ago

What are the coordinates of the point on the directed line segment from (-10, -3)(−10,−3) to (2, -3)(2,−3) that partitions the s

azamat

Answer:

the coordinates of the point would be (-2.5,3)

Step-by-step explanation:

We want to split the segment from (-10,-3) to (2,-3) into segments with a ratio of 5:3. Since the y-coordinate is -3 for both coordinates, the y-coordinate of the partitioning point will be -3. The ratio of 5:3 corresponds to 5/8 of the distance between the x-coordinates of the two points. So we would be moving 5/8 of the distance from -10 to 2 for the x-coordinate, so the x-coordinate would be -10 + 5/8 (12) = -2.5. So the coordinates of the point would be (-2.5,3)

6 0

3 years ago

A coffee shop needs to purchase custom cups. Company C charges a flat fee of $155 plus $15 perrl cup. Company D does not have a

bezimeni [28]

Answer:

11 cups

Step-by-step explanation:

Not really a method just started at 10 and then realised deal d was better so go up by one so 11 and c worked out better as I timed 30 by 11 and 15 by 1 + the flat fee.

8 0

3 years ago

The cost table shows the average cost, a(x), of producing x number of earbuds.

Tasya [4]

Answer:

a

Step-by-step explanation:

7 0

3 years ago

7 to the negative 3rd porwer times 7 to the negative 2nd porwer times 2 the 0 power divided by 7 to negative 1st power times 7 t

Jobisdone [24]

oof

remember some rules

(a/b)(c/d)=(ac)/(bd)

$\frac{x^a}{x^b}=x^{a-b}$

$x^0=1$

$(x^a)(x^b)=x^{a+b}$

$x^{-a}=\frac{1}{x^a}$

$\frac{(7^{-3})(7^{-2})(2^0)}{(7^{-1})(7^{-4})(2^3)}=$

$\frac{(7^{-3-2})(2^0)}{(7^{-1-4})(2^3)}=$

$\frac{(7^{-5})(2^0)}{(7^{-5})(2^3)}=$

$(\frac{7^{-5}}{7^{-5}})(\frac{2^0}{2^3})=$

$(7^{-5-(-5)})(2^{0-3})=$

$(7^0)(2^{-3})=$

$\frac{1}{2^3}=$

$\frac{1}{8}$

7 0

3 years ago