Please help out its due today :]

Fernando has gathered details about his finances to create a financial assets and liabilities record. Which represents one of hi

Bingel [31]

<span>
C.
His college savings account has a balance of $13,700.</span>

4 0

3 years ago

Jenna found three solution to a linear equation she graphed them, but they don't lie in a straight line. What errors could jenna

steposvetlana [31]

Answer:

A system of linear equation could only have 1 solution. This is because the straight lines will only have to meet, cross, or intersect each other once.

There are many different methods in arriving to the final answer. However, errors cannot be perfectly avoided. One of these errors to mistakenly identify equations as linear. It is important that we know that the equations we are dealing with are of exact or correct characteristics.

Also, if she had used substitution method, she might have mistakenly taken the value of one variable for the other.

5 0

3 years ago

Find the sum of 7a^3 + 14a +12 and -6a^3 +12a^2 -7

SOVA2 [1]

To find the sum:

${7a}^{3} + 14a + 12 + ( {- 6a }^{3} ) + {12a }^{2} - 7 \\ = {7a }^{3} - {6a}^{3} + {12a }^{2} + 14a + 12 - 7 \\ = {a}^{3} + {12a }^{2} + 14a + 5$
Therefore the answer is a^3+12a^2+14a+5.
Hope it helps!

8 0

4 years ago

Read 2 more answers

7/15 divided by 3/10

Yuliya22 [10]

Answer:

i think thats my answer

Step-by-step explanation:

1 5/8

5 0

3 years ago

99 POINTS!!! JUST ANSWER THIS QUESTION, PLEASE!!!

-Dominant- [34]

Hi!

I would approach this question by first looking at where the parallelogram is located, by graphing the image. I did so and drew a rough image on Paint (See attached image)

To get the part where the diagonals intersect, I would find the midpoint of the line between points (1, 3) and (5,-9) (or the other pair). The reason is a parallelogram's diagonals always bisect each other, meaning the point they intersect is always the middle of the two diagonals.

Therefore, you can find the midpoint of a diagonal, between (1, 3) and (5, -9). The midpoint theorem is ( $\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}$ .