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babymother [125]
3 years ago
7

Solve the system of equations. 7x=y=0 14x+2y=0

Mathematics
1 answer:
Gnoma [55]3 years ago
4 0
For the first one x and y equal 0 and for the second one the answer is y=-7x

Your welcome
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Help please
Vadim26 [7]
<span>✡ </span>Answer: 75; 100-25-5 ✡


- - To solve this you are going to subtract.
Because you are taking away apples, in this case eating them

- - So: 100-20-5

Answer: 75

✡Hope this helps!<span>✡</span>





7 0
3 years ago
Read 2 more answers
What is 69 over 24 as a terminating decimal
Dahasolnce [82]
2.875 is 69/24 as a terminating decimal
4 0
3 years ago
In KLM, LM=7 and angle K=45. Find KL. Leave your answer in simplest radical form.
Sphinxa [80]
Answer in simplest radical form would be 7.(:
5 0
3 years ago
The graph shows two lines, A and B.
Harlamova29_29 [7]

Answer with explanation:

<u>Part A: </u>

One solution.

The number of points of intersection represents the number of solutions. Since the two lines only intersect at one point, there is only one solution.

<u>Part B:</u>

(3, 4)

The point(s) of intersection marks the solution(s) to the lines. Since lines A and B intersect at the point (3, 4), the solution to the equation of their lines is (3, 4), or x=3, y=4, as coordinates are written as (x, y).

3 0
3 years ago
Suppose that $3000 is placed in an account that pays 16% interest compounded each year. Assume that no withdrawals are made from
Papessa [141]

Answer:

a) $3480

b) $4036.8

Step-by-step explanation:

The compound interest formula is given by:

A(t) = P(1 + \frac{r}{n})^{nt}

Where A(t) is the amount of money after t years, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per year and t is the time in years for which the money is invested or borrowed.

Suppose that $3000 is placed in an account that pays 16% interest compounded each year.

This means, respectively, that P = 3000, r = 0.16, n = 1

So

A(t) = P(1 + \frac{r}{n})^{nt}

A(t) = 3000(1 + \frac{0.16}{1})^{t}

A(t) = 3000(1.16)^{t}

(a) Find the amount in the account at the end of 1 year.

This is A(1).

A(t) = 3000(1.16)^{t}

A(1) = 3000(1.16)^{1} = 3480

(b) Find the amount in the account at the end of 2 years.

This is A(2).

A(2) = 3000(1.16)^{2} = 4036.8

4 0
3 years ago
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