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3 years ago

1. How many solutions can a linear system of equations have?

Mathematics

1 answer:

ASHA 777 [7]3 years ago

6 0

Answer:

<h2>D. All of the Above</h2>

Step-by-step explanation:

Since a linear system might have 0, 1 , or many solutions, the correct answer is d.

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16 ÷ 22 + (6 x 2) - 8

stepladder [879]

I did the PEMDAS method (P= parentheses, E= exponents, M= multiplication, D = Division, A= addition, S = subtraction.)

First I did what’s inside the parentheses, (6x2) and got 12.

Then I divided 16 by 22 and received 0.72.

I added 0.72 with 12 and the result was 12.72.

Last step was to subtract 12.72 - 8 and that equals 4.72

3 0

3 years ago

Read 2 more answers

DF=78, DE=5x-9, and EF-2x+10, find EF

lbvjy [14]

Answer:

As e is point between df. So we can write df as

df = de + ef

As given df = 78 , de = 5x - 9 , ef = 2x + 10.

So we can write

78 = 5x - 9 + 2x + 10

78 = 7x + 1

On subtracting 1 on booth side

78 - 1 = 7x +1 - 1

77 = 7x

On dividing both side by 7.

11 = x

So x = 11.

So ef = 2x + 10 = 2*11 + 10 = 22 + 10 = 32

Step-by-step explanation:

8 0

3 years ago

Use the first five terms of the trigonometric series to approximate the value of <img src="https://tex.z-dn.net/?f=sin%5Cfrac%7B

Schach [20]

Answer:b

Step-by-step explanation:

3 0

3 years ago

A certain college graduate borrows 5510 dollars to buy a car. The lender charges interest at an annual rate of 17%. Assuming tha

Diano4ka-milaya [45]

Answer:

His annual payment rate required to pay off the loan in 7 years is $k = $2362.41$ a year.

He has to pay $11026.85 in interest

Step-by-step explanation:

The first step to solve this problem is find how much the student will have to pay for the car. So, we have to find how much the present value of the car(in $) will be worth in 7 years. This is a compound interest problem:

The compound interest formula is given by:

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

Where A is the amount of money, 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 unit t and t is the time the money is invested or borrowed for.

In this exercise, we have:

A = The value we want to find

P = The initial value = 5510

r = 0.17

n = 1

t = 7

$A = 5510(1 + \frac{0.17}{1})^{7}$