Slope=5,y-intercept=-1

<img src="https://tex.z-dn.net/?f=%20%5Chuge%5Cmathfrak%5Cpink%7BS%7D%20%5Chuge%5Cmathfrak%5Cpurple%7Bo%7D%20%5Chuge%5Cmathfrak%

Ainat [17]

Answer:

Step-by-step explanation:

2+2=4 is the naswer 4

7 0

3 years ago

Read 2 more answers

For f(x) = 2−3x and g(x) = 3x – 2, find the indicated values. of (a

Neporo4naja [7]

I hope this helps you

3 0

3 years ago

-x+4y=4 -x+3y=1 is (8,3) a solution of the system?

svetoff [14.1K]

check by plugging into both equations

-8+4(3)=4

-8+12=4

4=4

-8+3(3)=1

-8+9=1

1=1

(8,3) is a solution of the system

7 0

4 years ago

Which of the following numbers is a rational but not a integer

Temka [501]

Rational numbers are numbers that can be turned into a fraction. They can be negative or positive.
Integers are whole numbers. They can be negative or positive.

Numbers such as 0.10 are rational because they can be turned into a fraction
0.10 = 1/10....and since it is not a whole number, it is not an integer.

8 0

3 years ago

Extended Problems, Extraneous Solutions

diamong [38]

Answer:

1a. 2 is the solution

1b. -3 is because square root of 1 doesn't equal -1.

1c.-3

Step-by-step explanation:

1.

$(x + 2 = \sqrt{3x + 10}$

Square both sides

$(x + 2) {}^{2} = 3x + 10$

Remeber perfect square Rule

$(x + y) {}^{2} = {x}^{2} + 2xy + {y}^{2}$

So now we got

$x {}^{2} + 4x + 4 = 3x + 10$

${x}^{2} + x + 4 = 10$

${x}^{2} + x - 6 = 0$

Factor into binomial.

$(x + 3)(x - 2) = 0$

So our solution is 3 or 2. However, let's check our work.

$- 3 + 2 = \sqrt{3( - 3) + 10}$

$- 1 = \sqrt{1}$

However square root of 1 is 1. It could be 1 but in Math, We only take the principal square root which is positive always. So -3 isn't a solution.

So let try 2.

 $2 + 2 = \sqrt{3(2) + 10} = 4 = \sqrt{16} = 4 = 4$ 

So this is indeed the solution.

Make up equation for 2 and 3. Ask me if you need help.

5 0

2 years ago