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2 years ago

State the domain and range of the relation { (-3,2),(4,1)(0.3),(5,2),(2,7)} .Then determine whether the relation is a function.

1 answer:

6 0

Answer:

Domain: {-3, 0, 2, 4, 5)
Range: {1, 2, 3, 7}

The relation is a function because for every x-value there is one y-value. In other words, for every input there is one output.

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Solve to find x 3/4 (5x-3) + 8 = 17

Talja [164]

Answer:

Step-by-step explanation:

3/4 (5x -3)+8=17

3/4 (5x -3) = 17 - 8 = 9

(5x -3) = 9 * (4/3)= 36/3 = 12

5x = 12 + 3 = 15

x = 3

6 0

3 years ago

Let point L be between M and N on MN. Given that MN=31, ML= h-15 and LN = 2h-8. Find LN

MrRa [10]

Since the distance of MN is 31 and we have a distance of ML of h -15 and LN of 2h – 8, therefore

ML + LN = h -15 + 2h – 8 = 31

h = 18

substituting the value of h to the expression for LN obtaining

LN = 2h – 8

LN = 2(18) – 8

LN = 28

Therefore the length of the segment LN is 28 units

7 0

2 years ago

FIRST TO ANSWER GET BRAINLEST <br> Solve for x. 2/3(x−7)=−2 Enter your answer in the box. x =

Alex_Xolod [135]

2/3(x−7)=−2

2(x - 7) = (-2) * (3)

2x - 14 = - 6

2x = 8

x = 4

3 0

2 years ago

3+-√(-3)^2 - 4(5)(-1)

Alona [7]

Answer:

Step-by-step explanation:

Easy way to do this is step by step. Your quadratic, from your entry, must be

$5x^2-3x-1$ .

Step by step looks like this, one thing at a time:

$x=\frac{3+\sqrt{(-3)^2-4(5)(-1)} }{2(5)}$ becomes

$x=\frac{3+\sqrt{9-(-20)} }{10}$ becomes

$x=\frac{3+\sqrt{9+20} }{10}$

and this of course is

$x=\frac{3+\sqrt{29} }{10}$

Do the same with the subtraction sign to get the other solution.

If you're unsure of how to enter it into your calculator, do it step by step so you don't mess up the sign. If you enter it incorrectly, you could end up with an imaginary number when it should be real, or a real one that should be imaginary.

Just my advice as a high school math teacher.

3 0

3 years ago

Which of the following is the endpoint of the altitude of a regular pyramid

larisa [96]

It’s either vertex or base

4 0

3 years ago