2 years ago

State the domain and range of the given relation

Mathematics

1 answer:

GuDViN [60]2 years ago

5 0

Answer:

work is shown and pictured

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You are in charge of finding a place to have your little sister's birthday party. You

kow [346]

Answer:

Y=x(5)+120

Step-by-step explanation:

Y= the money you have to pay for the party

X(5)= you multiply x by 5

+120= you add the fee

7 0

2 years ago

Please pls pls asap help

Tpy6a [65]

Answer:

1/2

Step-by-step explanation:

7 0

1 year ago

BRAINLIEST WHAT IS 2*2? WHAT IS 2+2? help

gladu [14]

2 * 2 = 4

2 + 2 = 4

Hope it helps! :)

4 0

2 years ago

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In △ABC,∠B is a right angle. The coordinates for each point are A(10, 7), B(5, 9), and C(3, 4). Rounded to the nearest tenth,

shepuryov [24]

Answer:

$A=14.5\ units^2$

Step-by-step explanation:

we know that

The area of the right triangle ABC is equal to

$A=\frac{1}{2}(AB)(BC)$

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have

A(10, 7), B(5, 9), and C(3, 4)

step 1

Find the distance AB

A(10, 7), B(5, 9)

substitute in the formula

$d=\sqrt{(9-7)^{2}+(5-10)^{2}}$

$d=\sqrt{(2)^{2}+(-5)^{2}}$

$d_A_B=\sqrt{29}\ unjts$

step 2

Find the distance BC

B(5, 9),C(3, 4)

substitute in the formula

$d=\sqrt{(4-9)^{2}+(3-5)^{2}}$

$d=\sqrt{(-5)^{2}+(-2)^{2}}$

$d_B_C=\sqrt{29}\ unjts$

step 3

Find the area

substitute the values

$A=\frac{1}{2}(\sqrt{29})(\sqrt{29})$

$A=14.5\ units^2$

5 0

2 years ago

Part A: Write an equation of a line that is perpendicular to the line y = 2/3x + 5 and passes through the point (4, 0).

Ivahew [28]

Step-by-step explanation:

First you need to make a line perpendicular to y=2/3x+5. To do that you have to know that, if we use y=ax+b, a1*a2=-1. That is if a1 and a2 are perpendicular. So we get 2/3*a2=-1. -1/ 2/3=a2. a2=-1.5. We get y=-1.5x+b. Fill in (4,0) on y=-1.5x+b abd we get 0=-1.5(4)+b. b=6

5 0

2 years ago

Read 2 more answers