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Otrada [13]
3 years ago
7

Given the following function, find f(-4), f(0), and f(3).

Mathematics
2 answers:
shutvik [7]3 years ago
8 0
You plug in the given value for x.

F(-4)=(-4)2-4
F(-4)=-8-4
F(-4)=-12

F(0)=(0)2-4
F(0)=0-4
F(0)=-4

F(3)=(3)2-4
F(3)=6-4
F(3)=2
Rudiy273 years ago
3 0
F(-4)= (-4)2-4
the answer is -12

f(0)=(0)2-4
the answer is -4

f(3)=(3)2-4
the answer is 2
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Solve the proportion using equivalent ratios. Explain the steps you used to solve the proportion, and include the answer in your
rodikova [14]

Answer: Your answer is X=6

Step-by-step explanation:

10/3 = 20/x

Determine the defined range

10/3 20/x, x=0

Cross-multiply

10x = 60

Divide both sides by 10

x=6,x=0

Check if the solution is in the defined range

SOLUTION x=6

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4 0
3 years ago
Suppose j varies jointly with g and​ v, and j=2 when g=4 and v=3. <br> Find j when g=8 and v=9.
vladimir2022 [97]

Answer:

j=12

Step-by-step explanation:

we know that

In a join variation, If j varies jointly with respect to g and v, the equation will be of the form j = kgv

where k is a constant

step 1

Find the value of k

we have

j=2,g=4,v=3

substitute and solve for k

2 = k(4)(3)

2 = 12k

k=\frac{1}{6}

The equation is equal to

j=\frac{1}{6}gv

step 2

Find the value of j when g=8,v=9

substitute the values in the equation and solve for j

j=\frac{1}{6}(8)(9)

j=12

3 0
3 years ago
Select TWO expressions that
8090 [49]

Answer:

A and D

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4×3=12

4×d=4d

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6 0
2 years ago
Read 2 more answers
How to solve x=6y-11 and 3x-2y=-1 by substitution
vagabundo [1.1K]
Substitute x into the second equation:
3(6y-11)-2y=-1
18y-33-2y=-1
16y=32
y=2
Then put the y value into the first equation to get x:
x=6(2)-11
x=1
There you go!

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7 0
3 years ago
Lesson 13.3 Checkpoint
Arte-miy333 [17]

Answer:

a. LK = 4√3, JL= 4√6

b.  PQ = 10√3, RP = 5√3

Step-by-step explanation:

a. LK =4√3,

LK=KJ= 4√3

sin(45°) = opposite/hypotenuse

hypotenuse (JL)= opposite/sin 45° = 4√3/√2/2 = 8√3/√2 = 8√3√2/2 = 4√6

b.

cos 30° = adjacent/hypotenuse = QR/PQ = 15/PQ

cos 30° = 15/PQ

PQ = 15/cos 30° = 15/√3/2)= 30/√3 = 30√3/3 = 10√3

tan 30° = opposite/adjacent = RP/QR

√3/3 = RP/QR

√3/3 = RP/15

RP = √3*15/3 = 5√3

3 0
3 years ago
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