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Burka [1]
2 years ago
6

Suppose y varies directly as x², and y = 160 when x = 4. Find y when x = 6.

Mathematics
1 answer:
labwork [276]2 years ago
7 0

<u>Finding the constant of proportionality</u> :

  • y = kx²
  • 160 = k(4)²
  • k = 10

<u>Finding y when x = 6</u> :

  • y = 10(6)²
  • y = 10(36)
  • y = 360

The value of y is <u>360</u> when x = 6.

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Which expression correctly describes the word praise "4 less than x" ?
HACTEHA [7]

x - 4

4 less than x means that we subtract 4 from x.

6 0
3 years ago
The vertex of the graph of f(x)=|X-3|+6 is located at
frosja888 [35]

The vertex of the graph of f(x)= |x-3|+6 is located at (3, 6)

<h3>How to determine the vertex?</h3>

The equation of the function is given as:

f(x) = |x - 3| + 6

The above function is an absolute value function.

An absolute value function is represented as:

f(x) = a|x - h| + k

Where:

Vertex = (h, k)

By comparison, we have:

Vertex = (3, 6)

Hence, the vertex of the graph of f(x)= |x-3|+6 is located at (3, 6)

Read more about vertex at:

brainly.com/question/16799565

#SPJ1

6 0
2 years ago
Find the fifth term in a sequence: 15,20,25...<br> A) 20<br> B) 45<br> C) 35 <br> D) 40
guajiro [1.7K]

Answer:

C. 35

Step-by-step explanation:

Use the equation: y= 10+5x

8 0
4 years ago
Please answer to receive ten points. I'll vote who answers first the brainliest if its correct. 
beks73 [17]

The distance between points P_1(x_1,y_1) and P_2(x_2,y_2) can be calculated using formula

P_1P_2=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.

1.

AB=\sqrt{(-7-(-11))^2+(8-4)^2}=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}\approx 5.657.

2.

AC=\sqrt{(-4-(-11))^2+(4-4)^2}=\sqrt{49+0}=\sqrt{49}=7.

3.

CB=\sqrt{(-7-(-4))^2+(8-4)^2}=\sqrt{9+16}=\sqrt{25}=5.

Then

P_{ABC}=5.657+7+5=17.657\approx 17.7

Answer: correct choice is C.

3 0
3 years ago
PLEASE HELP WILL GIVE BRAINLIEST!!!
suter [353]

Answer:

Vertical Angles Theorem

Step-by-step explanation:

The two triangles will be equal by SAS

3 0
3 years ago
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