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2 years ago

Which rule best represents the dilation that was applied to Figure S to create Figure T?

Mathematics

1 answer:

Anettt [7]2 years ago

6 0

Answer:

Step-by-step explanation:

Hgg

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Solve for x. (RIGHT ANSWER WILL GET BRAINLIEST)

puteri [66]

Answer:

Step-by-step explanation:

The angle with x in it is a vertical angle to the other one labeled & the right angle put together.

Therefore:

48 + 90 = 2x + 4

Solve for x:

138 = 2x + 4

138 - 4 = 2x + (4-4)

134 = 2x

134/2 = 2x/2

X = 67

Hope this helps! Have a great day!

7 0

3 years ago

, 80 POINTS Find k so that P(k,0) is equidistant from points A(1,2) and B(2,5).

blagie [28]

Answer:

P(1.5, 3.5)

Step-by-step explanation:

We need to find the average of the x and y coordinates. We do this by adding the two values, then dividing by two:

P(x1 + x2/ 2, y1 + y2/2) = P(1 + 2/ 2, 2 + 5/ 2) = P(3/2, 7/2) = P(1.5, 3.5)

These are our coordinates!

8 0

3 years ago

A water tank in the shape of an inverted circular cone has a base radius of 3 m and height of 7 m . If water is being pumped int

Wittaler [7]

The water level increases by 0.608 meters per minute when the water is 3.5 m deep

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

The given parameters are:

Radius, r = 3
Height, h = 7
Rate in, V' = 4.3m^3/min

The relationship between the radius and height is:

r/h = 3/7

Make r the subject

r = 3h/7

The volume of a cone is;

$V = \frac 13\pi r^2h$

This gives

$V = \frac 13\pi (\frac{3h}{7})^2h$

Expand

$V = \frac{3h^3}{49}\pi$

Differentiate

$V' = \frac{9h^2}{49}\pi h'$

Make h' the subject

$h' = \frac{49}{9\pi h^2}V'$

When the water level is 3.5.

We have:

$h' = \frac{49}{9\pi * 3.5^2}V'$

Also, we have:

V' = 4.3

So, the equation becomes

$h' = \frac{49}{9\pi * 3.5^2} * 4.3$

Evaluate the products

$h' = \frac{210.7}{346.36}$

Evaluate the quotient

h' = 0.608

Hence, the water level increases by 0.608 meters per minute