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3 years ago

Use the distributive property to write 3y - y as a product. Then simplify.

Mathematics

1 answer:

kvasek [131]3 years ago

4 0

Y(3-1)= 2y is the answer to the question

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Hi! I cant figure this out... could someone please help me?

Fed [463]

Answer:

Step-by-step explanation:

open circle on -4 and open circle on 4 and a line connecting them

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3 years ago

What number is 300% of six

nadya68 [22]

Answer:

Step-by-step explanation:

100% of 6 is 6 so 300% (which does not actually exist) would be 3 times as much. 6x3=18

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3 years ago

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A pool had 10 1/2 gallons of water in it Jake and his friend splashed g gallons of water out of the pool. There are 7 7/8 gallon

kolezko [41]

Answer: 6/625 gallons of water wa splashed out of the pool

Step-by-step explanation:

Total number of gallons of water in the pool was 10 1/2. Converting to improper fraction, it becomes

21/2 gallons.

Jake and his friend splashed g gallons of water out of the pool. There are 7 7/8 gallons still left in the pool. Converting 7 7/8 to improper fraction, it becomes

63/8 gallons

This means that

g + 63/8 = 21/2

g = 21/2 - 63/8 = (84 - 63)/8

g = 21/8 = 2.625 gallons

3 0

3 years ago

Type the correct answer in the box. Use numerals instead of words. If necessary, use / for the fraction bar. Yesterday, the snow

Ronch [10]

2 f = 100% 1.6f = x X=(1.6*100)/2=80% The percent change from yesterday to today is 20℅

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3 years ago

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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