Ask question

Ask question

All categories

3 years ago

15

At noon, a tank contained 5in. of water. After several hours, it contained 3in. of water. What is the percent decrease of wate

r in the tank?

2 answers:

Sonbull [250]3 years ago

5 0

I dont really know what this could be

Alex787 [66]3 years ago

3 0

The amount of decrease was 5 -3 = 2 inches,

Percent decrease = 2 /5 = 0.40 x 100 = 40%

You might be interested in

Order numbers 11/8,1.405, and1.45 smallest to largest

Diano4ka-milaya [45]

Step-by-step explanation:11/8 is 12.5 goes into one hundred 8 times so 12 x 11 is 137.5so then turn decimals to a percent 1.405 is 140.5 1.45 is 145 so the it is 137.5 140.5,145

3 0

3 years ago

Which of the following is not a literal equation?

Alchen [17]

C. nn = 10
it isn't a literal equation, because an unknown variable (n) cannot be next to itself

5 0

3 years ago

Read 2 more answers

Please help this is worth 15 points! Solve in y=mx+b from!

nignag [31]

Answer:

The coordinates of every point on the line will solve the equation if you substitute them in the equation for x and y. The equation of any straight line, called a linear equation, can be written as: y = mx + b, where m is the slope of the line and b is the y-intercept

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Area of a rectangle is 81 m^2. Express the perimeter of the rectangle as a function of the length (L) of one of its sides

LuckyWell [14K]

Answer:

$P = 2\bigg(\dfrac{l^2 + 81}{l}\bigg)~m$

Step-by-step explanation:

We are given the following in the question:

Area of rectangle = 81 square meter.

Let l be the length of the rectangle and b be the breadth of rectangle.

Then, we can write,

$\text{Area of rectangle} = \text{Length}\times \text{breadth}\\A = l\times b\\81 = l\times b\\\\b = \dfrac{81}{l}~m$

Perimeter of rectangle =

$2(\text{Lenght + Breadth})$

Putting values, we get,

$P = 2(l + b)\\P = 2(l + \dfrac{81}{l})\\\\P = 2\bigg(\dfrac{l^2 + 81}{l}\bigg)~m$

6 0

4 years ago

3x + y =11<br> 5x - y = 21<br> Solve each system of equations by elimination. Answer QUICK!!!!!!!

julia-pushkina [17]

Answer/Step-by-step Explanation:

1. 3x + y = 11 => equation 1

5x - y = 21 => equation 2

Subtract equation 2 from equation 1 to eliminate y variable

3x + y = 11

5x - y = 21

-2x = - 10

Divide both sides by -2

-2x/-2 = -10/-2

x = 5

Substitute x = 5 into equation 1 or 2:

3x + y = 11 => equation 1

3(5) + y = 11

15 + y = 11

y = 11 - 15 (subtraction 15 from each side)

y = -4

Solution: x = 5, y = -4

2. 5x + 12y = -34 => equation 1

2x + 4y = -12 => equation 2

Multiply equation 1 by 2, multiply equation 2 by 5.

2(5x + 12y = -34) => 10x + 24y = -68 => eqn 3

5(2x + 4y = -12) => 10x + 20y = -60 => eqn 4

Subtract eqn 4 from eqn 3

4y = -8

Divide both sides by 4

y = -2

Substitute y = -2 into equation 2

2x + 4y = -12 => equation 2

2x + 4(-2) = -12

2x - 8 = -12

2x = -12 + 8 (adding 8 to both sides)

2x = -4

Divide both sides by 2

x = -2

Solution: x = -2, y = -2

7 0

3 years ago