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anzhelika [568]
2 years ago
8

Suppose 15% of x equals 20% of y. What percentage of x is y ?

Mathematics
2 answers:
WINSTONCH [101]2 years ago
7 0

Answer:

x = 133\frac{1}{3} of y

Step-by-step explanation:

Given:

15% of x equals 20% of y

So,

0.15x = 0.20y

so

x = 0.20 y / 0.15

x = 1.3333y

x = 133\frac{1}{3} of y

Afina-wow [57]2 years ago
7 0

Answer:

Step-by-step explanation:

     

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Can someone help me with this
Morgarella [4.7K]

Answer:

FG=30

Step-by-step explanation:

Since we know that Point G is on the Segment FH, it doesn't really matter where G is, but we can know for certain that:

FH=FG+GH

We are given that FH is 4x, GH is x, and FG is 2x+10. Substitute:

4x=(2x+10)+x

Solve for x. On the right, combine like terms:

4x=3x+10

Subtract 3x from both sides:

x=10

So, the value of x is 10.

To find the value of FG, substitute 10 into its x:

FG=2x+10\\FG=2(10)+10

Multiply:

FG=20+10

Add:

FG=30

And we're done!

6 0
3 years ago
Find the number that makes the ratio equivalent 1:3 9:_
dimaraw [331]

Answer:

27

Step-by-step explanation:

If the ratio is 1:3 multiply that by 9:9 so you effectively get 9:9x3 = 9:27. Or, you can say to yourself if B (3) is 3 times greater than A (1) then B is 3 times greater than 9. So 9x3 = 27

3 0
3 years ago
Denise and Jacqueline are training for a swimming and running race.
bezimeni [28]

Answer:

Jacqueline runs at 6.25 minutes per mile

Step-by-step explanation:

Here we have that the total training time for Jacqueline = 54 minutes

The time for which Jacqueline swims = 30 minutes

Therefore, the time in which Jacqueline runs = 54 min. - 30 min. = 20 minutes

Therefore the equation for Jacqueline's speed = Distance/Time we have

Jacqueline's pace = 3.2 miles/20 minutes = 0.16 miles/minute

Therefore Jacqueline's pace = 0.16 miles/minute

The equation for Jacqueline's pace in miles/minute is presented as follows

Jacqueline's pace in miles/minute = Time of running/Distance

Jacqueline's pace in miles/minute  = 1/0.16 miles/minute = 6.25 minutes/mile.

5 0
3 years ago
Help ASAP please
Setler [38]
The length is 8 and the width is 4

P = L + L + W + W
P = 8 + 8 + 4 + 4
P = 16 + 4 + 4
P = 16 + 8
P = 24
5 0
3 years ago
Read 2 more answers
A 1.5-mm layer of paint is applied to one side of the following surface. Find the approximate volume of paint needed. Assume tha
Mariulka [41]

Answer:

V = 63π / 200  m^3

Step-by-step explanation:

Given:

- The function y = f(x) is revolved around the x-axis over the interval [1,6] to form a spherical surface:

                                 y = √(42*x - x^2)

- The surface is coated with paint with uniform layer thickness t = 1.5 mm

Find:

The volume of paint needed

Solution:

- Let f be a non-negative function with a continuous first derivative on the interval [1,6]. The Area of surface generated when y = f(x) is revolved around x-axis over the interval [1,6] is:

                           S = 2*\pi \int\limits^a_b { [f(x)*\sqrt{1 + f'(x)^2} }] \, dx

- The derivative of the function f'(x) is as follows:

                            f'(x) = \frac{21-x}{\sqrt{42x-x^2} }

- The square of derivative of f(x) is:

                            f'(x)^2 = \frac{(21-x)^2}{42x-x^2 }

- Now use the surface area formula:

                           S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2} *\sqrt{1 + \frac{(21-x)^2}{42x-x^2 } }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2+(21-x)^2} }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2+441-42x+x^2} }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{441} }] \, dx\\S = 2*\pi \int\limits^6_1 { 21} \, dx\\\\S = 42*\pi \int\limits^6_1 { dx} \,\\\\S = 42*\pi [ 6 - 1 ]\\\\S = 42*5*\pi \\\\S = 210\pi

- The Volume of the pain coating is:

                           V = S*t

                           V = 210*π*3/2000

                          V = 63π / 200 m^3

8 0
3 years ago
Read 2 more answers
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