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hram777 [196]
3 years ago
15

Solve 3/5x = 15

Mathematics
1 answer:
Pavel [41]3 years ago
4 0

Answer:

25

Step-by-step explanation:

Do 5 X times 15 = 75 x

Do 75x / 3 = 25

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Dakota has four tents kilogram of clay he divided the clay to make eight equal size pots how many grams of clay are in each pot
sukhopar [10]
Answer:
There are 2.22 grams of clay are in each pot.
Step-by-step explanation:
Given:
Dakota has 4/10 kg of clay. He divides the clay to make 888 equal-sized pots.
Now, to find the number of grams of clay are in each pot.
Dakota has clay =
So, by using conversion factor we convert it into grams:

Quantity of clay he has = 400 grams.
Number of equal-sized pots to make = 888.
Now, to get the quantity of grams of clay are in each pot we divide number of equal-sized pots to make by quantity of clay he has:
Therefore, there are 2.22 grams of clay are in each pot.
5 0
3 years ago
What is the period of the function y = 2 sinx
DanielleElmas [232]

Answer:

The period is 360° ( in radians, the period is 2π)

Step-by-step explanation:

For a sine/cosine (sinusoidal) function in the form

y = ASin(Bx-C) + D, we can say

A is the amplitude

360/B is the period

C is the phase shift

D is the vertical translation

<em>The function given is y = 2Sinx</em>

To find the period, we need 360/B. Since "B" is just "1", the period is

360/1 = 360

5 0
3 years ago
Read 2 more answers
Tom exercised 4/5 hour on monday and 5/6 hour on tuesday. Complete the calculations below to write equivalent fractions with a c
stepladder [879]
4/5= 24/30 5/6= 25/30 Equals in all= 1 19/30
7 0
4 years ago
Find an explicit solution to the Bernoulli equation. y'-1/3 y = 1/3 xe^xln(x)y^-2
NNADVOKAT [17]

y'-\dfrac13y=\dfrac13xe^x\ln x\,y^{-2}

Divide both sides by \dfrac13y^{-2}(x):

3y^2y'-y^3=xe^x\ln x

Substitute v(x)=y(x)^3, so that v'(x)=3y(x)^2y'(x).

v'-v=xe^x\ln x

Multiply both sides by e^{-x}:

e^{-x}v'-e^{-x}v=x\ln x

The left side can be condensed into the derivative of a product.

(e^{-x}v)'=x\ln x

Integrate both sides to get

e^{-x}v=\dfrac12x^2\ln x-\dfrac14x^2+C

Solve for v(x):

v=\dfrac12x^2e^x\ln x-\dfrac14x^2e^x+Ce^x

Solve for y(x):

y^3=\dfrac12x^2e^x\ln x-\dfrac14x^2e^x+Ce^x

\implies\boxed{y(x)=\sqrt[3]{\dfrac14x^2e^x(2\ln x-1)+Ce^x}}

4 0
3 years ago
Who has the greater rate of change?
vova2212 [387]
Your answer is the yellow one your welcome
6 0
3 years ago
Read 2 more answers
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