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3 years ago

If y=10 when x=15 find x when y=35

Mathematics

1 answer:

konstantin123 [22]3 years ago

7 0

Answer:

52.5

Step-by-step explanation:

You divide 35 by 10 and then multiply 15 by that number

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the radius of a circular Frisbee is 9.8 cm. Using rounding to the nearest whole number, what is an estimate for the circumferenc

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Answer:

60 cm

Step-by-step explanation:

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2 years ago

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How do you solve this someone help please

tekilochka [14]

The film lasted until 8:27.

This is true because 102 minutes is 1 hour and 42 minutes. If you add 1 hour and 42 minutes onto 6 hours and 45 minutes, the beginning time then. You would get 8 hours and 27 minutes.

This is my equation. ---->
6:45
+1:42
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7:87 <---- But that isn't correct because 87 minutes is over an hour.

So if you do the correct numbers you would get 8:27 as your answer!

HOPE I HELPED!!! :D

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3 years ago

Integrate cosx/sqrt(1+cosx)dx

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<span>Take the integral: integral (cos(x))/sqrt(cos(x)+1) dx
For the integrand (cos(x))/sqrt(1+cos(x)), substitute u = 1+cos(x) and du = -sin(x) dx: = integral (u-1)/(sqrt(2-u) u) du
For the integrand (-1+u)/(sqrt(2-u) u), substitute s = sqrt(2-u) and ds = -1/(2 sqrt(2-u)) du: = integral -(2 (1-s^2))/(2-s^2) ds
Factor out constants: = -2 integral (1-s^2)/(2-s^2) ds
For the integrand (1-s^2)/(2-s^2), cancel common terms in the numerator and denominator: = -2 integral (s^2-1)/(s^2-2) ds
For the integrand (-1+s^2)/(-2+s^2), do long division: = -2 integral (1/(s^2-2)+1) ds
Integrate the sum term by term: = -2 integral 1/(s^2-2) ds-2 integral 1 ds
Factor -2 from the denominator: = -2 integral -1/(2 (1-s^2/2)) ds-2 integral 1 ds
Factor out constants: = integral 1/(1-s^2/2) ds-2 integral 1 ds
For the integrand 1/(1-s^2/2), substitute p = s/sqrt(2) and dp = 1/sqrt(2) ds: = sqrt(2) integral 1/(1-p^2) dp-2 integral 1 ds
The integral of 1/(1-p^2) is tanh^(-1)(p): = sqrt(2) tanh^(-1)(p)-2 integral 1 ds
The integral of 1 is s: = sqrt(2) tanh^(-1)(p)-2 s+constant
Substitute back for p = s/sqrt(2): = sqrt(2) tanh^(-1)(s/sqrt(2))-2 s+constant
Substitute back for s = sqrt(2-u): = sqrt(2) tanh^(-1)(sqrt(1-u/2))-2 sqrt(2-u)+constant
Substitute back for u = 1+cos(x): = sqrt(2) tanh^(-1)(sqrt(sin^2(x/2)))-2 sqrt(1-cos(x))+constant
Factor the answer a different way: = sqrt(1-cos(x)) (csc(x/2) tanh^(-1)(sin(x/2))-2)+constant
Which is equivalent for restricted x values to:
Answer: | | = (2 cos(x/2) (2 sin(x/2)+log(cos(x/4)-sin(x/4))-log(sin(x/4)+cos(x/4))))/sqrt(cos(x)+1)+constant</span>

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3 years ago

To compute a student's Grade Point Average (GPA) for a term, the student's grades for each course are weighted by the number of

postnew [5]

Answer:

The student's GPA for that term is 3.07

Step-by-step explanation:

So this is a problem where we need to compute a weighted arithmetic mean.

To compute the weighted arithmetic mean of a set, we need to multiply each value of the set by their respective weight, add them, and then divide by the sum of the weights.

I will write the set in a {V,W} way, where V is the value(the grade) and W is the weight of V.

So your set G will be

G = {{3.8, 5}, {1.8,3}, {3.1,5}, {3.1,5}}.

Multiplying each value by it's respective weigth and then adding, we have:

3.8*5 + 1.8*3 + 3.1*5 + 3.1*5 = 19 + 5.4 + 15.5 + 15.5 = 55.4

The sum of the weigths is 5 + 3 + 5 + 5 = 18

So the student's GPA for that term, rounded to two decimal places, is: 55.4/18 = 3.07

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3 years ago

There is same number of pennies, nickels, and dimes in a wallet. Together they make $1.28. How many coins are there in the walle

Anarel [89]

A = amount

.01a + .05a + .1a = 1.28
Combine like terms.
.16a = 1.28
Divide both sides by .16.
a = 8

8 * 3 = 24

There are 24 coins in the wallet.

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3 years ago

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