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PIT_PIT [208]
3 years ago
8

Quinn had $4.00. He spent 1/2​ of his money on a baseball card, 1/4 of it on a drink, and 1/8 of it on bubble gum. How much mone

y does Quinn have left?
US dollars btw
Mathematics
2 answers:
Pepsi [2]3 years ago
4 0

Answer:

$0.50

Step-by-step explanation:

He spent 2 on a card 1 on a drink and 1/2 on a piece of gum.

4 - (2+1+1/2) = 1/2

I am not sure it's correct.

mart [117]3 years ago
3 0

Answer: Quinn has $0.50 left.

Step-by-step explanation:

This is because if you were to divide $4.00 into 1/2 it would equal $2.

Next, you would divide $4 into 1/4, this would give you $1.

Finally, you would divide $4 into 1/8, this would give you $0.50.

To get the answer you would add all of those quotients together, adding up to $3.50.

You would then take $4.00 - $3.50 = $0.50

(To get those numbers just divide by the denominator it is way easier. Example: To divide $4 into 1/8 you would divide by 8. aka 4 ÷ 8.)

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Step-by-step explanation:

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Find the missing side length in the image below​
Tcecarenko [31]

Answer:

Step-by-step explanation:

7 0
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<img src="https://tex.z-dn.net/?f=%20%20%5Cdisplaystyle%20%5Cint%20%5Climits_%7B0%7D%5E%7B%20%5Cfrac%7B%20%5Cpi%7D%7B2%7D%20%7D%
murzikaleks [220]

Let x = \arcsin(y), so that

\sin(x) = y

\tan(x)=\dfrac y{\sqrt{1-y^2}}

dx = \dfrac{dy}{\sqrt{1-y^2}}

Then the integral transforms to

\displaystyle \int_{x=0}^{x=\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \int_{y=\sin(0)}^{y=\sin\left(\frac\pi2\right)} \frac{y}{\sqrt{1-y^2}} \ln(y) \frac{dy}{\sqrt{1-y^2}}

\displaystyle \int_{x=0}^{x=\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy

Integrate by parts, taking

u = \ln(y) \implies du = \dfrac{dy}y

dv = \dfrac{y}{1-y^2} \, dy \implies v = -\dfrac12 \ln|1-y^2|

For 0 < y < 1, we have |1 - y²| = 1 - y², so

\displaystyle \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy = uv \bigg|_{y\to0^+}^{y\to1^-} + \frac12 \int_0^1 \frac{\ln(1-y^2)}{y} \, dy

It's easy to show that uv approaches 0 as y approaches either 0 or 1, so we just have

\displaystyle \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy = \frac12 \int_0^1 \frac{\ln(1-y^2)}{y} \, dy

Recall the Taylor series for ln(1 + y),

\displaystyle \ln(1+y) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}n y^n

Replacing y with -y² gives the Taylor series

\displaystyle \ln(1-y^2) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}n (-y^2)^n = - \sum_{n=1}^\infty \frac1n y^{2n}

and replacing ln(1 - y²) in the integral with its series representation gives

\displaystyle -\frac12 \int_0^1 \frac1y \sum_{n=1}^\infty \frac{y^{2n}}n \, dy = -\frac12 \int_0^1 \sum_{n=1}^\infty \frac{y^{2n-1}}n \, dy

Interchanging the integral and sum (see Fubini's theorem) gives

\displaystyle -\frac12 \sum_{n=1}^\infty \frac1n \int_0^1 y^{2n-1} \, dy

Compute the integral:

\displaystyle -\frac12 \sum_{n=1}^\infty \frac1n \int_0^1 y^{2n-1} \, dy = -\frac12 \sum_{n=1}^\infty \frac{y^{2n}}{2n^2} \bigg|_0^1 = -\frac14 \sum_{n=1}^\infty \frac1{n^2}

and we recognize the famous sum (see Basel's problem),

\displaystyle \sum_{n=1}^\infty \frac1{n^2} = \frac{\pi^2}6

So, the value of our integral is

\displaystyle \int_0^{\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \boxed{-\frac{\pi^2}{24}}

6 0
3 years ago
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tamaranim1 [39]
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6 0
3 years ago
Read 2 more answers
Note: Enter your answer and show all the steps that you use to solve this problem in the space provided.
Troyanec [42]

Answer: 300

Step-by-step explanation:

Let the number be represented by x.

Therefore, based on the information given in the question, this can be expressed as:

36% of x = 108

36/100 × x = 108

0.36 × x = 108

0.36x = 108

x = 108/0.36

x = 300

Therefore, the number is 300

8 0
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