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3 years ago

Please show your work

Mathematics

2 answers:

coldgirl [10]3 years ago

7 0

Answer:

m(h+p) - p÷h + m

h = -1, m = 3, and p = 4

3(-1+4) - 4÷(-1) + 3

= 3(3) - (-4) + 3

= 9 + 4 + 3

= 16

D. 16

notka56 [123]3 years ago

7 0

Answer:

3(-1+4)-4/-1+3

-3+12+4+3

12+4

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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

the original price for a set of golf clubs is $500 at the beginning of the seasons the price was marked up by 20% at the end of

Blizzard [7]

Answer: The final price is less than the original

Step-by-step explanation:

Let us break this down a bit.

The original price is $500

-> At the beginning, the price was marked up by 20%

-> At the end, the price at the beginning of the season is marked down by 20%

We are solving for: "How does the final price compare with the original price of $500"

First, we will find the price at the beginning.

-> Please note a percent divided by 100 becomes a decimal

20% / 100 = 0.2

$500 * 0.2 = $100

$500 + $100 = $600

Now, we will find the price at the end.

$600 * 0.2 = $120

$600 - $120 = <u>$480</u>

The final price is less than the original price of $500.

5 0

3 years ago

Read 2 more answers

Find the value of y for given value of x.

Scrat [10]

1.)x+5;x=3
(3) + 5
8

2.)7x;x=-5
7(-5)
35

3.)y=1-2x;x=9
y = 1 - 2(9)
y = 1 - 18
y = -17

4.)y=3x+2;x=0.5
y = 3(0.5) + 2
y = 1.5 + 2
y = 3.5

5.)y=2x^3;x=3
y = 2(3^3)
y = 2(3)(3)(3)
y = 2(27)
y = 54

<span>6.)y=x/2+9;x=-12
y = (-12)/2 + 9
y = -6 + 9
y = -3</span>

6 0

4 years ago

Plz help I need help

Gre4nikov [31]

Answer:

The first one: -1 1/4

Step-by-step explanation"

3 0

3 years ago

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kow [346]

Answer:

Step-by-step explanation:

6 0

3 years ago