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3 years ago

What is the absolute value of 5? a. 5 b. 5 c. 5 or 5 d. 0

1 answer:

6 0

Simply 5. Absolute value is most commonly used to show that even though what is in the brackets is negative [-7] the actual value for practical purposes needed is positive 7.

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What is larger0.04 or 0.2

Korolek [52]

I hope this helps you

0,2 close 1

0,2>0,04

5 0

3 years ago

Graph a line with a slope of -5 that contains the point -3,-4

guapka [62]

Answer:

The equation of the line is 5x + y + 19 = 0

Step-by-step explanation:

The equation of the line with slope 'm' and given a point (x₁, y₁) passing through it we use the Slope - one - point form which is given by:

y - y₁ = m(x - x₁)

The point given is: (-3, -4) and the slope is -5.

We get the equation of the line to be:

y - (-4) = -5(x - (-3))

⇒ y + 4 = -5(x + 3)

⇒ y + 4 = -5x - 15

⇒ 5x + y + 19 = 0. is the required equation of the line.

4 0

3 years ago

The equation for the pH of a substance is pH = –log[H+], where H+ is the concentration of hydrogen ions. Which equation models t

Zanzabum

Answer: [H+] = 10^-7.2

Step-by-step explanation:

Given that the PH of the solution = 7.2

Using the formula pH = –log[H+]. To get the H+ concentration from the pH, raise both sides by the base of 10. Then we have

10^ -pH = H+. with pH of 7.2,

Thus the answer to this problem is

[H+] = 10^-7.2

8 0

3 years ago

Pam likes to practice dancing while preparing for a math tournament. She spends 80 minutes every day practicing dance and math.

RSB [31]

Answer: 628

Step-by-step explanation:

I found it online

7 0

3 years ago

"find the reduction formula for the integral" sin^n(18x)

dexar [7]

Let

$I(n,a)=\displaystyle\int\sin^nax\,\mathrm dx$
For demonstration on how to tackle this sort of problem, I'll only work through the case where $n$ is odd. We can write

$\displaystyle\int\sin^nax\,\mathrm dx=\int\sin^{n-2}ax\sin^2ax\,\mathrm dx=\int\sin^{n-2}ax(1-\cos^2ax)\,\mathrm dx$
$\implies I(n,a)=I(n-2,a)-\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx$

For the remaining integral, we can integrate by parts, taking

$u=\sin^{n-3}ax\implies\mathrm du=a(n-3)\sin^{n-4}ax\cos ax\,\mathrm dx$ $\mathrm dv=\sin ax\cos^2ax\,\mathrm dx\implies v=-\dfrac1{3a}\cos^3ax$

$\implies\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx=-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{a(n-3)}{3a}\int\sin^{n-4}ax\cos^4ax\,\mathrm dx$

For this next integral, we rewrite the integrand

$\sin^{n-4}ax\cos^4ax=\sin^{n-4}ax(1-\sin^2ax)^2=\sin^{n-4}ax-2\sin^{n-2}ax+\sin^nax$
$\implies\displaystyle\int\sin^{n-4}ax\cos^4ax\,\mathrm dx=I(n-4,a)-2I(n-2,a)+I(n,a)$

So putting everything together, we found

$I(n,a)=I(n-2,a)-\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx$
$I(n,a)=I(n-2,a)-\left(-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{n-3}3\displaystyle\int\sin^{n-4}ax\cos^4ax\,\mathrm dx\right)$
$I(n,a)=I(n-2,a)-\dfrac{n-3}3\bigg(I(n-4,a)-2I(n-2,a)+I(n,a)\bigg)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax$
$\dfrac n3I(n,a)=\dfrac{2n-3}3I(n-2,a)-\dfrac{n-3}3I(n-4,a)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax$

$\implies I(n,a)=\dfrac{2n-3}nI(n-2,a)-\dfrac{n-3}nI(n-4,a)+\dfrac1{na}\sin^{n-3}ax\cos^3ax$

7 0

3 years ago