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2 years ago

The set of all whole numbers not greater than 5

Mathematics

1 answer:

k0ka [10]2 years ago

6 0

Answer:

-5, -4 , -3 , -2, -1, 1, 2 , 3, 4, 5

Step-by-step explanation:

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What is the surface area of this rectangular prism?

dybincka [34]

Answer:

256 ft²

Step-by-step explanation:

4 lateral faces + 2 bases

The total area

$=2 \times (12 \times 4) + 2 \times (12 \times 5) + 2 \times (5 \times 4)$

$= 96 + 120 + 40$

$= 256$

3 0

2 years ago

1/2r -3= 3 (4-3/2r) solve for r

s2008m [1.1K]

1/2r -3= 3 (4-3/2r) is to be solved for r.

I'll begin by making the assumptions that by 1/2r you actually meant (1/2)r and that by 3/2r you actually meant (3/2)r. When in doubt, please use parentheses to make your meaning clear.

Thus, 1/2r -3= 3 (4-3/2r) becomes (1/2)r -3= 3 (4-(3/2)r ) .

Simplify this by multiplying all 3 terms by 2. Doing this will eliminate the fractions:

r -6 = 3 (4*2-(3)r ) or r - 6 = 24 - 9r

Now expand the right side, using the distributive property of

r - 3 = 24 - 9r

Regrouping so as to combine like terms:

10r = 30

Solving for r: r = 30/10 = 3

The value of r that satisfies this equation is 3

8 0

3 years ago

What is the limit of f (×) as x approaches - infinty

tatyana61 [14]

If the degree of numerator and denominator are equal, then limit will be leading coefficient of numerator divided by the leading coefficient of denominator.

So then the limit would be 3/1 = 3.

Alternatively,

$\displaystyle \lim_{x\to\infty}\dfrac{3x^2+6}{x^2-4}=\displaystyle \lim_{x\to\infty}\dfrac{3x^2+6}{x^2-4}\cdot\dfrac{1/x^2}{1/x^2}=\lim_{x\to\infty}\dfrac{3+\frac6{x^2}}{1-\frac4{x^2}} = \dfrac{3+0}{1-0}=\boxed{3}$

Hope this helps.

5 0

2 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

2 years ago

What is the rectangle area

pentagon [3]

Length times the width

3 0

3 years ago

Read 2 more answers