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3 years ago

Want brainlyest? Please help me. :)))) and I swear I will!!!

Mathematics

2 answers:

guajiro [1.7K]3 years ago

4 0

Answer:

5/8 so, it is a fraction

Step-by-step explanation:

there is 5 out of 8 shaded

mihalych1998 [28]3 years ago

4 0

Answer:

Fraction: $\frac{5}{8}$

Decimal: 0.625

Percentage: 62.5%

Step-by-step explanation:

There are 8 sections of the square. 5 are filled in.

If you divide 5 by 8, you'll find the answer as a decimal: 0.625.

To change it to a percentage, simply move the decimal point back two spaces (62.5%).

For a fraction, the answer is simply $\frac{5}{8}$ .

I hope this helps.

Please let me know if you need anything else. :)

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Can u pls help me with this question

ziro4ka [17]

Step-by-step explanation:

I am not sure bout this question sry

you reap what you sow ;)

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3 years ago

A cylinder with a radius of 4 yd and a volume of 80yd (find the height)

Afina-wow [57]

Answer:

Height of the cylinder = 1.60yd

Step-by-step explanation:

Volume of a cylinder =

V = πr^2h

V - volume of cylinder

π - 3.142

r - radius of cylinder

h - height of cylinder

We are provided with

V - 80

r - 4

π - 3.142

h - ?

80 = 3.142 * 4^2 * h

80 = 50.272 * h

To find h, divide both sides by 50.272

80/50.272 = 50.272h/ 50.272

1.599 = h

h = 1.60yd

7 0

4 years ago

Read 2 more answers

Somebody please help with this problem

Mrrafil [7]

Step-by-step explanation:

We have been given that AE=BE and $\angle1\cong \angle2$ .

We can see that angle CEA is vertical angle of angle DEB, therefore, $m\angle CEA=m\angle DEB$ as vertical angles are congruent.

We can see in triangles CEA and DEF that two angles and included sides are congruent.

$\angle 1\cong \angle 2$

$AE=BE$

$\angle CEA\cong\angle DEB$ or $\angle 3\cong \angle 4$

Therefore, $\Delta CEA\cong \Delta DEB$ by ASA postulate.

Since corresponding parts of congruent triangles are congruent, therefore CE must be congruent to DE.

5 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

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

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

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3 years ago

Write the number 2.7 x 10^-3

spayn [35]

Answer:

0.0027 - zero-point-zero-zero-twenty-seven

Step-by-step explanation:

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3 years ago

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