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3 years ago

Which number results in a true sentence in 0.475 ＜ ______ ? a.) 0.473 c.) 0.475 b.) 0.474 d.) 0.476

Mathematics

1 answer:

patriot [66]3 years ago

7 0

Answer:

D.) 0.476

Step-by-step explanation:

The < sign means 'less than'. The only thing that .475 is less than is .476

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A premature infant weighing 3lbs is ordered to 1 milligram per gram of a drug. How many milligrams should you administer to the

Alona [7]

Answer:

1360.78 mg

Step-by-step explanation:

3 pounds in grams is 1360.78 grams.

so, 1 mg per gram would mean the baby should recieve 1360.78 mg of the drug.

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2 years ago

I will mark brainliestand 40 points! :)

frosja888 [35]

Answer is D) take 5 blocks away from the left side.

There are 8 blocks on the left and only 3 on the right, 8 - 3 = 5

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3 years ago

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Use the distributive property to fill in the blanks below.

elena-14-01-66 [18.8K]

Answer:

(5×7)-(5×6) = 5×(7-6)

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3 years ago

Jack bought 4 bagels for $3.00.How many bagels can he buy for $4.50?

nydimaria [60]

Answer:

He can buy 6 bagels.

Step-by-step explanation:

In order to figure out how much each bagel is, you need to divide $3.00 by 4. This gives you .75 because 3.00/4=75. Each bagel is therefore $0.75. Now, in order to find how many bagels you can buy with $4.50, you have to divide 4.50 by 0.75. The equation is 450/75=6. You can buy 6 bagels with $4.50.

Let me know if you need any more help. Have a nice day. :)

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3 years ago

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"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