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Damm [24]
2 years ago
5

The fraction 5/8 is located between what two fractions?

Mathematics
2 answers:
S_A_V [24]2 years ago
8 0
C is the answer I pretty sure of it
Digiron [165]2 years ago
4 0
4/10 = 16/40
5/8 = 25/40
2/5 = 16/40

1/2 = 12/24
5/8 = 15/24
2/3 = 16/24

3/7 = 48/112
5/8 = 70/112
10/16 = 70/112
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Use properties of equality to simplify each of the following expressions so that you can simplify them in your head. Identify th
Salsk061 [2.6K]
Properties of equality have nothing to do with it. The associative and commutative properties of multiplication are used (along with the distributive property and the fact of arithmetic: 9 = 10 - 1).

All of these problems make use of the strategy, "look at what you have before you start work."


1. = (4·5)·(-3) = 20·(-3) = -60 . . . . if you know factors of 60, you can do this any way you like. It is convenient to ignore the sign until the final result.

2. = (2.25·4)·23 = 9·23 = 23·10 -23 = 230 -23 = 207 . . . . multiplication by 4 can clear the fraction in 2 1/4, so we choose to do that first. Multiplication by 9 can be done with a subtraction that is often easier than using ×9 facts.

4. = (2·5)·12·(-1) = 10·12·(-1) = (-1)·120 = -120 . . . . multiplying by 10 is about the easiest, so it is convenient to identify the factors of 10 and use them first. Again, it is convenient to ignore the sign until the end.

5. = 0 . . . . when a factor is zero, the product is zero
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3 years ago
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What is the absolute value function?
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2 years ago
How many bags of rice weigh at least 1<br> 2 <br> Pound?
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The answer depends on the weight of each individual bag. Do you know exactly how heavy each of the bags are? Please be more clear in your question.
3 0
3 years ago
99 POINT QUESTION, PLUS BRAINLIEST!!!
VladimirAG [237]
First, we have to convert our function (of x) into a function of y (we revolve the curve around the y-axis). So:


y=100-x^2\\\\x^2=100-y\qquad\bold{(1)}\\\\\boxed{x=\sqrt{100-y}}\qquad\bold{(2)} \\\\\\0\leq x\leq10\\\\y=100-0^2=100\qquad\wedge\qquad y=100-10^2=100-100=0\\\\\boxed{0\leq y\leq100}

And the derivative of x:

x'=\left(\sqrt{100-y}\right)'=\Big((100-y)^\frac{1}{2}\Big)'=\dfrac{1}{2}(100-y)^{-\frac{1}{2}}\cdot(100-y)'=\\\\\\=\dfrac{1}{2\sqrt{100-y}}\cdot(-1)=\boxed{-\dfrac{1}{2\sqrt{100-y}}}\qquad\bold{(3)}

Now, we can calculate the area of the surface:

A=2\pi\int\limits_0^{100}\sqrt{100-y}\sqrt{1+\left(-\dfrac{1}{2\sqrt{100-y}}\right)^2}\,\,dy=\\\\\\= 2\pi\int\limits_0^{100}\sqrt{100-y}\sqrt{1+\dfrac{1}{4(100-y)}}\,\,dy=(\star)

We could calculate this integral (not very hard, but long), or use (1), (2) and (3) to get:

(\star)=2\pi\int\limits_0^{100}1\cdot\sqrt{100-y}\sqrt{1+\dfrac{1}{4(100-y)}}\,\,dy=\left|\begin{array}{c}1=\dfrac{-2\sqrt{100-y}}{-2\sqrt{100-y}}\end{array}\right|= \\\\\\= 2\pi\int\limits_0^{100}\dfrac{-2\sqrt{100-y}}{-2\sqrt{100-y}}\cdot\sqrt{100-y}\cdot\sqrt{1+\dfrac{1}{4(100-y)}}\,\,dy=\\\\\\ 2\pi\int\limits_0^{100}-2\sqrt{100-y}\cdot\sqrt{100-y}\cdot\sqrt{1+\dfrac{1}{4(100-y)}}\cdot\dfrac{dy}{-2\sqrt{100-y}}=\\\\\\

=2\pi\int\limits_0^{100}-2\big(100-y\big)\cdot\sqrt{1+\dfrac{1}{4(100-y)}}\cdot\left(-\dfrac{1}{2\sqrt{100-y}}\, dy\right)\stackrel{\bold{(1)}\bold{(2)}\bold{(3)}}{=}\\\\\\= \left|\begin{array}{c}x=\sqrt{100-y}\\\\x^2=100-y\\\\dx=-\dfrac{1}{2\sqrt{100-y}}\, \,dy\\\\a=0\implies a'=\sqrt{100-0}=10\\\\b=100\implies b'=\sqrt{100-100}=0\end{array}\right|=\\\\\\= 2\pi\int\limits_{10}^0-2x^2\cdot\sqrt{1+\dfrac{1}{4x^2}}\,\,dx=(\text{swap limits})=\\\\\\

=2\pi\int\limits_0^{10}2x^2\cdot\sqrt{1+\dfrac{1}{4x^2}}\,\,dx= 4\pi\int\limits_0^{10}\sqrt{x^4}\cdot\sqrt{1+\dfrac{1}{4x^2}}\,\,dx=\\\\\\= 4\pi\int\limits_0^{10}\sqrt{x^4+\dfrac{x^4}{4x^2}}\,\,dx= 4\pi\int\limits_0^{10}\sqrt{x^4+\dfrac{x^2}{4}}\,\,dx=\\\\\\= 4\pi\int\limits_0^{10}\sqrt{\dfrac{x^2}{4}\left(4x^2+1\right)}\,\,dx= 4\pi\int\limits_0^{10}\dfrac{x}{2}\sqrt{4x^2+1}\,\,dx=\\\\\\=\boxed{2\pi\int\limits_0^{10}x\sqrt{4x^2+1}\,dx}

Calculate indefinite integral:

\int x\sqrt{4x^2+1}\,dx=\int\sqrt{4x^2+1}\cdot x\,dx=\left|\begin{array}{c}t=4x^2+1\\\\dt=8x\,dx\\\\\dfrac{dt}{8}=x\,dx\end{array}\right|=\int\sqrt{t}\cdot\dfrac{dt}{8}=\\\\\\=\dfrac{1}{8}\int t^\frac{1}{2}\,dt=\dfrac{1}{8}\cdot\dfrac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}=\dfrac{1}{8}\cdot\dfrac{t^\frac{3}{2}}{\frac{3}{2}}=\dfrac{2}{8\cdot3}\cdot t^\frac{3}{2}=\boxed{\dfrac{1}{12}\left(4x^2+1\right)^\frac{3}{2}}

And the area:

A=2\pi\int\limits_0^{10}x\sqrt{4x^2+1}\,dx=2\pi\cdot\dfrac{1}{12}\bigg[\left(4x^2+1\right)^\frac{3}{2}\bigg]_0^{10}=\\\\\\= \dfrac{\pi}{6}\left[\big(4\cdot10^2+1\big)^\frac{3}{2}-\big(4\cdot0^2+1\big)^\frac{3}{2}\right]=\dfrac{\pi}{6}\Big(\big401^\frac{3}{2}-1^\frac{3}{2}\Big)=\boxed{\dfrac{401^\frac{3}{2}-1}{6}\pi}

Answer D.
6 0
3 years ago
Read 2 more answers
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