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3 years ago

Solve the proportion 2x/5 = 9/15

Mathematics

2 answers:

yuradex [85]3 years ago

7 0

The answer is x=4.5 because 9÷2=4.5 and 4.5×2=9

garri49 [273]3 years ago

5 0

You need to get x alone so multiply each side by 5 to get rid of the fractions the will get you to 2x=45/15 which equals 2x=3 so x equals 3/2 or 1.5

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What is the value of the 3 in 513

hodyreva [135]

Answer:

Step-by-step explanation:

513 = 500 + 10 + 3

The 5 is worth 500.

The 1 is worth 10.

And the 3 is worth 3 !

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4 years ago

Read 2 more answers

What is the y in 10.3=0.6y

ollegr [7]

Hello, you would divide 10.3 by 0.6 to get 17.166 repeating. So y equals 17.166.

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3 years ago

Find the flux of F = x^3 i + y^3 j + z^3k through the closed surface bounding the solid region x^2 + y^2 ≤ 4, 0 ≤ z ≤ 4

givi [52]

Use the divergence theorem. Let $R$ be the cylindrical region, then

$\displaystyle\iint_{\partial R}\mathbf F\cdot\mathbf n\,\mathrm dS=\iiint_R\nabla\cdot\mathbf F\,\mathrm dV$

(where $\mathbf n$ denotes the unit normal vector to $\partial R$ , but we don't need to worry about it now)

We have

$\mathrm{div }\mathbf F=(\nabla\cdot\mathbf F)(x,y,z)=\dfrac{\partial\mathbf F}{\partial x}+\dfrac{\partial\mathbf F}{\partial y}+\dfrac{\partial\mathbf F}{\partial z}$
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For the solid $R$ with boundary $\partial R$ , we can set up the following volume integral in cylindrical coordinates for ease:

$\displaystyle3\iiint_R(x^2+y^2+z^2)\,\mathrm dV=3\int_{z=0}^{z=4}\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=2}(r^2+z^2)r\,\mathrm dr\,\mathrm d\theta\,\mathrm dz$
$=\displaystyle6\pi\int_{z=0}^{z=4}\int_{r=0}^{r=2}(r^3+rz^2)\,\mathrm dr\,\mathrm dz$
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3 years ago

The quotient of 17 less than four times a number and seven is the same as three times the number?

Semenov [28]

Answer:

$x=-1$

Step-by-step explanation:

The equation they are giving is:

$\frac{4x-17}{7} =3x$

Multiply both sides by 7 to get:

$4x-17=21x$

Subtract 4x from both sides to reach:

$-17=17x$

Therefore, $x=-1$ .

Hope this helps :)

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3 years ago

Let R be the region bounded by

loris [4]

a. The area of $R$ is given by the integral

$\displaystyle \int_1^2 (x + 6) - 7\sin\left(\dfrac{\pi x}2\right) \, dx + \int_2^{22/7} (x+6) - 7(x-2)^2 \, dx \approx 9.36$

b. Use the shell method. Revolving $R$ about the $x$ -axis generates shells with height $h=x+6-7\sin\left(\frac{\pi x}2\right)$ when $1\le x\le 2$ , and $h=x+6-7(x-2)^2$ when $2\le x\le\frac{22}7$ . With radius $r=x$ , each shell of thickness $\Delta x$ contributes a volume of $2\pi r h \Delta x$ , so that as the number of shells gets larger and their thickness gets smaller, the total sum of their volumes converges to the definite integral

$\displaystyle 2\pi \int_1^2 x \left((x + 6) - 7\sin\left(\dfrac{\pi x}2\right)\right) \, dx + 2\pi \int_2^{22/7} x\left((x+6) - 7(x-2)^2\right) \, dx \approx 129.56$

c. Use the washer method. Revolving $R$ about the $y$ -axis generates washers with outer radius $r_{\rm out} = x+6$ , and inner radius $r_{\rm in}=7\sin\left(\frac{\pi x}2\right)$ if $1\le x\le2$ or $r_{\rm in} = 7(x-2)^2$ if $2\le x\le\frac{22}7$ . With thickness $\Delta x$ , each washer has volume $\pi (r_{\rm out}^2 - r_{\rm in}^2) \Delta x$ . As more and thinner washers get involved, the total volume converges to

$\displaystyle \pi \int_1^2 (x+6)^2 - \left(7\sin\left(\frac{\pi x}2\right)\right)^2 \, dx + \pi \int_2^{22/7} (x+6)^2 - \left(7(x-2)^2\right)^2 \, dx \approx 304.16$ <em />

d. The side length of each square cross section is $s=x+6 - 7\sin\left(\frac{\pi x}2\right)$ when $1\le x\le2$ , and $s=x+6-7(x-2)^2$ when $2\le x\le\frac{22}7$ . With thickness $\Delta x$ , each cross section contributes a volume of $s^2 \Delta x$ . More and thinner sections lead to a total volume of

$\displaystyle \int_1^2 \left(x+6-7\sin\left(\frac{\pi x}2\right)\right)^2 \, dx + \int_2^{22/7} \left(x+6-7(x-2)^2\right) ^2\, dx \approx 56.70$

7 0

2 years ago