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Savatey [412]
3 years ago
6

How to do 10 to 4 decrease

Mathematics
2 answers:
USPshnik [31]3 years ago
6 0
If you're talking about the ratio 10:4, you can decrease it by dividing both numbers by 2.

10:4=5:2

The ratio decreases to 5:2
chubhunter [2.5K]3 years ago
6 0
By subtracting or dividing.

Hoped this helped.

~Bob Ross®
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that means the nth term is equal to the first term plus n-1 times the common difference.

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Which symbol correctly compares the two fractions? Use > , < , or =. Enter your answer in the box. 2/4 3/8
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The correct answer is:  [A]:  >   .
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Note that:  "2/4"  <span>> "3/8" .
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Specifically, note that "2/4" </span>><span> "4/8" .
{i.e. "2/4 is GREATER THAN 3/8" }.
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     2/4 = ?/8  ;   4* ? = 8 ? ;  8 </span>÷ 4 = 2 .  So;  4 * 2 = 8 ;

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 So;  "2/4" = "4/8".  and: "4/8" is GREATER THAN "3/8" ;

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8 0
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How do you kno 2130 is greater than 2/3
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4 years ago
Evaluate the given integral by changing to polar coordinates. sin(x2 y2) dAR, where R is the region in the first quadrant betwee
rjkz [21]

Answer:

The result of the integral is \frac{\pi}{4}(-\cos{16} + \frac{9})

Step-by-step explanation:

Polar coordinates:

In polar coordinates, we have that:

x^2 + y^2 = r^2

And

\int \int_{dA} f(x,y) da = \int \int f(r) r dr d\theta

In this question:

\int \int_{dA} \sin{(x^2+y^2)} dA = \int \int_{dR} = \sin{r^2}r dr d\theta

Region in the first quadrant between the circles with center the origin and radii 3 and 4

First quadrant means that \theta ranges between 0 and \frac{\pi}{2}

Between these circles means that r ranges between 3 and 4. So

\int \int_{dR} = \sin{r^2}r dr d\theta = \int_{0}^{\frac{\pi}{2}} \int_{3}^{4} \sin{r^2} r dr d\theta

Applying the inner integral:

\int_{3}^{4} \sin{r^2} r dr

Using substitution, with u = r^2, du = 2rdr, dr = \frac{du}{2r}, and considering that the integral of the sine is minus cosine, we have:

-\frac{\cos{r^2}}{2}|_{3}{4} = \frac{1}{2}(-\cos{16} + \frac{9})

Applying the outer integral:

\int_{0}^{\frac{\pi}{2}} \frac{1}{2}(-\cos{16} + \frac{9}) d\theta

Has no factors of \theta, so the result is the constant multiplied by \theta, and then we apply the fundamental theorem.

\frac{\theta}{2}(-\cos{16} + \frac{9}) = \frac{\pi}{4}(-\cos{16} + \frac{9})

The result of the integral is \frac{\pi}{4}(-\cos{16} + \frac{9})

8 0
3 years ago
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