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1 year ago

Let's find2. 1+5 3First write the addition with a common denominator.Then add.12— +51-4-13Х5

Mathematics

1 answer:

emmasim [6.3K]1 year ago

4 0

Answer: $\frac{2}{5}+\frac{1}{3}=\frac{6}{15}+\frac{5}{15}=\frac{11}{15}$ Explanation:

The given addition exercise is:

$\frac{2}{5}+\frac{1}{3}$

The LCM of the denominator (5 and 3) = 15

Multiply 2/5 by 3/3

$\frac{2}{5}=\frac{2\times3}{5\times3}=\frac{6}{15}$

Multiply 1/3 by 5/5

$\frac{1}{3}=\frac{1\times5}{3\times5}=\frac{5}{15}$

The addition becomes

$\frac{6}{15}+\frac{5}{15}=\frac{11}{15}$

Therefore, we can fill in the vacant boxes as shown below:

$\frac{2}{5}+\frac{1}{3}=\frac{6}{15}+\frac{5}{15}=\frac{11}{15}$

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Answer:

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2 years ago

Campbell Middle school has 912 students.How many computers are in this schools computer lab?

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3 years ago

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Make a table with the domain of {2,3,4,5,6} and draw a graph of the absolute value function y = 2|x-4| + 3.

Jet001 [13]

Answer:

<h2>In the attachment.</h2>

Step-by-step explanation:

$|a|=\left\{\begin{array}{ccc}a&for\ a\geq0\\-a&for\ a$

Put each value of x from the set {2, 3, 4, 5, 6}

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x = 2 → y = 2|2 - 4| + 3 = 2|-2| + 3 = 2(2) + 3 = 4 + 3 = 7 → (2, 7)

x = 3 → y = 2|3 - 4| + 3 = 2|-1| + 3 = 2(1) + 3 = 2 + 3 = 5 → (3, 5)

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Mark the points in the coordinates system.

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3 years ago

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) Set up a double integral for calculating the flux of F=3xi+yj+zk through the part of the surface z=−5x−2y+2 above the triangle

Fynjy0 [20]

The surface (call it $S$ ) is a triangle with vertices at the points

$x=0,y=0\implies z=2\implies(0,0,2)$

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Parameterize $S$ by

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$\vec s_v\times\vec s_u=(20v,8v,4v)$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_S\vec F(x,y,z)\cdot\mathrm d\vec S=\int_0^1\int_0^1\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_v\times\vec s_u)\,\mathrm du\,\mathrm dv$

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$=\displaystyle8\int_0^1\int_0^1(11v-10v^2)\,\mathrm du\,\mathrm dv=\boxed{\frac{52}3}$

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