Adding and subtracting fractions with unlike denominators

What is 432*80? Show your work by multiplying with zeros

Dima020 [189]

Answer:

in the steps

Step-by-step explanation:

do 432*8 then just add a 0

3 0

3 years ago

Read 2 more answers

How do you solve: f(x) = -x2e-x +2xe-x , finding the zero's of f.

Maksim231197 [3]

Answer:

0 and 2

Step-by-step explanation:

$f(x) = -x^2e^{-x}+2xe^{-x}$

To find zeros set f(x)=0 and solve for x

$0 = -x^2e^{-x}+2xe^{-x}$

$0 = -e^{-x}(x^2-2x)$

now factor out x

$0 = -xe^{-x}(x-2)$

set each factor =0 and solve for x

$x=0$

$-e^{-x}=0$ no solution for x

$x-2=0, x=2$

zeros of f are 0 and 2

7 0

4 years ago

How to do this ? Don’t understand it

riadik2000 [5.3K]

X=y+3
-5x-4y=30
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Substitute y + 3 for x in -5x-4y=30
-5(y+3)-4y=30
-9y-15=30
----------------------------
Add 15 to each side
-9y-15+15=30+15
-9y = 45
-----------------------------------
Divide each side by -9
-9y ÷ -9 = 45 ÷ -9
y = -5
Now we solve for x
====================================================================
Substitute -5 for y in x = y + 3
x=-5+3
x = -2
----------------------
(-2,-5) is your answer

7 0

3 years ago

Help.........................

garri49 [273]

<h3>Answer: Choice A</h3>

$x^2\left(\sqrt[4]{x^2}\right)$

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Explanation:

The fourth root of x is the same as x^(1/4)

I.e,

$\sqrt[4]{x} = x^{1/4}$

The same applies to x^10 as well

$\sqrt[4]{x^{10}} = \left(x^{10}\right)^{1/4}$

Multiply the exponents 10 and 1/4 to get 10/4

$\sqrt[4]{x^{10}} = \left(x^{10}\right)^{1/4} = x^{10*1/4} = x^{10/4}$

$\sqrt[4]{x^{10}} = x^{10/4}$

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If we have an expression in the form x^(m/n), with m > n, then we can simplify it into an equivalent form as shown below

$x^{m/n} = x^a\sqrt[n]{x^b}$

The 'a' and 'b' are found through dividing m/n

m/n = a remainder b

'a' is the quotient, b is the remainder

-----------------------

The general formula can easily be confusing, so let's replace m and n with the proper numbers. In this case, m = 10 and n = 4

m/n = 10/4 = 2 remainder 2

We have a = 2 and b = 2

So

$x^{m/n} = x^a\sqrt[n]{x^b}$

turns into

$x^{10/4} = x^2\sqrt[4]{x^2}$

which means

$\sqrt[4]{x^{10}} = {x^2} \sqrt[4]{x^2}$

7 0

3 years ago

5,5,3 Is this a triangle? Explain.

Vlad1618 [11]

Problem 1

Answer: Yes a triangle can be formed with side lengths 5,5,3

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Explanation:

Take any two sides and add them up. If we get a sum larger than the third side, then a triangle is possible. This is the triangle inequality theorem.

5+5 = 10 is larger than 3

5+3 = 8 is larger than 5

This shows a triangle is possible. The triangle is isosceles because two sides are the same length.

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Problem 2

Answer: Yes a triangle can be formed with side lengths 8,8,8

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Explanation:

This triangle is equilateral because all sides are the same length.

Take any two sides, add them up, and you'll find the sum is larger than the third side.

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Problem 3

Answer: No a triangle cannot be formed with sides 7,8,15

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Explanation:

Yes it is true that

8+15 = 23 is larger than 7

7+15 = 22 is larger than 8

but

7+8 = 15 is not larger than 15

So this means a triangle is not possible.

Imagine you had 3 strips of paper that were 7 inches, 8 inches and 15 inches long. The 7 and 8 inch strips add to 15 inches, but those two smaller strips only form a straight line. We cannot pull them so a triangle forms. If we did, then the third side would be smaller than 15 inches. This is one way to get hands on to see why the triangle inequality theorem works.

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Problem 4

Answer: Yes a triangle is possible with side lengths 5,6,10

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Explanation:

Same idea as before. We can take any two sides, add them, and get a sum larger than the third side. A triangle is possible.

5+6 = 11 is larger than 10
6+10 = 16 is larger than 5
5+10 = 15 is larger than 6

4 0

3 years ago