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2 years ago

What is the sum of the fractions? 3/5+-7/5 Use the number line to help find the answer

Mathematics

2 answers:

masha68 [24]2 years ago

8 0

Answer:

Step-by-step explanation:

<em>The given equation is: </em><em>(3/5) + (-7/5)</em>

<em />

<em>To start off, we can make this equation simpler:</em>

= the negative value of 4/5

<h2><em>Thus, </em>the solution to the equation above is simply: <em>-4/5</em></h2>

ehidna [41]2 years ago

5 0

I think the answer is -4/5

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GalinKa [24]

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3 years ago

Can someone do this fast ? I need it for tomorrow

gregori [183]

Answer:

Step-by-step explanation:

hg(x) means you multiply h(x) times g(x) and then we will set it equal to $3x^2+x-1$ .

hg(x) = $x^2(x+2)$ which simplifies to

$x^3+2x^2$ . Now set that equal to $3x^2+x-1$ :

$x^3+2x^2=3x^2+x-1$ and get everything on the same side and factor:

$x^3-x^2-x+1=0$ . Factor by grouping:

$(x^3-x^2)-(x+1)=0$ and factor out what's common in each set of ( ):

$x^2(x-1)-1(x-1)=0$ which factors out to

$(x^2-1)(x-1)=0$ . But $x^2-1$ factors somoe to:

(x + 1)(x - 1)(x - 1) = 0 So the solutions for this are

x = -1, 1

7 0

2 years ago

Evaluate the integral by interpreting it in terms of areas Draw a picture of the region the integral

denis23 [38]

Answer:

Step-by-step explanation:

The picture is below of how to separate this into 2 different regions, which you have to because it's not continuous over the whole function. It "breaks" at x = 2. So the way to separate this is to take the integral from x = 0 to x = 2 and then add it to the integral for x = 2 to x = 3. In order to integrate each one of those "parts" of that absolute value function we have to determine the equation for each line that makes up that part.

For the integral from [0, 2], the equation of the line is -3x + 6;

For the integral from [2, 3], the equation of the line is 3x - 6.

We integrate then:

$\int\limits^2_0 {-3x+6} \, dx+\int\limits^3_2 {3x-6} \, dx$ and

$-\frac{3x^2}{2}+6x\left \} {{2} \atop {0}} \right. +\frac{3x^2}{2}-6x\left \} {{3} \atop {2}} \right.$ sorry for the odd representation; that's as good as it gets here!