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2 years ago

ASAP PLEASE HELP I BEEN WORKING ON IT FOR 5 HOURS

2 answers:

7 0

So in this case, the solution is where the graphs intersect. Put the equations in y-intercept form.
-4x+y=-6
y=4x-6

8x-2y=14
-2y=14-8x
y=4x-7

These lines have the same slope, which is 4. Therefore, they are parallel. Parallel lines never intersect, so there is NO SOLUTION.

babymother [125]2 years ago

7 0

Answer:

there is no solution

Step-by-step explanation:

i just did the test and it was no solution...and btw i think u meant -3/2

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Suppose that y varies directly with x and y=12 when x=-2. What is x when y=-6<br> Please help me!!

hodyreva [135]

Answer:

x = 1 because when you multiply negative 2 by negative six you get twelve. when you multiply positive one by negative six you get negative six.

8 0

3 years ago

Help plzzzzzz plz help this ? is realy hard Simplify the expression 7(a−3+4b).

velikii [3]

7(a−3+4b)

distribute

7a-7*3+7*4b

7a-21+28b

4 0

2 years ago

Read 2 more answers

Whats the intrgral of <img src="https://tex.z-dn.net/?f=%20%5Cint%20%20%5Cfrac%7Bx%5E2%2Bx-3%7D%7B%28x%5E3%2Bx%5E2-4x-4%29%5E2%7

rosijanka [135]

$\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx$

Notice that $x^3+x^2-4x-4=x^2(x+1)-4(x+1)=(x-2)(x+2)(x+1)$ . Decompose the integrand into partial fractions:

$\dfrac{x^2+x-3}{(x-2)^2(x+2)^2(x+1)^2}$
$=\dfrac1{3(x+1)}-\dfrac{11}{32(x+2)}-\dfrac1{3(x+1)^2}-\dfrac1{16(x+2)^2}+\dfrac1{96(x-2)}+\dfrac1{48(x-2)^2}$

Integrating term-by-term, you get

$\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx$
$=-\dfrac1{48(x-2)}+\dfrac1{3(x+1)}+\dfrac1{16(x+2)}+\dfrac1{96}\ln|x-2|+\dfrac13\ln|x+1|-\dfrac{11}{32}\ln|x+2|+C$

5 0

3 years ago

3/2 dived by 7/8 dived by 1/2

Leviafan [203]

Answer:

3.428571429 or approximately 3.43

Step-by-step explanation:

3/2

= 1.5

7/8

= 0.875

1.5/0.875

= 1.714285714

1.714285714/0.5

= 3.428571429

or approximately 3.43

6 0

3 years ago

Which function represents the area of the triangle h(x)=1/2f(x)g(x)

tresset_1 [31]

Answer:

h=1/2fg

Step-by-step explanation:

Solve for x, h=1/2fg

It is true for all x; h=1/2fg

h=1/2fg

Both sides are equal

It is true for all x; h=1/2fg

6 0

2 years ago