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2 years ago

Can someone help me out with these two problems and show work please !!

Mathematics

1 answer:

lara [203]2 years ago

8 0

Answer :

Step-by-step explanation:

(6 y^4+ 3 y^2-7) - (12 y^4 - y^2 + 5 )

=6 y^4 +3 y ^2 - 7 - 12 y^4 + y^2 - 5

=6 y ^4 - 12 y ^4 +3 y ^2 + y ^2 - 7 -5

= -6 y ^4 + 4 y ^2- 12

Hope this helps!

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In square $ABCD$, $E$ is the midpoint of $\overline{BC}$, and $F$ is the midpoint of $\overline{CD}$. Let $G$ be the intersectio

Vinvika [58]

Since ABCD is a square
BE = FD
BC = AB
Angle ABE = angle BCF
With these, triangle ABE = triangle BCF
And
angle BEA = angle CFB by CPCTC
And with this
DG = AB

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3 years ago

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gtnhenbr [62]

Step-by-step explanation:

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3 years ago

CAN SOMEONE PLEASE HELP, I DON'T WANT TO FAIL

larisa [96]

Answer:

Step-by-step explanation: Linda already has two bags of popcorn. She wants to buy x amount of popcorn but each one costs three dollars.

I hope this helps somewhat?

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3 years ago

Suppose that f is an odd function of x. Does knowing that modifyingbelow lim with x right arrow 0 superscript plus f left parent

algol13

Answer: we should have that:

$\lim_{x \to \--0} f(x) = -3$

Step-by-step explanation:

We know that f(x) is an odd function, this means that f(-x) = -f(x)

We know that:

$\lim_{x \to \++0} f(x) = 3$

this means that wen we aproximate to zero for the right (the positive side) we have that the value is.

First, this tell us that f(x) can not be a continue function, because of the fact that is odd we will have that when we aproximate the same lim but from the negative side, we will have that:

$\lim_{x \to \--0} f(x) = -3$

7 0

3 years ago

Read 2 more answers

3x+4y=-23 <br> 2y-x=-19 <br> What is the solution (x,y) to the systems of equations above?

Shkiper50 [21]

If we multiply the bottom equation by 2 and move x to the right, it becomes:

4y = 2x-38

Now we can substitute it for the 4y in the top equation:

3x + (2x-38) = -23 => 5x = -23+38 => 5x = 15 => x=3

Then 4y = 2*3-38 => y = -8

So the solution is (3,-8)

3 0

4 years ago