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Lina20 [59]
3 years ago
9

Can someone please help me with this

Mathematics
1 answer:
Kazeer [188]3 years ago
5 0

Answer:

Step-by-step explanation:

Its  A

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(01.03 MC)<br> Which of the following is a step in simplifying the expression<br> (Xy^4/x^-5y^5)^3
Sati [7]

\displaystyle\tt\left(\frac{xy^4}{x^{-5}y^5}\right)^{-3} =\frac{x^{-3}y^{4\cdot(-3)}}{x^{-5\cdot(-3)}y^{5\cdot(-3)}}=\boxed{\tt\frac{x^{-3}y^{-12}}{x^{15}y^{-15}}}

4 0
3 years ago
13 + 31 = 24 not 44 how on earth is this possible all help is great. Thanks.
Aneli [31]
It is 44. Can't argue about that
3 0
3 years ago
when directed to solve a quadratic equation by completing the square, Sam arrived at the equation (x-(5/2))^2= (13/4). which equ
Tasya [4]
We have the following expression:
 (x- (5/2)) ^ 2 = (13/4)
 Let's rewrite the given expression:
 x ^ 2 - 5x + 25/4 = (13/4)
 x ^ 2 - 5x + 25/4 - 13/4 = 0
 x ^ 2 - 5x + 12/4 = 0
 x ^ 2 - 5x + 3 = 0
 Answer:
 
the original equation given to Sam could have been:
 
x ^ 2 - 5x + 3 = 0
7 0
3 years ago
Help please! I’m giving brainliest to however gets this question right.
Arisa [49]

Answer:

They both have area 4

Step-by-step explanation:

Area of the square:

base*height = 2 * 2 = 4

Area of the triangle:

Using the left-side of the triangle as the base, and the height from the left-side to the bottom-right corner:

\frac{1}{2}base*height

\frac{1}{2}4\sqrt2*\sqrt2=4

We know the length of the diagonal is \sqrt2 as we are using a centimetre grid, so we can create an isosceles triangle with side lengths 1 and our unknown length, we can then use Pythagorean Theorem to work out our unknown side length.

a^2+b^2=c^2

1^2+1^2=c^2

2=c^2

\sqrt2=c

7 0
3 years ago
I need the answer or thos problem <br> -n+(-4)-(-4n)+6
cluponka [151]

−n−4−−4n+6

=−n+−4+4n+6

Combine Like Terms:

=−n+−4+4n+6

=(−n+4n)+(−4+6)

=3n+2

4 0
4 years ago
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