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Archy [21]
3 years ago
5

Plz help ASAP! I will mark brainliest if correct

Mathematics
2 answers:
MaRussiya [10]3 years ago
5 0
A = L^2
A = L^2 = 2^2 + 4^2 (Pythagorean’s theorem)
A = L^2 = 20
Therefore the area of the square is 20 units square.
Sergio [31]3 years ago
4 0

Answer:

The answer is 20

Step-by-step explanation:

To find the hypotenuse, you use the pythagorean theorem (A^{2}+ B^{2}= C^{2}).

We are given A and B so you insert.

2^{2}+ 4^{2} =C^{2}

4+16=C^{2}

C^{2}=20

Now you take the square root of C and since it is even, I will leave it in a square root sign.

C=\sqrt{20}

C is one side of a square and since all sides are the same, we don't have to find anymore. Now we insert C into the formula for finding the area of a square.

A=C^{2}

A=\sqrt{20} ^{2}

A=20

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Match the expressions with their equivalent simplified expressions.
Tasya [4]

Answer:

\sqrt[4]{\frac{16x^6y^4}{81x^2y^8}}\rightarrow\frac{2x}{3y}\\\sqrt[4]{\frac{81x^2y^{10}}{81x^6y^6}} \rightarrow\frac{3y}{2x}\\\sqrt[3]{\frac{64x^8y^7}{125x^2y^{10}}}\rightarrow\frac{4x^2}{5y}\\\sqrt[5]{\frac{243x^{17}y^{16}}{32x^7y^{21}}}\rightarrow\frac{3x^2}{2y}\\\sqrt[5]{\frac{32x^{12}y^{15}}{243x^7y^{10}}} \rightarrow\frac{2xy}{3}\\\sqrt[4]{\frac{16x^{10}y^{9}}{256x^2y^{17}}}\rightarrow\frac{x}{2y}


Step-by-step explanation:

\sqrt[4]{\frac{16x^6y^4}{81x^2y^8}} =\sqrt[4]{\frac{(2^4)(x^{6-2})(y^{4-8})}{(3^4)}} =\sqrt[4]{\frac{2^4x^4y^{-4}}{3^4}} =\frac{2xy^{-1}}{3}=\frac{2x}{3y}

\sqrt[4]{\frac{81x^2y^{10}}{81x^6y^6}} =\sqrt[4]{\frac{(3^4)(x^{2-6})(y^{10-6})}{(2^4)}} =\sqrt[4]{\frac{3^4x^{-4}y^{4}}{2^4}} =\frac{3x^{-1}y^1}{3}=\frac{3y}{2x}

\sqrt[3]{\frac{64x^8y^7}{125x^2y^{10}}} =\sqrt[3]{\frac{(4^3)(x^{8-2})(y^{7-10})}{(5^3)}} =\sqrt[3]{\frac{4^3x^6y^{-3}}{5^3}} =\frac{4x^2y^{-1}}{5}=\frac{4x^2}{5y}

\sqrt[5]{\frac{243x^{17}y^{16}}{32x^7y^{21}}} =\sqrt[5]{\frac{(3^5)(x^{17-7})(y^{16-21})}{(2^5)}} =\sqrt[5]{\frac{3^5x^{10}y^{-5}}{2^5}} =\frac{3x^2y^{-1}}{2}=\frac{3x^2}{2y}

\sqrt[5]{\frac{32x^{12}y^{15}}{243x^7y^{10}}} =\sqrt[5]{\frac{(2^5)(x^{12-7})(y^{15-10})}{(3^5)}} =\sqrt[5]{\frac{2^5x^{5}y^{5}}{3^5}} =\frac{2x^1y^{1}}{3}=\frac{2xy}{3}

\sqrt[4]{\frac{16x^{10}y^{9}}{256x^2y^{17}}} =\sqrt[4]{\frac{(2^4)(x^{10-2})(y^{9-17})}{(4^4)}} =\sqrt[4]{\frac{2^4x^{8}y^{-8}}{4^4}} =\frac{2x^{1}y^{-1}}{4}=\frac{x}{2y}

Thus,

\sqrt[4]{\frac{16x^6y^4}{81x^2y^8}}\rightarrow\frac{2x}{3y}\\\sqrt[4]{\frac{81x^2y^{10}}{81x^6y^6}} \rightarrow\frac{3y}{2x}\\\sqrt[3]{\frac{64x^8y^7}{125x^2y^{10}}}\rightarrow\frac{4x^2}{5y}\\\sqrt[5]{\frac{243x^{17}y^{16}}{32x^7y^{21}}}\rightarrow\frac{3x^2}{2y}\\\sqrt[5]{\frac{32x^{12}y^{15}}{243x^7y^{10}}} \rightarrow\frac{2xy}{3}\\\sqrt[4]{\frac{16x^{10}y^{9}}{256x^2y^{17}}}\rightarrow\frac{x}{2y}

3 0
3 years ago
Plss finish fast the question is linked thank you for ur help
kobusy [5.1K]

Answer: B. 2 2/5

Step-by-step explanation: Hope this help :D

3 0
3 years ago
Read 2 more answers
Which are right triangles that can be formed using a diagonal through the interior of the cube? Select all that apply.
photoshop1234 [79]

Answer:

triangle AEH, triangle CFG, triangle BFG, and triangle DEG.

so b,c,e,f

this is right for e2020.

8 0
3 years ago
Read 2 more answers
Blake and two friends equally shared the cups of water shown in the picture.
3241004551 [841]
The answer would be c- 2/6

there are six cups total and three (blake plus two others) friends

there are two ways you can solve this…

1. by making groups
if you look at the picture, you can split it into three groups because that is the number of people, then count how many cups there are in one group which will be your numerator for the denominator six which is the total

2. simple division
divide the total number of cups buy the number of people which is 3
you get two and follow the same process above

6 0
3 years ago
Find the domain and range of 1/t+2
kotegsom [21]

Answer:

Domain = (-∞, -2) ∪ (-2, ∞).

Range =  (-∞, 0) ∪ (0, ∞).

Step-by-step explanation:

t can't have the value -2 because that would make  1/(t + 2) = 1/0 which is undefined.

So the domain is All real values of t. except t = -2.

As t approaches infinity or negative infinity  f(t) approaches  zero, and as t approaches -2 from below or above f(t) approaches negative infinity or positive infinity.

8 0
3 years ago
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