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3 years ago

Triangle A'B'C' is a translation of Triangle ABC. What segment length is equal in

Mathematics

1 answer:

snow_tiger [21]3 years ago

7 0

It’s a’b Fosho Fosho I just got done with the test

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In circle L, points E and F lie on the circle such that E, F and L are not collinear. If LE, LF and EF are drawn then which of t

frozen [14]

Answer:

<u>option (2) it is and isosceles triangle. </u>

Step-by-step explanation:

In circle L, points E and F lie on the circle such that E, F and L are not collinear

If LE, LF and EF are drawn

So, LE = EF = the radius of the circle L

So, LEF is an isosceles triangle.

The answer is option (2) it is and isosceles triangle.

8 0

3 years ago

6x2y^3/3xy Use Laws of exponents. answer is 2xy^2 but I don't know how to get it. My initial answer was 4y^2

allsm [11]

Step-by-step explanation:

$\frac{6x^{2}y^{3}}{3xy}$

If it helps, write as a product of fractions, grouped by variable.

$\frac{6}{3} \frac{x^{2}}{x} \frac{y^{3}}{y}$

Now reduce using exponent rules.

$2xy^{2}$

4 0

3 years ago

What is the area of the given circle?

zhenek [66]

Answer:

50.26 is the area.

Hope it helps!!!

5 0

3 years ago

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,

3 0

3 years ago

Max got another paycheck for $476.15 on April 20th.

Rufina [12.5K]

Answer:

$182.60

Step-by-step explanation:

4 0

3 years ago