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ankoles [38]
3 years ago
13

This is for geometry and i keep getting it wrong​

Mathematics
2 answers:
Hoochie [10]3 years ago
8 0
Answer: I believe it is 23
vova2212 [387]3 years ago
7 0

Step-by-step explanation:

As < A and < B are vertical angles so

<A = < B

5x + 12 = 6x - 11

6x - 5x = 12 + 11

x = 23

Hope it will help :)❤

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The amount Denise charges to repair computers $50 an hour plus a $25 service fee write an expression to show how much she will c
Natalija [7]
The answer you are looking for is 50h+25
Hope that this helps!

4 0
3 years ago
Read 2 more answers
The area of a rectangle is 63m^2 and the length of the rectangle is 11m more than twice the width. Find the length and width.
AleksandrR [38]

Answer:

width = 3.5 m

length = 18 m

Step-by-step explanation:

area = length x width

x is the width then length = 2x + 11

x(2x + 11) = 63

2x^2 + 11x - 63 = 0

The formula to solve a quadratic equation of the form ax^2 + bx + c = 0 is equal to x = [-b +/-√(b^2 - 4ac)]/2a

with a = 2

       b = 11

       c = -63

substitute in the formula

x = [-11 +/- √(11^2 - 4(2)(-63))]2(2)

x = [-11 +/- √(121 + 504)]/4

x = [-11 +/- √625]/4

x = (-11 +/- 25)/4

x1 = (-11 + 25)/4 = 14/4 = 3.5

x2 = (-11 - 25)/4 = -36/4 = -9

since width can't be a negative number, x is 3.5

length = 2x + 11 = 2(3.5) + 11 = 18

7 0
2 years ago
Can you solve it, please
Nimfa-mama [501]
T
T
F
F
Yayyyy that’s the answer
5 0
3 years ago
Read 2 more answers
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
Can someone please help me with #12?
Genrish500 [490]

4x+12=90

4x=90-12

4x=8

x=8/4

x=2

5 0
3 years ago
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