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Paraphin [41]
3 years ago
6

3x^2 - x - 10

Mathematics
2 answers:
kramer3 years ago
7 0
One way is to factor it
trial and error, with experience
(3x+5)(x-2)

answer is C
bearhunter [10]3 years ago
6 0
Your answer is (3x+5 ) for one factor and the other is (x-2 )
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Graph the solution set 2(x-y)>6-y
Greeley [361]

Answer:

The area right side to the line

Step-by-step explanation:

2 (X-Y) > 6-Y

2X - 2Y > 6 - Y

2X -6 > Y

Y < 2X - 6

8 0
3 years ago
C an some one help me<br>​
Pavlova-9 [17]

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b

Step-by-step explanation:

8 0
3 years ago
A line parallel to a triangle's side splits AB into lengths of x - 5 and x + 2. The other side, AC, is split into lengths of x a
lorasvet [3.4K]
36 because the 36 is the answer
3 0
3 years ago
Read 2 more answers
Figure ABCD is a parallelogram. If point C lies on the line y= -1, what is the x-value or point C?
MrRa [10]

Answer:

The correct answer is option D.  4

Step-by-step explanation:

Points to remember

Opposite sides of the parallelogram are equal.

From the figure we can see that,AB and CD are opposite sides.

Therefore AB = CD

<u>To find the length of AB</u>

From the figure we get, AB = 4 units

Therefore CD = 4 units

<u>To find the x coordinate of point C</u>

y coordinate is -1

x coordinate is 4 units from point D

x coordinate of D is 0, therefore x coordinate of C = 0 + 4 = 4

The correct answer is option D. 4

4 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
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