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n200080 [17]
4 years ago
14

Find the other endpoint of the line segment with the given endpoint and midpoint. Endpoint: (6,-8) midpoint: (-1,7)

Mathematics
2 answers:
zaharov [31]4 years ago
6 0
Hope this helps. Look at attachment

sertanlavr [38]4 years ago
4 0
Midpoint formula : (x1+x2)/2 ; (y1+y2)/2
You have x1,y1 and the mid point
Where x1=6
Y1=-8
Midpoint (-1,7)

[(6+x2)/2]=-1
6+x2=-1*2
6+x2=-2
X2=(-2)-6
X2=-8

[(-8+y2)/2]=7
-8+y2=7*2
-8+y2=14
Y2=14+8
Y2=22


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Select the correct answer. Which expression is equivalent to 8x^2^3 sqrt 375x + 2^3 sqrt 3x^7, if x=0?
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Answer:

(8x^3)^ \frac{2}{3} \sqrt{75x^3}  + 2^3 \sqrt{ 3x^7} =28x^3\sqrt{3x}

Step-by-step explanation:

The question is poorly formatted. The original question is:

(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7

We have:

(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7

Open bracket

(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =(8^ \frac{2}{3} *x^{3* \frac{2}{3}}) \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}

(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =(8^ \frac{2}{3} *x^2) \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}

Express 8 as 2^3

(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =(2^{3* \frac{2}{3}} *x^2) \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}

(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =(2^2 *x^2) \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}

(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =4x^2 \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}

Express 2^3 as 8

(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}=4x^2 \sqrt{75x^3} + 8\sqrt{ 3x^7}

Expand each exponent

(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}=4x^2 \sqrt{25x^2 *3x} + 8\sqrt{ 3x * x^6}

Split

(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}=4x^2 \sqrt{25x^2} *\sqrt{3x} + 8\sqrt{3x} * \sqrt{x^6}

(8x^3)^ \frac{2}{3} \sqrt{75x^3}  + 2^3 \sqrt{ 3x^7} =4x^2 *5x *\sqrt{3x} + 8\sqrt{3x} * x^3

(8x^3)^ \frac{2}{3} \sqrt{75x^3}  + 2^3 \sqrt{ 3x^7} =20x^3\sqrt{3x} + 8x^3\sqrt{3x}

Factorize

(8x^3)^ \frac{2}{3} \sqrt{75x^3}  + 2^3 \sqrt{ 3x^7} =28x^3\sqrt{3x}

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