The correct answer would be D, because there’s no other leading common factors between the two besides one.
Answer:
![\boxed{5 \cdot \sqrt{2} \cdot \sqrt[6]{5} }](https://tex.z-dn.net/?f=%5Cboxed%7B5%20%5Ccdot%20%5Csqrt%7B2%7D%20%20%5Ccdot%20%5Csqrt%5B6%5D%7B5%7D%20%7D)
Step-by-step explanation:
![\sqrt[3]{250} \cdot \sqrt{\sqrt[3]{10} }](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B250%7D%20%5Ccdot%20%5Csqrt%7B%5Csqrt%5B3%5D%7B10%7D%20%7D)
![\sqrt{\sqrt[3]{10} } \implies (10^\frac{1}{3} )^\frac{1}{2} =10^\frac{1}{6} =\sqrt[6]{10}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Csqrt%5B3%5D%7B10%7D%20%7D%20%5Cimplies%20%2810%5E%5Cfrac%7B1%7D%7B3%7D%20%29%5E%5Cfrac%7B1%7D%7B2%7D%20%3D10%5E%5Cfrac%7B1%7D%7B6%7D%20%3D%5Csqrt%5B6%5D%7B10%7D)
![\therefore \sqrt{\sqrt[3]{10} }=\sqrt[6]{10}](https://tex.z-dn.net/?f=%5Ctherefore%20%5Csqrt%7B%5Csqrt%5B3%5D%7B10%7D%20%7D%3D%5Csqrt%5B6%5D%7B10%7D)
![\text{Solving }\sqrt[3]{250} \cdot \sqrt{\sqrt[3]{10} }](https://tex.z-dn.net/?f=%5Ctext%7BSolving%20%7D%5Csqrt%5B3%5D%7B250%7D%20%5Ccdot%20%5Csqrt%7B%5Csqrt%5B3%5D%7B10%7D%20%7D)

![\sqrt[3]{250}=\sqrt[3]{2\cdot 5^3}=5 \sqrt[3]{2}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B250%7D%3D%5Csqrt%5B3%5D%7B2%5Ccdot%205%5E3%7D%3D5%20%20%5Csqrt%5B3%5D%7B2%7D)
Once
![\sqrt[6]{2} \cdot \sqrt[6]{5} = \sqrt[6]{10}](https://tex.z-dn.net/?f=%5Csqrt%5B6%5D%7B2%7D%20%20%5Ccdot%20%5Csqrt%5B6%5D%7B5%7D%20%3D%20%5Csqrt%5B6%5D%7B10%7D)
We have
![5 \sqrt[3]{2} \cdot \sqrt[6]{2} \cdot \sqrt[6]{5}](https://tex.z-dn.net/?f=5%20%20%5Csqrt%5B3%5D%7B2%7D%20%5Ccdot%20%5Csqrt%5B6%5D%7B2%7D%20%20%5Ccdot%20%5Csqrt%5B6%5D%7B5%7D)
We can proceed considering the common base of exponentials
![\sqrt[3]{2} \cdot \sqrt[6]{2} = 2^{\frac{1}{3}} \cdot 2^{\frac{1}{6} } = 2^{\frac{3}{6} } = 2^{\frac{1}{2} }=\sqrt{2}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B2%7D%20%20%5Ccdot%20%5Csqrt%5B6%5D%7B2%7D%20%20%3D%20%202%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%20%5Ccdot%20%202%5E%7B%5Cfrac%7B1%7D%7B6%7D%20%7D%20%20%3D%202%5E%7B%5Cfrac%7B3%7D%7B6%7D%20%7D%20%3D%202%5E%7B%5Cfrac%7B1%7D%7B2%7D%20%7D%3D%5Csqrt%7B2%7D)
Therefore,
![5 \sqrt[3]{2} \cdot \sqrt[6]{2} \cdot \sqrt[6]{5} = 5 \cdot \sqrt{2} \cdot \sqrt[6]{5}](https://tex.z-dn.net/?f=5%20%20%5Csqrt%5B3%5D%7B2%7D%20%5Ccdot%20%5Csqrt%5B6%5D%7B2%7D%20%20%5Ccdot%20%5Csqrt%5B6%5D%7B5%7D%20%3D%205%20%5Ccdot%20%5Csqrt%7B2%7D%20%20%5Ccdot%20%5Csqrt%5B6%5D%7B5%7D)
Answer:
.18
Step-by-step explanation:
if ^2 is an exponent
(-16t*-16t)+8t+48
(258t+8t)+48
266t+48
48/266
.180
or .18
The student would earn %86. To calculate percentage you take the amount of questions correct, divide that by the amount of questions given, multiply by 100 and round to the nearest percent.
The answer is: [B]: " x + 3y + 10 = 0 " .
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Explanation:
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Note the equation for a line; in 'slope-intercept form': "y = mx + b" ;
in which "y" is isolated alone, as a single variable, on the left-hand side of the equation;
"m" = the slope of the line; and is the co-efficient of "x" ;
b = the "y-intercept"; (or the "y-coordinate of the point of the "y-intercept").
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So, given the information in this very question/problem:
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slope = m = (-1/3) ;
b = y-intercept = (10/3) ;
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And we can write the equation of the line; in "slope-intercept form"; that is:
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" y = mx + b " ; as:
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" y = (-1/3)x + (10/3) " ;
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Now, the problem asks for the equation of this line; in "general form"; or "standard format"; which is:
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"Ax + By + C = 0 " ;
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So; given:
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" y = (-1/3)x + (10/3) " ;
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We can multiply the ENTIRE EQUATION (both sides) by "3" ; to get rid of the "fractions" ;
→ 3* { y = (-1/3)x + (10/3) } ;
→ 3y = -1x + 10 ;
↔ -1x + 10 = 3y ;
Subtract "(3y)" from each side of the equation:
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-1x + 10 − 3y = 3y − 3y ;
to get:
-1x + 10 − 3y = 0 ;
↔ -1x − 3y − 10 = 0 ;
→ This is not one of the "3 (THREE) answer choices given" ;
→ So, multiply the ENTIRE EQUATION (both sides); by "-1" ; as follows:
-1 * {-1x − 3y − 10 = 0} ;
to get:
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→ " x + 3y + 10 = 0 " ; which is: "Answer choice: [B] ."
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Note that is equation is in the "standard format" ;
→ " Ax + By + C = 0 " .
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