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:
Ham costs: $8.249 repeating
Overall cost: 1.26 repeating
Step-by-step explanation:
$4.95 x 5/3= 8.249
2/3 + 3/5= (2 x 5) + (3 x 3)/ 3 x 5=
19/ 15 or 1 4/15
15 divided by 19= 1.26 repeating
<h3><u>4(2x^2 + 3x + 12)</u></h3>
We can factor a 4 out of this expression.
4(2x^2 + 3x + 12)
Now that the expression is simplified, we can see if splitting the middle term is possible.
<em><u>Display factors of 24.</u></em>
1 * 24
-1 * - 24
2 * 12
-2 * -12
3 * 8
-3 * - 8
4 * 6
-4 * -6
All factors of 24 have been listed, and none of these digits satisfy the criteria.
Trinomial cannot be further factored.
Answer:
The difference in slopes of
is = 3
We can say slope of
is positive and 3 more than slope of
while slope of
is negative.
Difference of y-intercepts of
is = -7
We can say the y-intercept of
is positive and 7 units above
while y-intercept of
is negative.
Step-by-step explanation:
Given equation:


We need to find the difference of slopes and y-intercepts of the given equations.
The standard form of a slope intercept equation of line is given by:

where
represents slope and
represents y-intercept of line.
Writing the given equations in standard form to find slope and y-intercept.

Slope = 2 and y-intercept =-2

Slope = -1 and y-intercept =5
The difference in slopes of
is = 
We can say slope of
is positive and 3 more than slope of
while slope of
is negative.
Difference of y-intercepts of
is = 
We can say the y-intercept of
is positive and 7 units above
while y-intercept of
is negative.
The distance between Branch A and Branch B is given by the Pythagoras theorem
AB² = (Vertical Distance)² + (Horizontal Distance)²
AB² = (4-1)² + (1 - -3)²
AB² = 3² + 4²
AB² = 9 + 16
AB² = 25
AB = 5
BC² = (Vertical distance)² + (Horizontal distance)²
BC² = (-2 - 4)² + (5-1)²
BC² = (-6)² + (4)²
BC² = 36 + 16
BC² = 52
BC = 7.21
Half way of BC = 7.21 ÷ 2 = 3.6 miles
Total distance travelled from A to B and then halfway from B to C is 3.6+5 = 8.6 miles