The answer is 2.22 if you divide it. Add the ingredients then divide on how many servings.
Answer:
1.5 ≤ 3.75x
Step-by-step explanation:
Given that:
Distance, D to jog is atleast 1.5 miles
D ≥ 1.5 miles
x = number of hours to jog
Jog rate = 3.75 mph
Actual distance covered if jog last for X hours
Actual Distance = speed * time
Actual Distance = 3.75 * x = 3.75x
Hence,
D ≤ Actual distance
1.5 ≤ 3.75x
Answer:
2.55
Step-by-step explanation:
To find the radius, the formula is:
R = C/2 x pi
pi = 3.14
16/(2 x 3.14) = 2.54648
= 2.55
Answer:
Option d) 5 to the power of negative 5 over 6 is correct.
![\dfrac{\sqrt[3]{\bf 5} \times \sqrt{\bf 5}}{\sqrt[3]{\bf 5^{\bf 5}}}= 5^{\frac{\bf -5}{\bf 6}}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%5B3%5D%7B%5Cbf%205%7D%20%5Ctimes%20%5Csqrt%7B%5Cbf%205%7D%7D%7B%5Csqrt%5B3%5D%7B%5Cbf%205%5E%7B%5Cbf%205%7D%7D%7D%3D%205%5E%7B%5Cfrac%7B%5Cbf%20-5%7D%7B%5Cbf%206%7D%7D)
Above equation can be written as 5 to the power of negative 5 over 6.
ie, 
Step-by-step explanation:
Given that cube root of 5 multiplied by square root of 5 over cube root of 5 to the power of 5.
It can be written as below
![\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%5B3%5D%7B5%7D%20%5Ctimes%20%5Csqrt%7B5%7D%7D%7B%5Csqrt%5B3%5D%7B5%5E5%7D%7D)
![\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{1}{3}} \times 5^{\frac{1}{2}}}{5^{\frac{5}{3}}}](https://tex.z-dn.net/?f=%20%5Cdfrac%7B%5Csqrt%5B3%5D%7B5%7D%20%5Ctimes%20%5Csqrt%7B5%7D%7D%7B%5Csqrt%5B3%5D%7B5%5E5%7D%7D%3D%20%5Cdfrac%7B5%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%20%5Ctimes%205%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%7D%7B5%5E%7B%5Cfrac%7B5%7D%7B3%7D%7D%7D)
![\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{1}{3}+\frac{1}{2}}}{5^{\frac{5}{3}}}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%5B3%5D%7B5%7D%20%5Ctimes%20%5Csqrt%7B5%7D%7D%7B%5Csqrt%5B3%5D%7B5%5E5%7D%7D%3D%20%5Cdfrac%7B5%5E%7B%5Cfrac%7B1%7D%7B3%7D%2B%5Cfrac%7B1%7D%7B2%7D%7D%7D%7B5%5E%7B%5Cfrac%7B5%7D%7B3%7D%7D%7D)
![\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{2+3}{6}}}{5^{\frac{5}{3}}}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%5B3%5D%7B5%7D%20%5Ctimes%20%5Csqrt%7B5%7D%7D%7B%5Csqrt%5B3%5D%7B5%5E5%7D%7D%3D%20%5Cdfrac%7B5%5E%7B%5Cfrac%7B2%2B3%7D%7B6%7D%7D%7D%7B5%5E%7B%5Cfrac%7B5%7D%7B3%7D%7D%7D)
![\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= 5^{\frac{5}{6}} \times 5^{\frac{-5}{3}}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%5B3%5D%7B5%7D%20%5Ctimes%20%5Csqrt%7B5%7D%7D%7B%5Csqrt%5B3%5D%7B5%5E5%7D%7D%3D%205%5E%7B%5Cfrac%7B5%7D%7B6%7D%7D%20%5Ctimes%205%5E%7B%5Cfrac%7B-5%7D%7B3%7D%7D)
![\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= 5^{\frac{5-10}{6}}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%5B3%5D%7B5%7D%20%5Ctimes%20%5Csqrt%7B5%7D%7D%7B%5Csqrt%5B3%5D%7B5%5E5%7D%7D%3D%205%5E%7B%5Cfrac%7B5-10%7D%7B6%7D%7D)
![\dfrac{\sqrt[3]{5} \times \sqrt{5}}{5^5}= 5^{\frac{-5}{6}}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%5B3%5D%7B5%7D%20%5Ctimes%20%5Csqrt%7B5%7D%7D%7B5%5E5%7D%3D%205%5E%7B%5Cfrac%7B-5%7D%7B6%7D%7D)
Above equation can be written as 5 to the power of negative 5 over 6.
Answer:
A vector's magnitude represents its length, so your answer is C, the length of a vector.
Step-by-step explanation:
