Answer:
well, there are 52 weeks in a year, so if you just multiply the pounds of cans by the amount of years, you get 1,300
Step-by-step explanation:
*sigh*
Answer:
3 3/8
Step-by-step explanation:
6 minus 2 5/8 equals 3 3/8
3 3/8 plus 2 5/8 equals 6
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.
<h3>2
Answers: Choice C and choice D</h3>
y = csc(x) and y = sec(x)
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Explanation:
The term "zeroes" in this case is the same as "roots" and "x intercepts". Any root is of the form (k, 0), where k is some real number. A root always occurs when y = 0.
Use GeoGebra, Desmos, or any graphing tool you prefer. If you graphed y = cos(x), you'll see that the curve crosses the x axis infinitely many times. Therefore, it has infinitely many roots. We can cross choice A off the list.
The same applies to...
- y = cot(x)
- y = sin(x)
- y = tan(x)
So we can rule out choices B, E and F.
Only choice C and D have graphs that do not have any x intercepts at all.
------------
If you're curious why csc doesn't have any roots, consider the fact that
csc(x) = 1/sin(x)
and ask yourself "when is that fraction equal to zero?". The answer is "never" because the numerator is always 1, and the denominator cannot be zero. If the denominator were zero, then we'd have a division by zero error. So that's why csc(x) can't ever be zero. The same applies to sec(x) as well.
sec(x) = 1/cos(x)
