Identify cosQ as a fraction and as a decimal rounded to the nearest hundredth. HELP ASAP! PLEASE!!
1 answer:
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One way is to factor and group and get every 3
729=3 times 3 times 3 times 3 times 3 times 3
so we group the ones that happen 3 times
729=(3*3*3) times (3*3*3)
we know that we can take the cube root of each group and multiply the result
729=![( \sqrt[3]{3*3*3})( \sqrt[3]{3*3*3})](https://tex.z-dn.net/?f=%28%20%5Csqrt%5B3%5D%7B3%2A3%2A3%7D%29%28%20%5Csqrt%5B3%5D%7B3%2A3%2A3%7D%29)
=(3)(3)=9
the answer is 9
729=3 times 3 times 3 times 3 times 3 times 3
so we group the ones that happen 3 times
729=(3*3*3) times (3*3*3)
we know that we can take the cube root of each group and multiply the result
729=
the answer is 9
Midpoint Formula: (
,
)
Midpoint = (
,
)
Midpoint = (
,
)
Midpoint = (, -4)
So, the midpoint of the endpoints (-4,-3) and (7,-5) is (, -4).
Have a nice day! ♪
Midpoint = (
Midpoint = (
Midpoint = (
So, the midpoint of the endpoints (-4,-3) and (7,-5) is (
Have a nice day! ♪