To draw a locus point you have to draw<span> the half way line through the two </span>points. You can do this by drawing<span> the line bisector through AB points. Just open your compass to ¾ of the distance AB and then </span>draw<span> two arcs from both sides of your line. Once this is done </span>draw the line passing through the two intersection points<span>.
Hope this helps :)</span>
Answer:
Step-by-step explanation:
<u> </u>
3.4 or 3.4 repeating (3.4444444444444444444444∞)
and
3.44 rounded to the hundredth
Answer:
4x^5\sqrt[3]{3x}
Step-by-step explanation:
Given
![\sqrt[3]{16x^7}\left(\sqrt[3]{12x^9}\right)\\\sqrt[3]{4^{2} x^7}\sqrt[3]{4(3)x^9}\\ 4^\frac{2}{3} . x^\frac{7}{3} . 4^\frac{1}{3} . 3^\frac{1}{3} . x^\frac{9}{3}\\ 4^\frac{2+1}{3} . 3^\frac{1}{3}. x^\frac{9+7}{3} \\4^\frac{3}{3} . 3^\frac{1}{3}. x^\frac{16}{3}\\ 4\sqrt[3]{3x^16}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B16x%5E7%7D%5Cleft%28%5Csqrt%5B3%5D%7B12x%5E9%7D%5Cright%29%5C%5C%5Csqrt%5B3%5D%7B4%5E%7B2%7D%20x%5E7%7D%5Csqrt%5B3%5D%7B4%283%29x%5E9%7D%5C%5C%20%204%5E%5Cfrac%7B2%7D%7B3%7D%20.%20x%5E%5Cfrac%7B7%7D%7B3%7D%20.%204%5E%5Cfrac%7B1%7D%7B3%7D%20.%203%5E%5Cfrac%7B1%7D%7B3%7D%20.%20x%5E%5Cfrac%7B9%7D%7B3%7D%5C%5C%20%204%5E%5Cfrac%7B2%2B1%7D%7B3%7D%20.%20%203%5E%5Cfrac%7B1%7D%7B3%7D.%20x%5E%5Cfrac%7B9%2B7%7D%7B3%7D%20%5C%5C4%5E%5Cfrac%7B3%7D%7B3%7D%20.%203%5E%5Cfrac%7B1%7D%7B3%7D.%20x%5E%5Cfrac%7B16%7D%7B3%7D%5C%5C%204%5Csqrt%5B3%5D%7B3x%5E16%7D)
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I think the answer is c. hope that helps
There are 5 pennies in a nickel, and 10 pennies in a dime. Now in $3.90 there are 390 pennies.
n = number of nickels in the pile
d = number of dimes in the pile
we know the pile has 55 coins, therefore whatever "n" and "d" are, we know that
n + d = 55.
now, "n" is how many nickel coins are there, how many pennies is it in total? well, 5 pennies to a nickel, for "n" nickels, will then be 5(n) or
5n.
and "d" is how many dime coins are there, so therefore in "d" coins, there are 10(d) pennies, or
10d.
we also know the total amount of pennies is 390, therefore
5n + 10d = 390.

how many nickel coins are there anyway? well, n = 55 - d.