Answer:
Rounding to nearest hundredths gives us r=0.06.
So r is about 6%.
Step-by-step explanation:
So we are given:

where


.


Divide both sides by 1600:

Simplify:

Take the 6th root of both sides:
![\sqrt[6]{\frac{23}{16}}=1+r](https://tex.z-dn.net/?f=%5Csqrt%5B6%5D%7B%5Cfrac%7B23%7D%7B16%7D%7D%3D1%2Br)
Subtract 1 on both sides:
![\sqrt[6]{\frac{23}{16}}-1=r](https://tex.z-dn.net/?f=%5Csqrt%5B6%5D%7B%5Cfrac%7B23%7D%7B16%7D%7D-1%3Dr)
So the exact solution is ![r=\sqrt[6]{\frac{23}{16}}-1](https://tex.z-dn.net/?f=r%3D%5Csqrt%5B6%5D%7B%5Cfrac%7B23%7D%7B16%7D%7D-1)
Most likely we are asked to round to a certain place value.
I'm going to put my value for r into my calculator.
r=0.062350864
Rounding to nearest hundredths gives us r=0.06.
Answer:
5 points
Step-by-step explanation:
Alice, Bianca, Camila, and Dasha each scored 14 points. 14 x 4 = 56
56 + 15 = 71
3 x 5 = 15
The three remaining girls scored 5 points each.
Answer:
The attachment shows ΔBAC ~ ΔBDA
Step-by-step explanation:
You want segment AB to be part of two similar, but not congruent, triangles. One way to do that is to make AB the hypotenuse of one triangle and the leg of another.
It is convenient to construct these triangles using point M as the arbitrary midpoint of the hypotenuse of the larger triangle. (We don't know the coordinates of M—we just know it is on the perpendicular bisector of AB.) BC is a diameter of circle M, and AD is the altitude of ΔABC.