let's first off take a look at the <u>tickmarks</u>, three <u>side tickmarks</u>, so all those 3 sides are equal, all have a length of y - 25, so is an equilateral triangle.
there are two <u>angle tickmarks</u>, meaning those two angles are equal, wait a second! if those two angles are equal, that means is an isosceles triangle.
now, in an equilateral triangle, all sides are equal, but also all angles are equal, since the sum of all interior angles is 180°, then each angle is really 60°.
let's notice that angle on the upper-left-corner, is a right-angle, but 60° are on the equilateral triangle, and so the remaining 30° must be on the isosceles triangle.
the isosceles triangle has then a vertex of 30°, and twin angles, the twin angles let's say are each a° so then
30° + a° + a° = 180°
30 + 2a = 180
2a = 150
a = 75° = y
now, let's recall, the isosceles triangle has twin angles but it also has twin sides, so the side "x" and the side with the tickmark are equal.
well, we know that y = 75, so the sides with the tickmark are then (75) - 25 = 50 = x.
About 8,456.17849
Rounded: 8,456.18
Answer:
Exact form: -7/10
Decimal form: -0.7
Step-by-step explanation:
Isolate the variable by dividing each side by by factors that don't contain the variable.
Answer:
A.This is the number of combinations of 6 from 15
= 15C6
= 15! / (15-6)! 6!
= 5,005 ways.
B. This is the number of permutaions of 6 from 15:
= 15! / (15-6)!
= 3,603,600 ways.
Step-by-step explanation:
This was someone else's work not mine sorry here's creditssss :))
brainly.com/question/15145413
![\bf \begin{cases} f(x)=\sqrt[3]{7x-2}\\\\ g(x)=\cfrac{x^3+2}{7} \end{cases}\\\\ -----------------------------\\\\ now \\\\ f[\ g(x)\ ]\implies f\left[ \frac{x^3+2}{7} \right]\implies \sqrt[3]{7\left[ \frac{x^3+2}{7} \right]-2}\implies \sqrt[3]{x^3+2-2} \\\\\\ \sqrt[3]{x^3}\implies x\\\\ -----------------------------\\\\ or \\\\ g[\ f(x)\ ]\implies g\left[\sqrt[3]{7x-2}\right]\implies \cfrac{\left[\sqrt[3]{7x-2}\right]^3+2}{7} \\\\\\ \cfrac{7x-2+2}{7}\implies \cfrac{7x}{7}\implies x](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%0Af%28x%29%3D%5Csqrt%5B3%5D%7B7x-2%7D%5C%5C%5C%5C%0Ag%28x%29%3D%5Ccfrac%7Bx%5E3%2B2%7D%7B7%7D%0A%5Cend%7Bcases%7D%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0Anow%0A%5C%5C%5C%5C%0Af%5B%5C%20g%28x%29%5C%20%5D%5Cimplies%20f%5Cleft%5B%20%5Cfrac%7Bx%5E3%2B2%7D%7B7%7D%20%5Cright%5D%5Cimplies%20%5Csqrt%5B3%5D%7B7%5Cleft%5B%20%5Cfrac%7Bx%5E3%2B2%7D%7B7%7D%20%5Cright%5D-2%7D%5Cimplies%20%5Csqrt%5B3%5D%7Bx%5E3%2B2-2%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Csqrt%5B3%5D%7Bx%5E3%7D%5Cimplies%20x%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0Aor%0A%5C%5C%5C%5C%0Ag%5B%5C%20f%28x%29%5C%20%5D%5Cimplies%20g%5Cleft%5B%5Csqrt%5B3%5D%7B7x-2%7D%5Cright%5D%5Cimplies%20%5Ccfrac%7B%5Cleft%5B%5Csqrt%5B3%5D%7B7x-2%7D%5Cright%5D%5E3%2B2%7D%7B7%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B7x-2%2B2%7D%7B7%7D%5Cimplies%20%5Ccfrac%7B7x%7D%7B7%7D%5Cimplies%20x)
thus f[ g(x) ] = x indeed, or g[ f(x) ] =x, thus they're indeed inverse of each other
