In order to prove that this quadrilateral is a parallelogram, we need to get the opposite sides to be parallel.
The answer would be D because it shows 2 triangles to be equal. If they are equal then they must have corresponding sides.
Answer:
x = 60°
Step-by-step explanation:
The circle is 360°, thus
2y + y = 360
3y = 360 ( divide both sides by 3 )
y = 120°
The tangent- secant angle ABT is one half the measure of the intercepted arc AB, thus
x = 0.5 × 120° = 60°
Answer:
To work this problem out you would do it as if it were just a simple equation
7x-19<16 (you would first have to add 19 to both sides)
7x<35 (then you would divide by 7 to get the variable by itself)
x<5 ( you answer would then be 5)
Step-by-step explanation:
The answer is 0 < x < infinity
Answer:
Step-by-step explanation:
Given two upward facing parabolas with equations
The two intersect at
=
x=
area enclosed by them is given by
A=![\int_{-\sqrt{\frac{2}{5}}}^{\sqrt{\frac{2}{5}}}\left [ \left ( x^2+2\right )-\left ( 6x^2\right ) \right ]dx](https://tex.z-dn.net/?f=%5Cint_%7B-%5Csqrt%7B%5Cfrac%7B2%7D%7B5%7D%7D%7D%5E%7B%5Csqrt%7B%5Cfrac%7B2%7D%7B5%7D%7D%7D%5Cleft%20%5B%20%5Cleft%20%28%20x%5E2%2B2%5Cright%20%29-%5Cleft%20%28%206x%5E2%5Cright%20%29%20%5Cright%20%5Ddx)
A=
A=![4\left [ \sqrt{\frac{2}{5}} \right ]-\frac{5}{3}\left [ \left ( \frac{2}{5}\right )^\frac{3}{2}-\left ( -\frac{2}{5}\right )^\frac{3}{2} \right ]](https://tex.z-dn.net/?f=4%5Cleft%20%5B%20%5Csqrt%7B%5Cfrac%7B2%7D%7B5%7D%7D%20%5Cright%20%5D-%5Cfrac%7B5%7D%7B3%7D%5Cleft%20%5B%20%5Cleft%20%28%20%5Cfrac%7B2%7D%7B5%7D%5Cright%20%29%5E%5Cfrac%7B3%7D%7B2%7D-%5Cleft%20%28%20-%5Cfrac%7B2%7D%7B5%7D%5Cright%20%29%5E%5Cfrac%7B3%7D%7B2%7D%20%5Cright%20%5D)
A=
