Step-by-step explanation:
Here
BC=P=10
AC=B=b
AB=H=15
then using fourmula
h^2= p^2+ b^2
b^2= 15×15 - 10× 10
b^2= 125
b=11.2
Think: You're treating the numerator and the denom. in precisely the same way. In doing so you are NOT changing the value of the fraction, only the appearance.
Example: start with 2/3. Mult num. and den. both by 7: 14/21.
2/3 and 14/21 result in precisely the same decimal fraction, showing that the latter set of fractions is equivalent to the former set.
Answer: to find the mean add up all the scores and then divide it by the number of scores that are there and to find the mode order the numbers lowest to highest and see which number appears the most
Circle: x^2+y^2=121=11^2 => circle with radius 11 and centred on origin.
g(x)=-2x+12 (from given table, find slope and y-intercept)
We can see from the graphics that g(x) will be almost tangent to the circle at (0,11), and that both intersection points will be at x>=11.
To show that this is the case,
substitute g(x) into the circle
x^2+(-2x+12)^2=121
x^2+4x^2-2*2*12x+144-121=0
5x^2-48x+23=0
Solve using the quadratic formula,
x=(48 ± √ (48^2-4*5*23) )/10
=0.5058 or 9.0942
So both solutions are real and both have positive x-values.
