You can solve for sides of a triangle using the following equation.
A = arccos (b^2 + c^2 - a^2/2bc)
And likewise for other variables by moving the letters.
Ultimately, in this problem you get the following angle measures:
15.87
23.97
140.16
Answer:
area of sector BCD =60/360×π×10²=52.36ft²
a area of sector ADC =90/360×π×√50²=39.27ft²
actual <u>area</u><u> </u><u>=</u><u>5</u><u>2</u><u>.</u><u>3</u><u>6</u><u>ft²-39</u><u>.</u><u>2</u><u>7</u><u>f</u><u>t</u><u>²</u><u>=</u><u>1</u><u>3</u><u>.</u><u>0</u><u>9</u><u>=</u><u>1</u><u>3</u><u>ft²</u>
I don't get what you're asking. sorry
Answer:
B) ( x + 3y)²
Step-by-step explanation:
u = x + y ; v = x + 5y
uv = ( x + y ) * ( x + 5y)
= x * ( x + 5y) + y * (x + 5y) [ x * ( x + 5y) = x*x + x*5y = x² + 5xy]
= x² + 5xy + xy + 5y²
uv = x² + 6xy + 5y²----------equ 1
v - u = x + 5y - ( x+y)
= x + 5y - x - y
= 4y
(v-u)² = (4y)² = 16y²
[(v-u)/2]² = 16y² / 2²
= 16y² / 4 = 4y²
[(v-u)/2]² = 4y² ----------- equ 2
uv + [(v-u)/2]² = x² + 6xy + 5y² + 4y² ( from equ 1 and equ 2)
= x² + 6xy + 9y²
= ( x + 3y)²
