Answer:
Yes by SSS because you have the same hypotenuse and same hatch marks on other sides
Answer:
see explanation
Step-by-step explanation:
Under a reflection in the line y = - x
a point (x, y ) → (- y, - x ), thus
T(- 1, 3 ) → T'(- 3, 1 )
U(- 1, 10 ) → U'(- 10, 1 )
V(- 2, 4 ) → V'(- 4, 2 )
A point has D) zero dimensions
9514 1404 393
Answer:
(c) 27x^11 +51x^7 +9x^6 -60x^5 +17x^2 -20
Step-by-step explanation:
As with many multiple-choice questions, you only need to look at something that will discriminate the correct answer from the wrong one.
The highest-degree product term is the product of the highest-degree terms in the factors:
(3x^5)(9x^6) = 27x^11
This matches choice C only.
_____
In case you're interested in actually performing the rest of the multiplication, the distributive property applies.
(1 +3x^5)(17x^2 +9x^6 -20)
= 1(17x^2 +9x^6 -20) +3x^5(17x^2 +9x^6 -20)
= 17x^2 +9x^6 -20 +51x^7 +27x^11 -60x^5
Writing these terms in order of decreasing exponents gives ...
= 27x^11 +51x^7 +9x^6 -60x^5 +17x^2 -20
Answer:
area of circle having radius 3
is πr²=π*3²=9π square units
area of circle having radius 2
=πr²=π×2²=4π square units
again
area of big circle having radius [3+2]=5 units
=πr²=π×5²=25π square units
Now
<u>area</u><u> </u><u>of</u><u> </u><u>shaded</u><u> </u><u>region</u><u>=</u><u>2</u><u>5</u><u>π</u><u>-</u><u>4</u><u>π</u><u>-</u><u>9</u><u>π</u><u>=</u><u>1</u><u>2</u><u>π</u><u> </u><u>or</u><u> </u><u>3</u><u>7</u><u>.</u><u>7</u><u>s</u><u>q</u><u>u</u><u>a</u><u>r</u><u>e</u><u> </u><u>units</u><u>.</u>
