Answer:
Answer
sqrt(108)
6 sqrt(3)
10.3923
I don't know which answer you want.
Step-by-step explanation:
- Drop a perpendicular from the top angle to the base.
- The base is cut into 2 equal parts.
- Each part is 12/2 = 6
- The perpendicular, as its name implies, meets the base at 90o.
- You can use the Pythagorean Theorem to find the height.
h^2 = side^2 - (1/2 b)^2
h^2 = 12^2 - 6^2
h^2 = 144 - 36
h^2 = 108
Take the square root of both sides
h = sqrt(108)
h = sqrt(2*2 * 3 * 3 * 3)
h = 2 * 3 sqrt(3)
h = 6sqrt(3)
h = 6 * 1.7321
h = 10.3923
Answer:
D
Step-by-step explanation:
10/7 = 1.4285
9/y = 1.4285
Multiply both sides by y.
9 = 1.4285y
Divide both sides by y.
6.3 = y
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:
Step-by-step explanation:
<em><u>Given</u></em><u>:</u> A line m is perpendicular to the angle bisector of ∠A. We call this
intersecting point as D. Hence, in figure ∠ADM=∠ADN =90°.
AD is angle bisector of ∠A. Hence, ∠MAD=∠NAD.
<u><em>To Prove</em></u>: <em><u>ΔAMN is an isosceles triangle. i.e any two sides in ΔAMN are</u></em>
<em> </em><em><u>equal. </u></em>
<em><u>Solution</u></em>: Now, In ΔADM and ΔADN
∠MAD=∠NAD ...(1) (∵Given)
AD=AD ...(2) (∵common side)
∠ADM=∠ADN ...(3) (∵Given)
<u><em> Hence, from equation (1),(2),(3) ΔADM ≅ ΔADN</em></u>
( ∵ ASA congruence rule)
⇒<u><em> AM=AN</em></u>
Now, In Δ AMN
AM=AN (∵ Proved)
Hence, ΔAMN is an isosceles triangle.
