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artcher [175]
2 years ago
6

QUESTION!!!! WHEN TRYING TO PROVE SIMILER TRIANGLES, IF YOUR USINF THE ANGLE ANGLE SIMILARITY POSTUALET HOW MANY ANGLES DO YOU N

EED TO PROVE IN YOUR STATEMENTS? AND IS THERE A WAY TO MAKE SURE THAT THE WAY YOU PROVED YOUR TRIANGLES SIMILAR IS RIGHT?
Mathematics
1 answer:
Leto [7]2 years ago
8 0

Answer: 2 pairs of angles

As the name "angle angle" suggests, we need 2 pairs of angles proven congruent in order to prove that two triangles are similar. Each "angle" refers to a pair of angles between the two triangles, which are corresponding angles. We could use the third pair, but it's extra unnecessary work. This theorem is often written as "AA similarity" or something along those lines.

One way to check for similar triangles is to compare the ratios of the sides. If the corresponding sides divide to the same value, then you have similar triangles. This method only works of course if you know the side lengths.

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What should be the b-value in the equation shown below if the equation is a perfect square trinomial?
masha68 [24]

Answer:

B=10

Step-by-step explanation:

It's a perfect square of equation ax²+2abx+b²

I'm the given equation a is 1 and b is 5 so 2(a)(b)=2(1)(5)=10

(ps: when I say b in the explanation I mean another b not the one in question as to not be confused)

4 0
2 years ago
Find the measure Of K
Mariana [72]

Answer:

K = 10

Step-by-step explanation:

Since this is a right triangle, we can use trig functions

sin theta = opp/ hyp

sin K = 11 / 61

Taking the inverse sin of each side

sin ^-1 ( sin K) = sin ^-1 (11/61)

K =10.38885782

K = 10

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If you roll a single six-sided die, what is the probability of rolling an odd number?
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Step-by-step explanation:

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6 0
2 years ago
Read 2 more answers
a. How wide must a wall footing be if the load is 9,500 pounds per foot of wall length, and the footing rests on a sandy gravel?
Phantasy [73]

Answer:

a) 38"

b) L = 5.7'

Step-by-step explanation:

a) Given

P = 9,500 lb/ft

Thick = 18"

We assume an allowable bearing pressure of σ = 3000 lb/sf

If Stress = Force / Area

σ = P/A

Solving for area

A = P/σ  ⇒  A = 9,500 lb/ft/3000 lb/sf

⇒  A = 3.167 sf,  or  3.167 ft wide, per foot of length.

⇒  A = 38" wide

We can see the pic 1 in order to understand the answer.

b) Given

P = 65,000 pounds = 65,000 lb

Thick = 18" = 1.5'

We assume an allowable bearing pressure of σ = 2000 lb/sf

If Stress = Force / Area

σ = P/A

Solving for area

A = P/σ  ⇒  A = 65,000 lb/2000 lb/sf

⇒  A = 32.5 sf

then A = L²  ⇒  L = √A = √(32.5 sf) = 5.7 ft

Finally L = 5.7'

We can see the pic 2 in order to understand the answer.

3 0
3 years ago
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