All categories

3 years ago

Help please I’m confused

Mathematics

1 answer:

Alex787 [66]3 years ago

4 0

Answer:

it is the 2nd option

Step-by-step explanation:

You might be interested in

Based on the information given say whether or not △ABC∼△FED. Explain your reasoning.

GalinKa [24]

Answer:

Yes, △ABC ∼ △FED by AA postulate.

Step-by-step explanation:

Given:

Two triangles ABC and FED.

m∠A = m∠B

m∠C = m∠A + 30°

m∠E = m∠F = $x$

m∠D = $2x-20$ °.

Now, let m∠A = m∠B = $y$

So, m∠C = m∠A + 30° = $y+30$

Now, sum of all interior angles of a triangle is 180°. Therefore,

m∠A + m∠B + m∠C = 180

$y+y+y+30=180\\3y=180-30\\3y=150\\y=\frac{150}{3}=50$

Therefore, m∠A = 50°, m∠B = 50° and m∠C = m∠A + 30° = 50 + 30 = 80°.

Now, consider triangle FED,

m∠D+ m∠E + m∠F = 180

$2x-20+x+x=180\\4x=180+20\\4x=200\\x=\frac{200}{4}=50$

Therefore, m∠F = 50°

m∠E = 50° and

m∠D = $2x-20=2(50)-20=100-20=80\°$

So, both the triangles have congruent corresponding angle measures.

m∠A = m∠F = 50°

m∠B = m∠E = 50°

m∠C = m∠D = 80°

Therefore, the two triangles are similar by AA postulate.

5 0

3 years ago

13. The least common multiple of two non-zero integers a and b is the unique positive integer m such that (i) m is a common mult

Vlad [161]

Answer:

[a,b] divides n

Step-by-step explanation:

Let us denote the least common multiple of a and b [a,b]=m.

We want to prove that m divides n, where n is a multiple of a and b.

We suppose m does not divide n, then by the Division Theorem, there exists q and r integers such that:

(1) ... n=mq+r, where 0<r<m

As n is a multiple of a and b, there exists s and t integers such that:

sa=n and tb=n

Same thing happens to m as it is the least common multiple, there exists u and v such that:

ua=m and vb=m

So (1) has the following form:

n=mq+r ⇒ sa=uaq+r ⇒sa-uaq=r⇒(s-uq)a=r and

n=mq+r ⇒ tb=vbq+r ⇒ tb-vbq=r⇒ (t-vq)b=r

So r is a multiple of a and b, but r<m which is a contradiction as, m is the least common multiple of a and b. So this concludes the proof.

So this means that $\frac{ab}{m}$ is and integer.

As m= vb, then $\frac{m}{b}$ is an integer, lets say $\frac{m}{b}$ =v; and as m=ua, then $\frac{m}{a}$ =u.

So $\frac{ab}{m}$ v= $\frac{ab}{m}$ $\frac{m}{b}$ =a, so $\frac{ab}{m}$ divides a; on the other hand, $\frac{ab}{m}$ u= $\frac{ab}{m}$ $\frac{m}{a}$ =b, so $\frac{ab}{m}$ divides b. From this we can conclude that $\frac{ab}{m}$ is a common divisor of a and b.