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3 years ago

Someone help me on this one pls!!

Mathematics

1 answer:

Nataly [62]3 years ago

7 0

The formula of a midpoint of AB:

$M_{AB}\left(\dfrac{x_A+x_B}{2};\ \dfrac{y_A+y_B}{2}\right)$

We have:

G(b; 0); F(3b; 2b)

substitute:

$M_{GF}\left(\dfrac{b+3b}{2};\ \dfrac{0+2b}{2}\right)\to M_{GF}(2b;\ b)$

Answer: (2b, b)

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Factor 4(3x + 1)2 - 5(3x + 1) + 1 completely.

lord [1]

C is the correct answer

5 0

3 years ago

ABC has coordinates of A(-8,-8) B(4,-2) C(2,2). Find the coordinates of its image after the dilation centered at the origin with

garri49 [273]

Answer:

A' ( -12 , -12 )

B' ( 6 , -3 )

C' ( 3 , 3 )

Step-by-step explanation:

To find the coordinates of a point after a dilation simply multiply the x and y values of the points by the scale factor

Points: A(-8,-8) B(4,-2) C(2,2)

Scale factor: 1.5

Coordinates after the dilation

A' = (-8,-8) --------> (-8 * 1.5 , -8 * 1.5 ) ------------> (-12 , -12)

B' = (4,-2) ---------> (4 * 1.5 , -2 * 1.5) -----------> (6 , -3 )

C' = (2 , 2) ----------> (2 * 1.5 , 2 * 1.5) -----------> (3, 3)

So inclusion the coordinates of ABC after a dilation centered at the origin with a scale factor of 1.5 are A' ( -12 , -12 ) B' ( 6 , -3 ) C' ( 3 , 3 )

4 0

2 years ago

The segment drawn from a vertex in a triangle to the midpoint of the opposite side.

AVprozaik [17]

check the picture below.

3 0

3 years ago

Read 2 more answers

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.

4 0

3 years ago

Hey loves, plz help if you have time:)

aksik [14]

Answer:

ΔABX and ΔEDX

Step-by-step explanation:

Given: AX ≅ EX

BX ≅ DX

SAS implies a congruent relation with respect to Side-Angle-Side between the two required triangle.

AX ≅ EX (given)

BX ≅ DX (given)

<BXA ≅ <DXE (vertical opposite angles)

Therefore,

ΔABX ≅ ΔEDX (Side-Angle-Side, SAS, congruence property)

3 0

3 years ago