Please help I can't upload a picture bc it's on my laptop and it wont let me take a screen shot, BUTTT

Line segment AB has endpoints A(-1.5, 0) and B(4.5 8). Point C is on line segment AB located at (0, 2)

Pie

Answer:

The ratio of AC/CB=1/3

Step-by-step explanation:

We use the section formula:

$(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n})$

We substitute the point A(-1.5,0) and B(4.5,8).

We plug in the points to get:

$(\frac{4.5m - 1.5n}{m+n},\frac{8m+n \times 0}{m+n})$

This is supposed to be equal to: (0,2).

$(\frac{4.5m - 1.5n}{m+n},\frac{8m+n \times 0}{m+n}) = (0,2)$

We can equate, corresponding coordinates to find m and n.

$\frac{4.5m - 1.5n}{m + n} = 0$

This implies that:

$4.5m - 1.5n = 0$

$4.5m = 1.5n$

$\frac{m}{n} = \frac{1.5}{4.5}$

$\frac{m}{n} = \frac{1}{3}$

Therefore the ratio is m:n=1:3

4 0

3 years ago

The quadratic function g(x) = a.ca + bx+c has the

Mumz [18]

<em>The value of b is 14 and the value of c is 65</em>

<h2>Explanation:</h2>

The quadratic function is a function of the form:

$f(x)=ax^2+bx+c$

Here we know that the leading coefficient $a=1$ so we reduce our equation to:

$g(x)=x^2+bx+c$

The roots are those values at which $g(x)=0$

So:

$x^2+bx+c=0 \\ \\ First \ root: \\ \\ (-7+4i)^2+b(-7+4i)+c=0 \\ \\ (-7)^2-2(7)(4i)+(4i)^2-7b+4bi+c=0 \\ \\ 49-56i+16i^2-7b+4bi+c=0 \\ \\ \\ Simplifying: \\ \\ 49-56i+16(-1)-7b+4bi+c=0 \\ \\ 49-56i-16-7b+4bi+c=0 \\ \\ 33-56i-7b+4bi+c=0 \\ \\ \\$

$Second \ root: \\ \\ (-7-4i)^2+b(-7-4i)+c=0 \\ \\ (-1)^2(7+4i)^2+b(-7-4i)+c=0 \\ \\ (7)^2+2(7)(4i)+(4i)^2-7b-4bi+c=0 \\ \\ 49+56i+16i^2-7b-4bi+c=0 \\ \\ \\ Simplifying: \\ \\ 49+56i+16(-1)-7b-4bi+c=0 \\ \\ 49+56i-16-7b-4bi+c=0 \\ \\ 33+56i-7b-4bi+c=0$

So we have:

$(1) \ 33-56i-7b+4bi+c=0 \\ \\ (2) \ 33+56i-7b-4bi+c=0 \\ \\ \\ Subtract \ 2 \ from: \\ \\ 33-56i-7b+4bi+c-(33+56i-7b-4bi+c)=0 \\ \\ 33-56i-7b+4bi+c-33-56i+7b+4bi-c=0 \\ \\ \\ Combine \ like \ terms: \\ \\ 33-33-56i-56i-7b+7b+4bi+4bi+c-c=0 \\ \\ -112i+8bi=0 \\ \\ Isolating \ b: \\ \\ b=\frac{112i}{8i} \\ \\ \boxed{b=14}$

Finding c from (1):

$33-56i-7b+4bi+c=0 \\ \\ \\ Substituting \ b: \\ \\ 33-56i-7(14)+4(14)i+c=0 \\ \\ 33-56i-98+56i+c=0 \\ \\ -65+c=0 \\ \\ \boxed{c=65}$

<h2>Learn more:</h2>

Complex conjugate: brainly.com/question/2137496

#LearnWithBrainly

5 0

3 years ago

How to factor this expression <br><br>x^3 - 4x^2 + 4x - 16

Natalija [7]

X²(x - 4) +4 (x - 4)

(x² + 4) (x - 4)

First find the common terms that can enter into both x³ and 4x² then write its down in this case it’s x² that can enter x³ leaving only x _since x³/x² = subtract of the indices. x² will also enter 4x² leaving only four hence you having x² (x - 4)
then do the same for the next pair of terms giving you 4 that can enter into both 4 and 16
Leaving you with +4 (x - 4)

Now you can put the common terms together like so (x² + 4) and choose get one of the other two which are the same= (x - 4)
= (x² + 4) (x - 4)

8 0

2 years ago

There are 9 socks in a bag. 1 of those socks is blue.

inessss [21]

1/9 or 11/100

Hope that helps

4 0

3 years ago

How can using a number line aid in ordering integers? Explain in a full paragraph. (SIX SENTENCES)

ludmilkaskok [199]

Answer:

is that u in the pic?

Step-by-step explanation:

7 0

3 years ago