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3 years ago

8

Consider points A (3, 6) and B (8, 4). Find point C on the x-axis so AC + BC is a

1 answer:

scoray [572]3 years ago

6 0

Answer:

(1,2)

Step-by-step explanation:

the st. lines 3x+4y=5 and 4x-3y=15 interrect at a point A(3, -1. On these linepoints B and C are chosen so that AB= AC. find the posible eqns of the lineBC pathrough the point

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Factorize 225-49<br><img src="https://tex.z-dn.net/?f=%20%7Bx%7D%5E%7B2%7D%20" id="TexFormula1" title=" {x}^{2} " alt=" {x}^{2}

erica [24]

$225-49x^2$

$-49x^2+225$

$=(7x+15)(-7x+15)$

7 0

3 years ago

Read 2 more answers

This table of values represents a linear function.

Vesna [10]

4.5 to 6.5 i think not really shure

3 0

3 years ago

What is y+5=-3(x-4)

schepotkina [342]

Answer:y=-3x+7

Step-by-step explanation:y+5=-3(x-4)

=>y+5=-3x+12

=>y=-3x+12-5

•°•y=-3x+7.

8 0

3 years ago

Will mark brainliest!!!plz helppp

muminat

Answer:

(5,-6)

Step-by-step explanation:

ONE WAY:

If $f(x)=x^2-6x+3$ , then $f(x-2)=(x-2)^2-6(x-2)+3$ .

Let's simplify that.

Distribute with $-6(x-2)$ :

$f(x-2)=(x-2)^2-6x+12+3$

Combine the end like terms $12+3$ :

$f(x-2)=(x-2)^2-6x+15$

Use $(x-b)^2=x^2-2bx+b^2$ identity for $(x-2)^2$ :

$f(x-2)=x^2-4x+4-6x+15$

Combine like terms $-4x-6x$ and $4+15$ :

$f(x-2)=x^2-10x+19$

We are given $g(x)=f(x-2)$ .

So we have that $g(x)=x^2-10x+19$ .

The vertex happens at $x=\frac{-b}{2a}$ .

Compare $x^2-10x+19$ to $ax^2+bx+c$ to determine $a,b,\text{ and } c$ .

$a=1$

$b=-10$

$c=19$

Let's plug it in.

$\frac{-b}{2a}$

$\frac{-(-10)}{2(1)}$

$\frac{10}{2}$

$5$

So the $x-$ coordinate is 5.

Let's find the corresponding $y-$ coordinate by evaluating our expression named $g$ at $x=5$ :

$5^2-10(5)+19$

$25-50+19$

$-25+19$

$-6$

So the ordered pair of the vertex is (5,-6).

ANOTHER WAY:

The vertex form of a quadratic is $a(x-h)^2+k$ where the vertex is $(h,k)$ .

Let's put $f$ into this form.

We are given $f(x)=x^2-6x+3$ .

We will need to complete the square.

I like to use the identity $x^2+kx+(\frac{k}{2})^2=(x+\frac{k}{2})^2$ .

So If you add something in, you will have to take it out (and vice versa).

$x^2-6x+3$

$x^2-6x+(\frac{6}{2})^2+3-(\frac{6}{2})^2$

$(x+\frac{-6}{2})^2+3-3^2$

$(x+-3)^2+3-9$

$(x-3)^2+-6$

So we have in vertex form $f$ is:

$f(x)=(x-3)^2+-6$ .

The vertex is (3,-6).

So if we are dealing with the function $g(x)=f(x-2)$ .

This means we are going to move the vertex of $f$ right 2 units to figure out the vertex of $g$ which puts us at (3+2,-6)=(5,-6).

The $y-$ coordinate was not effected here because we were only moving horizontally not up/down.

3 0

3 years ago

Given that f=cd3, f=450 and d =10 what is c?

Brut [27]

Answer:

15

Step-by-step explanation:

450=c*10*3 divide by 10

45=c*3 divide by 3

c=15

4 0

3 years ago

Read 2 more answers