Question is on picture Please help Zoom in if needed :D

Alecsey [184]

First question

$line:~~~~y=x+3$

$curve:~~~~x^2+y^2=29$

now we have to replace the y of curve by the y of line, therefore

$x^2+(x+3)^2=29$

$x^2+x^2+6x+9=29$

$2x^2+6x+9-29=0$

$2x^2+6x-20=0$

we can multiply each member by $\frac{1}{2}$

$\boxed{\boxed{x^2+3x-10=0}}$

Now we have to find the roots of this funtion

$x^2+3x-10=0$

Sum and produc or Bhaskara

Then we find two axis

$\boxed{x_1=2~~and~~x_2=-5}$

now we replace this values to find the Y, we can replace in curve or line equation, I'll prefer to replace it in line equation.

$y=x+3$

$y_1=x_1+3$

$y_1=2+3$

$\boxed{y_1=5}$

$y_2=x_2+3$

$y_2=-5+3$

$\boxed{y_2=-2}$

Therefore

$\boxed{\boxed{P_1(2,5)~~and~~P_2(-5,-2)}}$
_______________________________________________________________

The second question give to us

$y=ax+b$

$P_1(2,13)$

$P_2(-1,-11)$

We just have to replace the value then we'll get a linear system.

point 1

$13=2a+b$

point 2

$-11=-a+b$

then our linear system will be

$\begin{Bmatrix}2a+b&=&13\\-a+b&=&-11\end{matrix}$

I'll multiply the second line by -1 and I'll add to first one

$\begin{Bmatrix}2a+(a)+b+(-b)&=&13+(-11)\\-a+b&=&-11\end{matrix}$

$\begin{Bmatrix}3a&=&24\\-a+b&=&-11\end{matrix}$

$\begin{Bmatrix}a&=&8\\-a+b&=&-11\end{matrix}$

therefore we can replace the value of a, at second line

$\begin{Bmatrix}a&=&8\\-8+b&=&-11\end{matrix}$

$\begin{Bmatrix}a&=&8\\b&=&-11+8\end{matrix}$

$\boxed{\boxed{\begin{Bmatrix}a&=&8\\b&=&-3\end{matrix}}}$

then our function will be

$\boxed{\boxed{y=8x-3}}$

_________________________________________________________________

The third one, we have

$line:~~~~y=3x-4$

$curve:~~~~y=x^2-2x-4$

This resolution will be the same of our first question.

Let's replace the y of curve by the y of line

$3x-4=x^2-2x-4$

$0=x^2-2x-4-3x+4$

therefore

$\boxed{\boxed{x^2-5x=0}}$

now we have to find the roots of this function.

$x^2-5x=0$

put x in evidence

$x*(x-5)=0$

$\boxed{x_1=0~~and~~x_2=5}$

then

$y=3x-4$

$y_1=3x_1-4$

$y_1=3*0-4$

$\boxed{y_1=-4}$

$y_2=3x_2-4$

$y_2=3*5-4$

$y_2=15-4$

$\boxed{y_2=11}$

our points will be

$\boxed{\boxed{P_1(0,-4)~~and~~P_2(5,11)}}$

I hope you enjoy it ;)

5 0

3 years ago

Write the fraction or mixed number as a decimal. Use a bar to show a repeating decimal. 3/8

Pepsi [2]

The decimal from the fraction is .375

8 0

3 years ago

In which quadrant does the point that is graphed lie? A) I B) II C) III D) IV

Tpy6a [65]

B is the correct amswer

4 0

3 years ago

Andre wants to put a carpet down on a circular room that has a diameter of 14 feet

Fantom [35]

Answer:

49 pi

Step-by-step explanation:

I'll assume you want to know how big the carpet is. You need to find the area of the room, which is a circle. The area of a circle is given as pi*r^2, where r is the radius. The diameter is twice the radius, so as the problem gives us a diameter of 14, the radius would be 7. Plug in 7 for r, and we get 49pi.

3 0

3 years ago

PLEASEEEEEE HELP ME WITH THIS PLEASE

Maksim231197 [3]

Question 1

$r(q(-2))=r(3)=11$

Question 2

$q(r(-2))=q(6)=-13$

8 0

1 year ago