16

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

4 years ago

What is the slope of the line that passes through the points (–20, 18) and (30, 14)?

natali 33 [55]

The slope of the line that passes through the points (-20,18) and (30,14) is

m (slope) = -2/25

5 0

4 years ago

Read 2 more answers

PLEASE HELP FOR 87 POINTS!!!!!!!!!!!!!!!!!!!!!!!!!!

Anton [14]

1. You can expand the coefficients to yield 6, and add the powers since the bases are the same (ie A)

2. y⁻² can also be rewritten as 1/y², so it becomes 1/y² · 1/y⁵ = 1/y⁷ (ie C)

3. Only the x is affected by the power.

4. y = 3·5ˣ doesn't change in slope, so it essentially is similar to a 5ˣ graph.

5. Should be pretty straight forward.
6. Should be pretty straight forward.
7. Should be pretty straight forward.

8. Since x is the length of the border, calculate the area of the frame and subtract that from the area of the square painting. (Remember that the length of the square painting is x - 8, not x - 4)

9. Should be pretty straight forward.
11 - 25 are pretty much similar to each other.
If you have trouble, let me know and I'll walk you through it.

6 0

3 years ago

Read 2 more answers

If can babysits 2 hours and she earns $14 then she works 4 hours and gets $28 what how much does she make if she does 6 hours

mr Goodwill [35]

Answer:

He will earn 42$

Step-by-step explanation:

7 0

3 years ago

An eccentric philanthropist undertakes to give away $100,000. He is eccentric because he insists that each of his gifts be a num

gavmur [86]

Answer:

Step-by-step explanation:

An eccentric philanthropist undertakes to give away $100,000. He is eccentric because he insists that each of his gifts be a number of dollars that is a power of two, and he will give no more than one gift of any amount. How does he distribute the money?

To solve this problem, we will simply write 1,00,000 in base 2.

2^17 gives 131072, 2^16 gives 65536.

So, the first gift is 65536, we then need to repeat this process for 34,464 ((100,000-65536 = 34,464), to get all gifts in succession.

6 gifts in all: one each of $32, $128, $512, $1024, $32768, and $65536.

3 0

3 years ago