suppose y varies directly with x, and y=8 when x=-6. What direction variation equation relates x and y? What is the value of y when x=-2

Answers

Answer 1

Answer: B. y = –1.33x; 2.67

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Shiloh has to earn at least $200 to meet her fundraising goal. She has only 100 cakes that she plans to sell at 5 dollars each. Which inequality shows the number of cakes, x, Shiloh can sell to meet her goal?

Answers

Answer:

$40 \leq x \leq 100$

Step-by-step explanation:

Cost of 1 cake = $5

Let x be the no. of cakes

So, Cost of x cakes = 5x

Since we are given that Shiloh has to earn at least $200

So, $5x\geq 200$

$x\geq 40$

So, the no. of the cakes should be sold more than or equal to 40

We are also given that She has only 100 cakes

So, the maximum number of cakes she can sold is 100 i.e. $x\leq 100$

So, Inequality becomes : $40 \leq x \leq 100$

Hence inequality shows the number of cakes, x, Shiloh can sell to meet her goal is $40 \leq x \leq 100$

she will only have to sell 40 cakes to earn $200

Write the equation for the vertical line that contains point E(–7, 7).Choose one answer.
a. y = –7
b. y = 7
c. x = 7
d. x = –7

Answers

Answer:

x=-7

Step-by-step explanation:

the equation for the vertical line that contains point E(–7, 7)

General equation of a vertical line is x= some number

The given point is (x,y)

point E (-7,7) where $x=-7 , y=7$

To get the equation of the vertical line

x= value of x in the given point

$x=-7$ is the equation of the vertical line

It would be B. y = 7 because it says write the equation for the VERTICAL LINE which is the y axis. Since the points are (-7, 7), it is the same thing as (x, y) the positive 7 is where the y is, so it would be y = 7

A clock shows 5:00. What time is it after the minute hand rotates 90 degrees?

Answers

5:15 Because Rotating 90 Degrees would be at the 3 so. 5:15

Each 90° rotation of the minute hand correspond to 15 minutes.
So will be 5.15 hs

7(m-12)=5m
a) -7
b) 6
c) 42
d) -42

Answers

$7(m-12)=5m \n \n 7m - 84 = 5m \n \n -84 = 5m - 7m \n \n -84 = -2m \n \n (-84)/(-2) = m \n \n (84)/(2) = m \n \n 42 = m \n \n m = 42 \n \n Answer: \fbox {m = 42}$

7(m-12) = 5m

7m - 84 = 5m

2m - 84 = 0

2m = 84

m = 42

The answer is C) 42

What is the distance from (−2, 4) to (0, −6)? Round your answer to the nearest hundredth.

Answers

Answer:

10.2

Use the formula:

d=√(x - x) + (y - y)

2 1 2 1

Word form :

d equals the square root of x2 - x1 plus y2 - y1

Help with line equations?

Answers

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 $(1)/(2)$

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

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

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

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

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

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

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

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

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

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

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

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 ;)

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