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3 years ago

-21x+3y=6 How do you solve for y?

1 answer:

3 0

All you've got to do is some simple algebra! When you solve for a variable, you are trying to get it alone on one side of the equation. So, we want something that looks like y = ...

First, add 21x to both sides of the equation. This will cancel out the -21x on the left side.
-21x + 3y = 6
-21x + 3y + 21x = 6 + 21x
3y = 21x + 6

Now, to get y alone, divide both sides of the equation by 3. This will cancel out the coefficient of 3 on the left side.
3y = 21x + 6
(3y)/3 = (21x + 6)/3
y = 7x + 2

There's the answer. Hope that helps! :)

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Segment PH is graphed on the coordinate grid where P is (-31,8) and H is (29,-10). Point E is the midpoint of segment PH. If poi

maks197457 [2]

Answer:

The coordinates of point A are (-11 , 2)

Step-by-step explanation:

- The coordinates of point P are (-31 , 8)

- The coordinates of point H are (29 , -10)

- Point E is the mid-point of segment PH

- The coordinates of the the mid-point are:

→ $x=\frac{x_{1}+x_{2}}{2}$ and $y=\frac{y_{1}+y_{2}}{2}$

∵ E is the mid-point of segment PH

∴ The x-coordinate of point E is $x=\frac{-31+29}{2}=-1$

∴ The x-coordinate of point E is -1

∴ The y-coordinate of point E is $y=\frac{8+-10}{2}=-1$

∴ The y-coordinate of point E is -1

∴ The coordinates of point E are (-1 , -1)

- Point A is graphed $\frac{2}{3}$ of the way along the segment from P to E

- That mean the distances from points P to A is 2 parts and from P to E

is 3 parts, then the distance from A to E is "3 - 2" = 1 part

- So point A divides the line segment PE at ratio 2 : 1 from P

- The coordinates of the division point are:

→ $x=\frac{x_{1}m_{2}+x_{2}m_{1}}{m_{1}+m_{2}}$ and $y=\frac{y_{1}m_{2}+y_{2}m_{1}}{m_{1}+m_{2}}$

∵ Point A divides PE at ratio 2 : 1

∵ The coordinates of point P are (-31 , 8)

∵ The coordinates of point E are (-1 , -1)

∴ The x-coordinate of point A is $x=\frac{(-31)(1)+(-1)(2)}{2+1}= -11$

∴ The x-coordinate of point A is -11

∴ The y-coordinate of point A is $x=\frac{(8)(1)+(-1)(2)}{2+1}=2$

∴ The y-coordinate of point A is 2

∴ The coordinates of point A are (-11 , 2)

* The coordinates of point A are (-11 , 2)

3 0

3 years ago

Evaluate the function f(x) = 3x^2 + 4x + 19 Find f(-7)

yanalaym [24]

Answer:

Step-by-step explanation:

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(

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3 0

1 year ago

PART TWO MATH TEST :) Please Help! I will give brainliest and donuts <33

ra1l [238]

Hey there,

Question #1

The answer would be in the attachment below.
_____________________________________________________________

Question #2

The answer would be in the attachment below.
_____________________________________________________________

Question #3

The answer would be in the attachment below.
_____________________________________________________________

Question 4#

The last one was kind of tricky. But, as I saw this attachment, I noticed on how the rectangle was actually 3/4 on the base and for the height, it was 1/2. So by doing this,we need to find the area, and we would multiply these both. 1/2 x 3/4 = 3/8 but by looking at your options, those are not simplified so . . .your answer would be 6/16 because 3x2=6 & 8x2=16.
_____________________________________________________________

I really hope this can help you Amanda.

Have a great day! =)

~Jurgen

6 0

4 years ago

What is the length of the right triangle below in inches?

k0ka [10]

Answer:

6in

Step-by-step explanation:

Using the pythagerous thereom,

10^2 = 8^2 + b^2

b^2 = 10^2-8^2

b^2 = 100-64

b^2 = 36

b = 6

Feel free to mark this as brainliest :D

8 0

3 years ago

Read 2 more answers

If the mean weight of 4 backfield members on the football team is 221 lb and the mean weight of the 7 other players is 202 lb, w

MatroZZZ [7]

Let $x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8, x_9, x_{10}, x_{11}$ be the weight of i-th player.

1. If the mean weight of 4 backfield members on the football team is 221 lb, then

$\dfrac{x_1+x_2+x_3+x_4}{4}=221\ lb.$

2. If the mean weight of the 7 other players is 202 lb, then

$\dfrac{x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}}{7}=202\ lb.$

3. From the previous statements you have that

$x_1+x_2+x_3+x_4=221\cdot 4=884 \lb,\\ \\x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}=202\cdot 7=1414\ lb.$

Add these two equalities and then divide by 11:

$x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}=884+1414=2298\ lb,\\ \\\dfrac{x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}}{11}=\dfrac{2298}{11}=208\dfrac{10}{11}\ lb.$

Answer: the mean weight of the 11-person team is $208\dfrac{10}{11}\ lb.$

4 0

3 years ago