I need help! With algebra

Juan bought 12 pounds of dog food for $15.72. What is the price per pound of the dog food?

Greeley [361]

I believe it is $1.31

3 0

2 years ago

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

Which graph represents the function f(x)=

zhenek [66]

Step-by-step explanation:

you don't show us the choices.

anyway, the real graph must be similar to the one you are showing, as it goes to - and + infinity for x = 2.

because the denominator "x-2" will be 0 for x = 2.

but the horizontal limits are both y = +3 (and not 0).

because (3x-2)/(x-2) goes more and more to 3/1 the larger (or smaller in the - direction) x gets.

7 0

1 year ago

Fill in the blanks below using the word bank.

Charra [1.4K]

M is the slope.

B is the y-intercept

The constant of proportionality is the ratio between two directly proportional quantities.

7 0

3 years ago

Read 2 more answers

Two points on L1 and two points on L2 are given. Use the slope formula to determine if lines L1 and L2 are parallel, perpendicul

vlada-n [284]

Answer:

The lines L1 and L2 neither parallel nor perpendicular

Step-by-step explanation:

* Lets revise how to find a slope of a line

- If a line passes through points (x1 , y1) and (x2 , y2), then the slope

of the line is $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

- Parallel lines have same slopes

- Perpendicular lines have additive, multiplicative slopes

( the product of their slopes is -1)

* Lets solve the problem

∵ L1 passes through the point (1 , 10) and (-1 , 7)

- Let (1 , 10) is (x1 , y1) and (-1 , 7) is (x2 , y2)

∴ x1 = 1 , x2 = -1 and y1 = 10 , y2 = 7

∴ The slope of L1 is $m1 = \frac{7-10}{-1-1}=\frac{-3}{-2}=\frac{3}{2}$

∵ L2 passes through the point (0 , 3) and (1 , 5)

- Let (0 , 3) is (x1 , y1) and (1 , 5) is (x2 , y2)

∴ x1 = 0 , x2 = 1 and y1 = 3 , y2 = 5

∴ The slope of L2 is $m2=\frac{5-3}{1-0}=\frac{2}{1}=2$

∵ m1 = 3/2 and m2 = 2

- The two lines have different slopes and their product not equal -1

∴ The lines L1 and L2 neither parallel nor perpendicular

7 0

2 years ago