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wlad13 [49]
2 years ago
7

After moving 5 units up, 3 units right, and 4 units down, the ending position is (9, 6). What was the starting position.

Mathematics
1 answer:
Anika [276]2 years ago
6 0
You reverse what you did, so first you go 5 units down which makes the coordinates (9,1) then you move 3 units left which then is (6,1). Then you go 4 units up which lets you end out at (6,5). 
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Factorise a x + b x minus A Y minus b y​
shutvik [7]
X(a+b)-y(a+b)
*factor out (a+b)*
=(a+b)(x-y)
8 0
3 years ago
Line segment NY has endpoints N(-11, 5) and Y(3,-3).
777dan777 [17]

Given:

Line segment NY has endpoints N(-11, 5) and Y(3,-3).

To find:

The equation of the perpendicular bisector of NY.

Solution:

Midpoint point of NY is

Midpoint=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)

Midpoint=\left(\dfrac{-11+3}{2},\dfrac{5-3}{2}\right)

Midpoint=\left(\dfrac{-8}{2},\dfrac{2}{2}\right)

Midpoint=\left(-4,1\right)

Slope of lines NY is

m=\dfrac{y_2-y_1}{x_2-x_1}

m=\dfrac{-3-5}{3-(-11)}

m=\dfrac{-8}{14}

m=\dfrac{-4}{7}

Product of slopes of two perpendicular lines is -1. So,

m_1\times \dfrac{-4}{7}=-1

m_1=\dfrac{7}{4}

The perpendicular bisector of NY passes through (-4,1) with slope \dfrac{7}{4}. So, the equation of perpendicular bisector of NY is

y-y_1=m_1(x-x_1)

y-1=\dfrac{7}{4}(x-(-4))

y-1=\dfrac{7}{4}(x+4)

y-1=\dfrac{7}{4}x+7

Add 1 on both sides.

y=\dfrac{7}{4}x+8

Therefore, the equation of perpendicular bisector of NY is y=\dfrac{7}{4}x+8.

6 0
2 years ago
HELP NEEDED PLEASE!! WILL GIVE BRAINIEST
marysya [2.9K]

Answer:

12 20 22

Step-by-step explanation:

4 0
3 years ago
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
2 years ago
What is the value of x?
kotegsom [21]

Answer:

X=15

Step-by-step explanation:

A +B + C = 180

2x +4x+ 6x = 180

12x =180

X= 15

4 0
3 years ago
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