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Svetllana [295]
3 years ago
8

What is your question?

Mathematics
1 answer:
Elenna [48]3 years ago
4 0
Chisaka space walks for 2.5S
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The endpoints of a side of rectangle ABCD in the coordinate plane are at A(1,4) and
andrew-mc [135]

Answer:

<em>y = - x + 5 </em>

Step-by-step explanation:

A(1, 4)

B(4, 1)

m_{AB} = \frac{1-4}{4-1} = - 1

y - 1 = - 1 ( x - 4 )

<em>y = - x + 5</em>

6 0
3 years ago
Whats the lowest common multiple of 120 and 19600​
Levart [38]

Answer:

Multiples of 120 are 120, 240, 360, 480, 600, 720, 840 etc; Multiples of 150 are 150, 300, 450, 600, 750, 900 etc; Therefore, the least common multiple of 120 and 150 is 600.

Least common multiple (LCM) of 19600 and 19619 is 384532400.

4 0
3 years ago
Read 2 more answers
Which of the following describes the correct process for solving the equation 2x − 4 = 20 and arrives at the correct solution?
Alexxandr [17]

Answer:

a) Add 4 to both sides, and then divide by 2. The solution is x=12.

Step-by-step explanation:

Question:

Which of the following describes the correct process for solving the equation 2x − 4 = 20 and arrives at the correct solution?

a) Add 4 to both sides, and then divide by 2. The solution is x = 12.

b) Divide both sides by −4, and then subtract 2. The solution is x = −7.

c) Subtract 4 from both sides, and then divide by 2. The solution is x = −12.

d) Multiply both sides by −4, and then divide by 2. The solution is x = −40.

Solution:

Given equation:

2x-4=20

We need to determine the steps in order to solve for x.

Solving for x.

Step 1: Adding 4 both sides.

2x-4+4=20+4

2x=24

Step 2: Dividing both sides by 2.

\frac{2x}{2}=\frac{24}{2}

∴ x=12

Thus, we added 4 both sides and then divided both sides by 2 to get solution = 12

7 0
4 years ago
Do you know the answer? HELP?
Georgia [21]
V=LWH
if legnth AND width AND height are dobuled
V=2L2W2H
V=2*2*2LWH
V=8LWH

weight increases by 8 times
0.3 time s8=2.4

answer is 2.4lb
8 0
3 years ago
What is Simplify 3^-2 • 3^5
galina1969 [7]

Answer: 3^5 / 3^(-2) = 3^(3)

4 0
4 years ago
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