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2 years ago

Can someone be kind enough and help me I've posted my question many times and no one has answered

Mathematics

1 answer:

ahrayia [7]2 years ago

3 0

Answer:

The symbol that should be used is >

Step-by-step explanation:

In the question it says that half of a number is greater than 1 1/3 of a number, which only leaves that as an option

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Is 8/12 greater than 10/15? and how do you know?

valentina_108 [34]

Answer:

no, they are equal. we know this because when reducing both they are equal.

Step-by-step explanation:

first find the least common denominator

that's 3

reduce fractions until the denominator is both 3

we get 2/3 and 2/3. the proportions are equal.

8 0

2 years ago

Read 2 more answers

I need to know the answer to those two equations

sladkih [1.3K]

6(3x-5)-7x=25
18x-30-7x=25
11x=55
x=11
-2(5+6m)+16=-90
-10-12m+16=-90
-12m+6=-90
-12m=-96
m=8

3 0

3 years ago

I don't really understand this??....

Mila [183]

Answer:

Part 1

its asking you to find the slope of the given lines by using the points that they have plotted on the lines

first you need to figure out what the points are, then insert it into the formula

the formula for this would be $\frac{Y2- Y1}{X2-X1}$

Part 2:

its asking you to find the slope using the points that are written in each box

again, just substitute it into the formula

the formula for this would be $\frac{Y2- Y1}{X2-X1}$

5 0

2 years ago

1) Mike's sister, Kellie, is 4 years older than he is. Let m be Mike's age and let k be Kellie's age. ) Complete the table to sh

sveta [45]

Answer:

10 -> 14

15-> 19

20-> 24

25-> 29

Step-by-step explanation:

Add 4 each time

5 0

2 years ago

What is the solution set? (0, -2) (2, 0) (7, 0) (5, 3)

Dvinal [7]

Answer:

(5,3)

Step-by-step explanation:

Look for the coordinates of the point of intersection where the 2 graphs cross. This gives the solution.

In this case, they cross at (5,3) and they intersect at only one location (i.e there is only 1 solution)

7 0

3 years ago

Read 2 more answers

Can someone be kind enough and help me I've posted my question many times and no one has answered​

Can someone be kind enough and help me I've posted my question many times and no one has answered