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Ierofanga [76]
3 years ago
13

The least common multiple of two numbers is 60 and one of the numbers is 7 than the other numbers what are the numbers justify t

he answers
Mathematics
1 answer:
I am Lyosha [343]3 years ago
8 0

The correct numbers are 12 and 5. This is because these two are the only numbers out of all the factors of 60 that fit this description.

I really hope my answer helped you out! c:
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Which of the following inequalities is represented by the following graph?
garik1379 [7]
I'll call the answer options A, B, C, and D, in that order from top to bottom. The line is a dotted line, so the inequality cannot be a greater/less than or equal to. Thus the answer choices B and D are eliminated. According to the graph, the point (0,0) is one of the possible solutions of the inequality. So, just substitute (0,0) into the inequalities for A and C. For C, you get 0 < -3, which is not true, so C is wrong. For A, you get 0 > -3. This is true, so the correct answer must be A, or y > -2x - 3
5 0
3 years ago
What is the answer for f 6n-3(2n-5)?
wariber [46]

Answer:

f 6 n +−6 n +15

Step-by-step explanation:

f 6 n+(−3)(2 n)+(−3)(−5)

f 6 n +−6 n +15

4 0
3 years ago
41 tens×4hundreds=<br> 410×400=
weqwewe [10]

410x400 is correct and your answer would be 164,000.

7 0
3 years ago
This is the variable, usually shown on the y-axis, that gets its value from the one on the x-axis.
Furkat [3]
You need to have a little more detail in your answer, sir or madam.
8 0
4 years ago
All questions to be solved using linear combination.
monitta
1)
I:x-y=-7
II:x+y=7

add both equations together to eliminate y:
x-y+(x+y)=-7+7
2x=0
x=0

insert x=0 into II:
0+y=7
y=7

the solution is (0,7)

2)
I: 3x+y=4
II: 2x+y=5

add I+(-1*II) together to eliminate y:
3x+y+(-2x-y)=4+(-5)
x=-1

insert x=-1 into I:
3*-1+y=4
y=7

the solution is (-1,7)

3)

I: 2e-3f=-9
II: e+3f=18

add both equations together to eliminate f:
2e-3f+(e+3f)=-9+18
3e=9
e=3

insert e=3 into I:
2*3-3f=-9
-3f=-9-6
-3f=-15
3f=15
f=5

the solution is (3,5)

4)
I: 3d-e=7
II: d+e=5

add both equations together to eliminate e:
3d-e+(d+e)=7+5
4d=12
d=3

insert d=3 into II:
3+e=5
e=2

the solution is (3,2)

5)
I: 8x+y=14
II: 3x+y=4

add I+(-1*II) together to eliminate y
8x+y+(-3x-y)=14-4
5x=10
x=2

insert x=2 into II:
3*2+y=4
y=4-6
y=-2

the solution is (2,-2)
8 0
3 years ago
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