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3 years ago

Can two numbers have a GCF that is bigger than their LCM? Explain how you know?

1 answer:

7 0

The answer to this question would be: No, they can't

Greatest common factor biggest possible value is the number value itself. There is no factor that bigger than the number.

But the least common multiple smallest possible value is the number. There is no multiple that smaller than the number.

So it is not possible to have GCF bigger than LCM.

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Question 5 (1 point)

dexar [7]

Answer:

4.5

Step-by-step explanation:

8 + 7 + 6 + 5 + 4 + 3 + 2 + 1/8

36/8

=4.5

3 0

2 years ago

Help me please I'm terrible at math

mr_godi [17]

Answer:

Step-by-step explanation:

ok first you make a equation which is 2x+1+5x-14

then you combine like terms so 2x+5x= 7x and -14+1=-13

so now the equation is 7x-13

but i need to know the full length of AC to finish it

3 0

2 years ago

2y = x + 3 5y = x - 7 What is the solution set of the given system?

olga_2 [115]

-3=2x+10= -13=2x = -0.153

5 0

2 years ago

Show work only do those 32,34,36,38,40

Free_Kalibri [48]

Answer: see below

Step-by-step explanation:

$32)\quad \dfrac{3}{5}(x-12)>x-24$

3(x - 12) > 5(x - 24)

3x - 36 > 5x - 120

-5x -5x

-2x - 36 > -120

+36 +36

-2x > -84

÷ -2 ↓ ÷ -2

x < 42

Graph: ←------------o

34) 6[5y - (3y - 1)] ≥ 4(3y - 7)

6[5y - 3y + 1] ≥ 4(3y - 7)

6{2y + 1] ≥ 4(3y - 7)

12y + 6 ≥ 12y - 28

-12y -12y

6 ≥ -28

TRUE so the solution is All Real Numbers

Graph: ←-----------------------→

36) BC + AC > AB

4 + 8 - AB > AB

12 - AB > AB

+AB +AB

12 > 2AB

÷2 ÷2

6 > AB

AB < 6

$38)\quad \dfrac{1}{2}(y-16)\geq y+2\quad ;\text{claim}\ y\leq 20$

Check: let y = 16

then $\frac{1}{2}$ (16 - 16) ≥ 16 + 2

0 ≥ 18

FALSE so the claim is wrong

40) question not provided in the image so I cannot give a solution.

7 0

3 years ago

Help me ASAP ASAP plzzzzz

kykrilka [37]

Answer:

Step-by-step explanation:

22 is the answer cppyphxhpcpyx

7 0

2 years ago