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ra1l [238]
2 years ago
8

PLEASE HELPPP. I’ve been stuck on this for so long

Mathematics
1 answer:
nydimaria [60]2 years ago
3 0

Answer:

D. 2.5

I hope it's help.......

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What is the solution to this inequality: -4(3x-5)<2(3x+4)​
Delicious77 [7]

Answer:

x > 2/3

Step-by-step explanation:

-4(3x - 5) < 2(3x + 4)​

Divide both sides by 2.

-2(3x - 5) < 3x + 4

Distribute on the left side.

-6x + 10 < 3x + 4

Subtract 3x from both sides.

-9x + 10 < 4

Subtract 10 from both sides.

-9x < - 6

Divide both sides by -9. Remember that when you divide both sides of an inequality by a negative number, the inequality sign changes direction.

x > (-6)/(-9)

x > 6/9

x > 2/3

7 0
3 years ago
In the proportion 2:3 = 16:24 ,the product of the two middle values is __.
klio [65]

Answer:

48 and 48

Step-by-step explanation:

product means multiplication

two middle values are 3 and 16

3*16 = 48

two outside values are 2 and 24

2*24 = 48

4 0
3 years ago
HELP PLEASE! What does it mean if the probability is 20% that a spinner will stop in a red section? (I think it's C.)
Svetradugi [14.3K]
80% not stop on red so the answer would be c for that
8 0
3 years ago
What is the distance between point A and point B?
DIA [1.3K]
The answer is possibly 10
4 0
3 years ago
Read 2 more answers
A factory produce bicycles at a rate of 110+0.5t^2-0.9t bicycles per week (t in weeks)
klasskru [66]

Answer:

221

Step-by-step explanation:

Day 8 is the first day of the second week.

Day 21 is the last day of week 3.

We need to know the n umber of bicycles made from t = 1 to t = 3

The function is b(t) = 110 + 0.5t^2 - 0.9t, where t is in weeks.

We need to integrate the function with the limits of 1 to 3.

\int_{1}^{3} (110 + 0.5t^2 - 0.9t) dt

\int_{1}^{3} (110 + \dfrac{t^2}{2} - \dfrac{9t}{10}) dt

= 110t + \dfrac{t^3}{6} - \dfrac{9t^2}{20} \Biggr|_{1}^{3}

= 330.45 - 109.72

= 220.7333

Answer: 221

3 0
3 years ago
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