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aalyn [17]
3 years ago
11

Please please help me I’ve tried soooo many times

Mathematics
2 answers:
alexdok [17]3 years ago
7 0

Answer:

u = 2

Step-by-step explanation:

6u + 7 = 29 - 5u

6u + 5u + 7 = 29 - 5u + 5u

11u + 7 = 29

11u + 7 - 7 = 29 - 7

11u = 22

11u ÷ 11 = 22 ÷ 11

u = 2

scoundrel [369]3 years ago
5 0

value of u = 2

Hope it helps you...

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Please help me out!!!
Tasya [4]

Answer:

you hgave a 2 and you use 5 and you have to add it and it will be 12 and your answer is 15

Step-by-step explanation:

6 0
3 years ago
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The music teacher receives 9 boxes of recorders there are six recorders in each box she hands them out to third graders and has
Stels [109]

Answer:

49 third graders

Step-by-step explanation:

From the question:

The music teacher receives 9 boxes of recorders and there are six recorders in each box

The total number of recorders the teacher received =

1 box = 6 recorders

9 boxes = x

Cross Multiply

x = 9 × 6 recorders

x = 54 recorders.

She hands them out to third graders and has 5recorders left .

How many third graders received a recorder?

The amount of recorders shared =

54 recorders - 5 recorders

= 49 recorders

If 1 recorders was given to a third grader,

Therefore, the amount of third graders that received a recorder is 49 third graders

5 0
3 years ago
will's hair was originally 7 inches long. He asked her hair dresser to cut 2/8 of it off. How many inches did he have cut off?
saw5 [17]
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4 0
4 years ago
3 times negative 2 subtract 5 over negative 4
Nonamiya [84]
1/4 should be the answer :)
5 0
3 years ago
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May be used to model radioactive decay. Q represents the quantity remaining after t years; k is the decay constant for plutonium
MArishka [77]

Answer:

<em>The half life of the sample is </em><em>6301 years.</em>

Step-by-step explanation:

The function used to model radioactive decay or exponential decay is,

Q(t)=Q_0e^{-k\cdot t}

Where,

Q(t) = Quantity after t time

Q₀ = Initial

k = decay constant

t = time period

As after an half life, the amount becomes half so,

\Rightarrow \dfrac{Q_0}{2}=Q_0e^{-0.00011\cdot t}

\Rightarrow \dfrac{1}{2}=e^{-0.00011\cdot t}

Taking natural log of both sides,

\Rightarrow \ln \dfrac{1}{2}=\ln e^{-0.00011\cdot t}

\Rightarrow \ln \dfrac{1}{2}={-0.00011\cdot t}\times \ln e

\Rightarrow \ln \dfrac{1}{2}={-0.00011\cdot t}\times 1

\Rightarrow {-0.00011\cdot t}=\ln \dfrac{1}{2}

\Rightarrow t=\dfrac{\ln \dfrac{1}{2}}{-0.00011}

\Rightarrow t=6301.3\approx 6301\ years

8 0
4 years ago
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