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Nostrana [21]
2 years ago
9

Solve for x: (2x/3) + 10 = (x/5) + (36/5)

Mathematics
1 answer:
Reil [10]2 years ago
4 0

Answer:

x = -6

Step-by-step explanation:

(2x/3) + 10 = (x/5) + (36/5)

15*[(2x/3) + 10 = (x/5) + (36/5)]   <em>[Multiply both sides by 15 in order to convert the fractions to whole numbers]</em>

(15/3)*2x +15*10 = (15/5)x + (15/5)*36

5*2x + 150 = 3x + 3*36

10x +150 = 3x + 108

7x = -42

x = -6

===

<u>Check:</u>

Does (2x/3) + 10 = (x/5) + (36/5) work for x = -6?

(2(-6)/3) + 10 = (-6/5) + (36/5) ?

-4 + 10 = (30/5) ?

6 = 6  YES

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In hopes of encouraging healthier snacks at school, Leah brought in a tray of carrot sticks
aleksandr82 [10.1K]

Answer:

40 %  of the snacks were carrot sticks.

Step-by-step explanation:

Total number of carrot sticks in lunch = 18

Total number of apple slices in lunch = 27

So, Total Snacks in the Lunch = Carrot Sticks + Apple Slices

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Now, \textrm{Percentage of Carrot Sticks}  = \frac{\textrm{The number of Carrot sticks in lunch}}{\textrm{Total snacks in the lunch}}  \times 100

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6 0
3 years ago
The half-life of cesium-137 is 30 years. Suppose we have a 180-mg sample. (a) Find the mass that remains after t years. (b) How
DedPeter [7]

Answer:

a) Q(t) = 180e^{-0.023t}

b) 11.4mg of cesium-137 remains after 120 years.

c) 225.8 years.

Step-by-step explanation:

The following equation is used to calculate the amount of cesium-137:

Q(t) = Q(0)e^{-rt}

In which Q(t) is the amount after t years, Q(0) is the initial amount, and r is the rate at which the amount decreses.

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The half-life of cesium-137 is 30 years.

This means that Q(30) = 0.5Q(0). We apply this information to the equation to find the value of r.

Q(t) = Q(0)e^{-rt}

0.5Q(0) = Q(0)e^{-30r}

e^{-30r} = 0.5

Applying ln to both sides of the equality.

\ln{e^{-30r}} = \ln{0.5}

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r = \frac{\ln{0.5}}{-30}

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Q(t) = Q(0)e^{-0.023t}

180-mg sample, so Q(0) = 180

Q(t) = 180e^{-0.023t}

(b) How much of the sample remains after 120 years?

This is Q(120).

Q(t) = 180e^{-0.023t}

Q(120) = 180e^{-0.023*120}

Q(120) = 11.4

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(c) After how long will only 1 mg remain?

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t = \frac{\ln{0.00556}}{-0.023}

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