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liraira [26]
3 years ago
13

if a number is increased by 5 and the result is then divided by three,the result is seven. write an equation that models this ve

rbal description and solve the equation for the number described
Mathematics
2 answers:
gregori [183]3 years ago
6 0
A number (we don’t know the number so I’ll use X) plus five. So we can start the equation of with X+5 then divided by 3=7 so X+5➗3=7 so to solve what X is we can just go in reverse. 7 TIMES 3= 21-5=16 so X=16
Gekata [30.6K]3 years ago
5 0

Answer:

The required equation is \frac{x+5}{3}=7

The value of x is 16.

Step-by-step explanation:

Consider the provided information.

if a number is increased by 5 and the result is then divided by three,the result is seven.

Let the number is represented by x.

Increase the number by 5: x+5

Now divide the result by 3, and result is seven.

This can be written as:

\frac{x+5}{3}=7

Now solve the equation for x as shown.

\frac{x+5}{3}=7

x+5=21

Subtract 5 from both sides.

x=16

Hence the value of x is 16.

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olga55 [171]

The problem here is:  Too many words.

If you boil the problem down to the bare facts, the question is:

             "$210 is 15% of what number ?"

I'm pretty sure you can take it from there,
but here's the rest of it anyway:

                                                $210  =  15% of (March amount)

                                                $210  =  0.15 x (March amount)

Divide each side by  0.15 :     $1400 = March amount .

Check:
1400 + 210 =  1610
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5 0
3 years ago
The Pacific halibut fishery has been modeled by the differential equation dy dt = ky 1 − y K where y(t) is the biomass (the tota
Dafna1 [17]

Answer:

a. The biomass weighs 2.30 * 10^7 kg after a year

b. It'll take 2.56 years to get to 4*10^7kg

Step-by-step explanation:

a.

k = 0.78,K = 6E7 kg

Given

dy/dt = ky(1- y/K)

Make ky dt the subject of formula

ky dt = dy/(1-y/K) --- make k dt the subject of formula

k dt = dy/(y(1-y/K))

k dt = K dy / y(K-y)

k dt = ((1/y) + (1/(K-y)))dy ---- integrate both sides

kt + c = ln(y/(K-y))

Ce^(kt) = y/(K-y)

Substitute the values of k and K

Ce^(0.78t) = y/(6*10^7 - y) ----- (1)

Given that y(0) = 2 * 10^7kg

(1) becomes

Ce^(0.78*0) = (2 * 10^7)/(6*10^7 - 2*10^7)

Ce° = (2*10^7)/(4*10^7

C = 2/7

Substitute 2/7 for C in (1)

2/7e^0.78t = y/(6*10^7 - y) ---(2)

We're to find the biomass a year later

So, t = 1

2/7e^0.78 = y/(6*10^7 - y)

0.62 = y/(6*10^7 - y)

y = 0.62(6*10^7 - y)

y = 0.62*6*10^7 - 0.62y

y + 0.62y = 0.62*6*10^7

1.62y = 0.62*6*10^7

1.62y = 3.72 * 10^7

y = 2.30 * 10^7kg.

Hence, the biomass weighs 2.30 * 10^7 kg after a year

b.

Here, we're to calculate the time it'll take the biomass to get to 4*10^7 kg

Substitute 4*10^7 for y in (2)

2/7e^0.78t = 4*10^7/(6*10^7 - 4*10^7)

2/7e^0.78t = 4*10^7/2*10^7

2/7e^0.78t = 2

e^0.78t = (2*7)/2

e^0.78t = 2

t = 2 * 1/0.78

t = 2.56 years

Hence, it'll take 2.56 years to get to 4*10^7kg

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Now, we can compare the values, 

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Step-by-step explanation:

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