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3 years ago

Translate each expression

Mathematics

1 answer:

melomori [17]3 years ago

4 0

Answer:

<em>Solving the first 6</em>
<em>Assumed the number is x</em>

4x - 1

(2/3)x + 7

m - n

x² - 9

2x/5

(1/4)x + 27

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And outdoor deck is 7 feet wide. The perimeter of the deck is 64 feet. What is the length of the deck?

vova2212 [387]

Well perimeter is lengthx2 and widthx2 so 7x2=14
then 64-14=50
50÷2= 25

length=25

4 0

3 years ago

25>0.99x+3.99 show all work to solve the inequality

Aleonysh [2.5K]

25>99/100x+399/100
2500>99x+399
2101>99x
2101/99>x

8 0

3 years ago

Help please <br> And thank you so much

Shkiper50 [21]

Answer:

$0.1 /lb.

Step-by-step explanation:

First, divide 1.99 by 19. You get 0.1047368421. To round it to the nearest hundred, look to the number 3 points away from the decimal, which is number 4. Since it is not 5 or greater, we round it down, and it would round down to 0.1.

8 0

4 years ago

A tank originally contains 100 gallon of fresh water. Then water containing 0.5 Lb of salt per gallon is pourd into the tank at

Assoli18 [71]

Let $S(t)$ denote the amount of salt (in lbs) in the tank at time $t$ min up to the 10th minute. The tank starts with 100 gal of fresh water, so $S(0)=0$ .

Salt flows into the tank at a rate of

$\left(0.5\dfrac{\rm lb}{\rm gal}\right) \left(2\dfrac{\rm gal}{\rm min}\right) = 1\dfrac{\rm lb}{\rm min}$

and flows out with rate

$\left(\dfrac{S(t)\,\rm lb}{100\,\mathrm{gal} + \left(2\frac{\rm gal}{\rm min} - 2\frac{\rm gal}{\rm min}\right)t}\right) \left(2\dfrac{\rm gal}{\rm min}\right) = \dfrac{S(t)}{50} \dfrac{\rm lb}{\rm min}$

Then the net rate of change in the salt content of the mixture is governed by the linear differential equation

$\dfrac{dS}{dt} = 1 - \dfrac S{50}$

Solving with an integrating factor, we have

$\dfrac{dS}{dt} + \dfrac S{50} = 1$

$\dfrac{dS}{dt} e^{t/50}+ \dfrac1{50}Se^{t/50} = e^{t/50}$

$\dfrac{d}{dt} \left(S e^{t/50}\right) = e^{t/50}$

By the fundamental theorem of calculus, integrating both sides yields

$\displaystyle S e^{t/50} = Se^{t/50}\bigg|_{t=0} + \int_0^t e^{u/50}\, du$

$S e^{t/50} = S(0) + 50(e^{t/50} - 1)$

$S = 50 - 50e^{-t/50}$

After 10 min, the tank contains

$S(10) = 50 - 50e^{-10/50} = 50 \dfrac{e^{1/5}-1}{e^{1/5}} \approx 9.063 \,\rm lb$

of salt.

Now let $\hat S(t)$ denote the amount of salt in the tank at time $t$ min after the first 10 minutes have elapsed, with initial value $\hat S(0)=S(10)$ .

Fresh water is poured into the tank, so there is no salt inflow. The salt that remains in the tank flows out at a rate of

$\left(\dfrac{\hat S(t)\,\rm lb}{100\,\mathrm{gal}+\left(2\frac{\rm gal}{\rm min}-2\frac{\rm gal}{\rm min}\right)t}\right) \left(2\dfrac{\rm gal}{\rm min}\right) = \dfrac{\hat S(t)}{50} \dfrac{\rm lb}{\rm min}$