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3 years ago

What is 2x - 1 < 7 and 5x + 3 < 3.

Mathematics

2 answers:

schepotkina [342]3 years ago

6 0

1. x<4
2. x<0

Hope this helps.

lara [203]3 years ago

4 0

It would be 2x<8 and 5x<0

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I don’t know how to solve for m/ the slope and b/ the y-intercept on a horizontal line.

pav-90 [236]

Answer:

the y-intercept is where u start on a graph like where it crosses through the middle line and the slope it the rise and run so rise is how much it goes up and run is how much u go over.

hope this helps.

8 0

3 years ago

8<br> Evaluate 3n+ 7.<br> n = 3<br> Find the indicated sum? Help!

lubasha [3.4K]

Answer:

<h3>The answer is 16</h3>

Step-by-step explanation:

3n + 7

n = 3

Substitute the value of n into the expression

That's

3(3) + 7

= 9 + 7

Hope this helps you

6 0

4 years ago

Read 2 more answers

Three decreased by eleven times a number

Vadim26 [7]

Step-by-step explanation:

3 - 11x

x is "a number".

so, the request is to multiply it by 11 and subtract the result from 3.

6 0

2 years ago

A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so

Debora [2.8K]

Let $A(t)$ = amount of salt (in pounds) in the tank at time $t$ (in minutes). Then $A(0) = 11$ .

Salt flows in at a rate

$\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}$

and flows out at a rate

$\left(\dfrac{A(t)\,\rm lb}{75\,\rm gal + \left(3\frac{\rm gal}{\rm min} - 3.25\frac{\rm gal}{\rm min}\right)t}\right) \left(3.25\dfrac{\rm gal}{\rm min}\right) = \dfrac{13A(t)}{300-t} \dfrac{\rm lb}{\rm min}$

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.

Then the net rate of salt flow is given by the differential equation

$\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}$

which I'll solve with the integrating factor method.

$\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95$

$-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}$

$\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}$

Integrate both sides. By the fundamental theorem of calculus,

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}$

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)$

$\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}$

$\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}$

$\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}$

After 1 hour = 60 minutes, the tank will contain

$A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131$

pounds of salt.

7 0

2 years ago

The tablet below represents ordered pairs that satisfy the functions f(x) and g(x). If f(x)=4x, which statements are true of g(x

dexar [7]

B and D are the correct answers.

8 0

3 years ago

Read 2 more answers