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2 years ago

Evaluate math pls help algebra

Mathematics

2 answers:

Ne4ueva [31]2 years ago

4 0

Answer:

a. is equal to -10

b. is equal to -1

c. is equal to 5

Step-by-step explanation:

a :

3(-2)-4

-6-4

-10

b :

3(1)-4

3-4

-1

c :

3(3)-4

9-4

These are the correct answers,please mark brainliest.

Free_Kalibri [48]2 years ago

3 0

Answer:

A. -10

B. -1

C. 5

Step-by-step explanation:

A. 3x-4 = 3(-2)-4 = -10

B. 3x-4 = 3(1)-4 = -1

C. 3x-4 = 3(3)-4 = 5

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Solve the system of equations by substitution. 7+2y=8 3x-2y=0

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7+2y=8
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7+2y=8
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3x-2(8/3)=0
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3 years ago

What is the sum of the interior angels of a regular polygon with 7 sides. choices are

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Sum of Interior Angles of any poligon, where n is the number of sides: =

(n-2) × 180°

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3 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

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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

1 year ago

5 An army cook wants to make pudding

photoshop1234 [79]

Answer:

Step-by-step explanation:

First you dvide 864 by 4 to get 216. Next you multiply the 216 by 2 to find that there are 432 cups in 864 servings. Finally you divide by 16 to turn 432 cups into 27 gallons.

I hope this helps!

3 0

2 years ago

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Rasek [7]

Answer:

The hypotheses used in this situation

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