All categories

3 years ago

(05.03 MC)What is the area, in square units, of the polygon shown here?

Mathematics

2 answers:

goldenfox [79]3 years ago

6 0

Answer:

13 square units

Step-by-step explanation:

Look at image attached

Divide shape into two rectangles

Rectangle 1:

width: 2 units

length: 5 - 1 = 4 units

AREA: 2 x 4 = 8 sq. units

Rectangle 2:

width: 1 units

length: 5 units

AREA: 1 x 5 = 5 sq. Units

Combine area of rectangles:

8 + 5 = 13 sq. Units

That is the area of the whole polygon.

djverab [1.8K]3 years ago

5 0

To resolve the proposed issue, an explanation is needed in which the subject is addressed

You might be interested in

Anybody know the answers to this frl it’s due in 15 mins

hammer [34]

Answer:

4. j(11) = 8

I think it is linear

5. g(-3) = 18

No, it's a parabola since x is to the second power

6. f(-7) = 5

No, I think it's an absolute value function

7. h(0) = 13

I think it is linear, just the slope is 0

Step-by-step explanation:

3 0

4 years ago

Read 2 more answers

Question 1 of 10<br>Which of the following functions is graphed below?

Sedaia [141]

Attach picture, where is below

3 0

3 years ago

Can you please provide me step by step instructions on how to work this problem?

IgorC [24]

Whats the problem if u forgot type it over and type the problem

8 0

3 years ago

The answer is 585 R4 and could also be written as 585 4/8. Simplify the answer.

Digiron [165]

Answer:

585 1/2 ?

Step-by-step explanation:

i dont know if you put the whole question but i simplified 4/8 and got 1/2.

so i guess the answer your looking for is 585 1/2 ...

4 0

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

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