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3 years ago

Can some help me with this I gotta get to 50 im on 48 this the last one

Mathematics

1 answer:

kaheart [24]3 years ago

7 0

Answer:

Step-by-step explanation:

K (-5, 1)

L (0, 1)

M (0, 3)

N (-5, 3)

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What can I multiply 4/3 by to make it 1⅓, jk what's up

Simora [160]

1.<u>3 </u>x 1= 1.<u>3 </u>hope this helped

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3 years ago

What is 4/11 in decimal form?

Rudiy27

4/11 is 0.363636363636
And "36" just keeps repeating

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3 years ago

Find the associated reflex angle measure for an angle of 60°

sergij07 [2.7K]

The reflex angle is the other angle
see image below

so
reflex=360-angle
reflex=360-60=300
reflex angle=300

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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

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.

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1 year ago

A 20-sided regular polygon has an angle measure represented as 3gº.

Andre45 [30]

The value of g in the 20 sided regular polygon is 54.

<h3>How to find the angles of a regular polygon?</h3>

If all the polygon sides and interior angles are equal, then they are known as regular polygons.

The polygon given is a 20 sided regular polygon and the measure of each angle is 3g degrees.

Therefore, let's find g.

The sum of interior angles of a 20 sided polygon is as follows:

180(n - 2) = 180(20 - 2) = 180(18) = 3240

Therefore,

each angle = 3240 / 20 = 162

Hence,

162 = 3g

g = 162 / 3

Therefore,

g = 54

learn more on regular polygon here: brainly.com/question/16992545

#SPJ1

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1 year ago