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3 years ago

Plz help ASAP Plz plz plz help ASAP the pic is there so you can see the question

Mathematics

1 answer:

velikii [3]3 years ago

5 0

Answer:

If going from top to bottom.

The 2nd And 5th choices are correct.

Explamation:

2nd choice: The shortest feather's length is 1 7/8 inches. The Longest feather's length is 4 2/8 inches. If you subtract 1 7/8 from 4 2/8 you get 2 3/8.

5th choice: The shortest feather's length is 1 7/8 inches. The Longest feather's length is 4 2/8 inches. If you add 4 2/8 & 1 7/8 you get 6 1/8.

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I need to find the area of the crater

expeople1 [14]

First you need to find the r, if C=2pieR, than 2.4=2pieR now you divide 2.4 by 2 so you got 1.2=pieR now you divide 1.2 by pie and obtained aproximately 0.38 miles, that is your R, now you do pie multiply by 0.38 exposant 2 and there you go, your answer is aproximately 0.45 miles2 :)

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4 years ago

Help me please!!!!!

nadya68 [22]

Man tbh wth is this at this point just trow that mug out the window

5 0

3 years ago

A rectangle has a perimeter of 18 cm.its length is 5 cm greater than its width.find the dimensions.

Marat540 [252]

P=18
l=5 cm greater than the width
w=5 less than the length

USE YOUR NUMBER SENSE!!! - just plug in different number below 4 to figure out the width

3=w 3+5=8 8+8=16+6=22 WILL NOT WORK

2=w 2+5=7 7+7=14 14+2+2= 18 WE HAVE A WINNER!!!!

FInal Answer: width: 2 cm and length: 7 cm

5 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

I need help please?!!!!

sergejj [24]

A. i’m pretty sure !!

6 0

3 years ago

Read 2 more answers