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3 years ago

Could someone please explain surface area to me?Thanks will be rewarded

2 answers:

7 0

Surface area of a solid is the sum of all areas of the faces of the shape so if you have the net set up you will just have to find the area of each face and add them up!

solong [7]3 years ago

5 0

Ok its like for rectangular prisim its the volume here is a formula V=lwh. There you go use that for anything i do k12 so message me if you need anything.

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A red light

ANEK [815]

Answer:

Both lights would flash together after every 30 seconds.

Step-by-step explanation:

6 = 2 * 3 and 10 = 2 * 5.

LCM (Least Common Multiple) of 6 and 10 = 2 * 3 * 5 = 30.

Therefore, both lights wold flash together after every 30 seconds.

3 0

3 years ago

Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br

aliya0001 [1]

Answer:

$A=1500-1450e^{-\dfrac{t}{250}}$

Step-by-step explanation:

The large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved.

Volume = 500 gallons

Initial Amount of Salt, A(0)=50 pounds

Brine solution with concentration of 2 lb/gal is pumped into the tank at a rate of 3 gal/min

$R_{in}$ =(concentration of salt in inflow)(input rate of brine)

$=(2\frac{lbs}{gal})( 3\frac{gal}{min})\\R_{in}=6\frac{lbs}{min}$

When the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

Concentration c(t) of the salt in the tank at time t

Concentration, $C(t)=\dfrac{Amount}{Volume}=\dfrac{A(t)}{500}$

$R_{out}$ =(concentration of salt in outflow)(output rate of brine)

$=(\frac{A(t)}{500})( 2\frac{gal}{min})\\R_{out}=\dfrac{A}{250}$

Now, the rate of change of the amount of salt in the tank

$\dfrac{dA}{dt}=R_{in}-R_{out}$

$\dfrac{dA}{dt}=6-\dfrac{A}{250}$

We solve the resulting differential equation by separation of variables.

$\dfrac{dA}{dt}+\dfrac{A}{250}=6\\$The integrating factor: e^{\int \frac{1}{250}dt} =e^{\frac{t}{250}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{250}}+\dfrac{A}{250}e^{\frac{t}{250}}=6e^{\frac{t}{250}}\\(Ae^{\frac{t}{250}})'=6e^{\frac{t}{250}}$

Taking the integral of both sides

$\int(Ae^{\frac{t}{250}})'=\int 6e^{\frac{t}{250}} dt\\Ae^{\frac{t}{250}}=6*250e^{\frac{t}{250}}+C, $(C a constant of integration)\\Ae^{\frac{t}{250}}=1500e^{\frac{t}{250}}+C\\$Divide all through by e^{\frac{t}{250}}\\A(t)=1500+Ce^{-\frac{t}{250}}$

Recall that when t=0, A(t)=50 (our initial condition)

$50=1500+Ce^{-\frac{0}{250}}50=1500+Ce^{0}\\C=-1450\\$Therefore the amount of salt in the tank at any time t is:\\A=1500-1450e^{-\dfrac{t}{250}}$

4 0

3 years ago

Suppose you make coffee and it starts off just too hot to drink, so you put it in the refrigerator temporarily to cool it off. I

Mariana [72]

Uh..... Cold enough to drink?

8 0

3 years ago

Please no links or files I need a direct answer.

Hatshy [7]

Answer:

$\text{a. }128\:\mathrm{cm^3},\\\text{b. }4:1,\\\text{c. }179.4\:\mathrm{cm^2};\:44.8\:\mathrm{cm^2}$

Step-by-step explanation:

The volume of a square pyramid is given by $V=\frac{1}{3}\cdot s^2\cdot h$

Since pyramids A and B are similar, their corresponding side lengths are proportion. Since the base edge of B is half that of A's, each dimension of B will be half of A. Since the formula for volume requires the multiplication of three dimensions, the volume of pyramid A will be $2^3=8$ times larger than the volume of pyramid B.

Thus, the volume of pyramid A is equal to $16\cdot 8=\boxed{128\:\mathrm{in^3}}$ . You can also find the dimensions of pyramid A and use the formula.

The surface area consists of the sum of all areas of the 2D shapes that make the figure. Since all dimensions of pyramid A are twice the dimensions of pyramid B, the ratio of the surface area of pyramid A to pyramid B is $2^2:1=\boxed{4:1}$

The surface area of a square pyramid can be found by adding the areas of the three triangles and one square that make it up. As one long messy formula, that becomes $2s\sqrt{\left(\frac{s}{2}\right)^2+h^2}+s^2$ for a square pyramid with base edge $s$ and height $h$ .