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Luda [366]
2 years ago
5

What is the surface area of the right cone below?

Mathematics
1 answer:
maksim [4K]2 years ago
6 0

Answer:

SA=54\pi\ units^2

Step-by-step explanation:

we know that

The surface area of the cone is given by the formula

SA=\pi r^{2}+\pi rl

where

r is the radius of the base

l is the slant height

we have

r=3\ units\\l=15\ units

substitute

SA=\pi (3)^{2}+\pi (3)(15)

SA=54\pi\ units^2

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In a triangle ABC,AB=9 and BC =12 which of the following Cannot be the length of AC.
galben [10]

Step-by-step explanation:

mark it as the brainliest

5 0
2 years ago
Translate and solve: seventy-eight times n is at least 312.
RideAnS [48]

It translates to 78n \ge 312 which solves to n \ge 4 after dividing both sides by 78. This means n is equal to 4 or it can be larger than 4.

7 0
1 year ago
EF is located at E (x,y) and F (7,-10) and was rotated around the origin. EF became E’F’ with coordinates E’ (4,5) and For (10,7
monitta

Answer:

  90° CCW

Step-by-step explanation:

If we assume you intend F'(10, 7), point F has been transformed by the rule ...

  (x, y) ⇒ (-y, x)

The transformation rule (x, y) ⇒ (-y, x) represents a 90° CCW rotation.

5 0
1 year ago
A large tank is partially filled with 100 gallons of fluid in which 20 pounds of salt is dissolved. Brine containing 1 2 pound o
Valentin [98]

Answer:

47.25 pounds

Step-by-step explanation:

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

<u>First, we determine the Rate In</u>

Rate In=(concentration of salt in inflow)(input rate of brine)

=(0.5\frac{lbs}{gal})( 6\frac{gal}{min})\\R_{in}=3\frac{lbs}{min}

Change In Volume of the tank, \frac{dV}{dt}=6\frac{gal}{min}-4\frac{gal}{min}=2\frac{gal}{min}

Therefore, after t minutes, the volume of fluid in the tank will be: 100+2t

<u>Rate Out</u>

Rate Out=(concentration of salt in outflow)(output rate of brine)

R_{out}=(\frac{A(t)}{100+2t})( 4\frac{gal}{min})\\\\R_{out}=\frac{4A(t)}{100+2t}

Therefore:

\dfrac{dA}{dt}=3-\dfrac{4A(t)}{100+2t}\\\\\dfrac{dA}{dt}=3-\dfrac{4A(t)}{2(50+t)}\\\\\dfrac{dA}{dt}=3-\dfrac{2A(t)}{50+t}\\\\\dfrac{dA}{dt}+\dfrac{2A(t)}{50+t}=3

This is a linear differential equation in standard form, therefore the integrating factor:

e^{\int \frac{2}{50+t}dt}=e^{2\ln|50+t|}=e^{\ln(50+t)^2}=(50+t)^2

Multiplying the DE by the integrating factor, we have:

(50+t)^2\dfrac{dA}{dt}+(50+t)^2\dfrac{2A(t)}{50+t}=3(50+t)^2\\\{(50+t)^2A(t)\}'=3(50+t)^2\\$Taking the integral of both sides\\\int \{(50+t)^2A(t)\}'= \int 3(50+t)^2\\(50+t)^2A(t)=(50+t)^3+C $ (C a constant of integration)\\Therefore:\\A(t)=(50+t)+C(50+t)^{-2}

Initially, 20 pounds of salt was dissolved in the tank, therefore: A(0)=20

20=(50+0)+C(50+0)^{-2}\\20-50=C(50)^{-2}\\C=-\dfrac{30}{(50)^{-2}} =-30X50^2=-75000

Therefore, the amount of salt in the tank at any time t is:

A(t)=(50+t)-75000(50+t)^{-2}

After 15 minutes, the amount of salt in the tank is:

A(15)=(50+15)-75000(50+15)^{-2}\\=47.25$ pounds

8 0
1 year ago
The quotient of 38 times a number and -4
nikdorinn [45]

Answer:

140y

Step-by-step explanation:

35 times a number times -4 is 35×y×(-4) which is -140y

4 0
1 year ago
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