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sladkih [1.3K]
3 years ago
11

Find the value of the expression 3m2 + 2p2 − 15 when m = 3 and p = 10.

Mathematics
2 answers:
photoshop1234 [79]3 years ago
6 0
The answer would be 212 :)
Marianna [84]3 years ago
3 0

Answer:

212

Step-by-step explanation:

The given expression is 3m^2+2p^2-15

Substituting m = 3 and p = 10 in the expression

3(3)^2+2(10)^2-15

Now, simplify this expression

3\times9+2\times100-15\\\\=27+200-15\\\\=227-15\\\\=212

Therefore, the value of the expression is 212.

Third option is correct.

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Misha Larkins [42]

Answer:

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Step-by-step explanation:

By geometric mean theorem and Pythagoras theorem:

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3 0
3 years ago
A 200-gal tank contains 100 gal of pure water. At time t = 0, a salt-water solution containing 0.5 lb/gal of salt enters the tan
Artyom0805 [142]

Answer:

1) \frac{dy}{dt}=2.5-\frac{3y}{2t+100}

2) y(t)=(50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}

3) 98.23lbs

4) The salt concentration will increase without bound.

Step-by-step explanation:

1) Let y represent the amount of salt in the tank at time t, where t is given in minutes.

Recall that: \frac{dy}{dt}=rate\:in-rate\:out

The amount coming in is 0.5\frac{lb}{gal}\times 5\frac{gal}{min}=2.5\frac{lb}{min}

The rate going out depends on the concentration of salt in the tank at time t.

If there is y(t) pounds of  salt and there are 100+2t gallons at time t, then the concentration is: \frac{y(t)}{2t+100}

The rate of liquid leaving is is 3gal\min, so rate out is =\frac{3y(t)}{2t+100}

The required differential equation becomes:

\frac{dy}{dt}=2.5-\frac{3y}{2t+100}

2) We rewrite to obtain:

\frac{dy}{dt}+\frac{3}{2t+100}y=2.5

We multiply through by the integrating factor: e^{\int \frac{3}{2t+100}dt }=e^{\frac{3}{2} \int \frac{1}{t+50}dt }=(50+t)^{\frac{3}{2} }

to get:

(50+t)^{\frac{3}{2} }\frac{dy}{dt}+(50+t)^{\frac{3}{2} }\cdot \frac{3}{2t+100}y=2.5(50+t)^{\frac{3}{2} }

This gives us:

((50+t)^{\frac{3}{2} }y)'=2.5(50+t)^{\frac{3}{2} }

We integrate both sides with respect to t to get:

(50+t)^{\frac{3}{2} }y=(50+t)^{\frac{5}{2} }+ C

Multiply through by: (50+t)^{-\frac{3}{2}} to get:

y=(50+t)^{\frac{5}{2} }(50+t)^{-\frac{3}{2} }+ C(50+t)^{-\frac{3}{2} }

y(t)=(50+t)+ \frac{C}{(50+t)^{\frac{3}{2} }}

We apply the initial condition: y(0)=0

0=(50+0)+ \frac{C}{(50+0)^{\frac{3}{2} }}

C=-12500\sqrt{2}

The amount of salt in the tank at time t is:

y(t)=(50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}

3) The tank will be full after 50 mins.

We put t=50 to find how pounds of salt it will contain:

y(50)=(50+50)- \frac{12500\sqrt{2} }{(50+50)^{\frac{3}{2} }}

y(50)=98.23

There will be 98.23 pounds of salt.

4) The limiting concentration of salt is given by:

\lim_{t \to \infty}y(t)={ \lim_{t \to \infty} ( (50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }})

As t\to \infty, 50+t\to \infty and \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}\to 0

This implies that:

\lim_{t \to \infty}y(t)=\infty- 0=\infty

If the tank had infinity capacity, there will be absolutely high(infinite) concentration of salt.

The salt concentration will increase without bound.

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3 years ago
Need help with slope practice assignment eight greade
Viktor [21]

Answer: Option D is correct. The function equation y= -16x^2 + 9x + 4

is non-linear.

Step-by-step explanation:

It is not linear because it is a quadratic function.

You can see this on the graph in the photo below:

,  Hope this helps :)

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Isaac played a video game 20 times and wont about 70% of the games. how many games did he win
skelet666 [1.2K]
Hello!

We know that Isaac played 20 games and won about 70% of them. Now, we must find what 70% of 20 is, so that we know how many games Isaac won.

First, convert the percentage to a decimal by dividing it by 100.

70% ÷ 100 = 0.70

Next, multiply the total amount of games Isaac played (20) by about how many he won (0.70):

20 × 0.70 = 14

ANSWER:

Isaac won about 14 games.
4 0
3 years ago
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What is the angle of rotation about the center that will map the regular pentagon onto itself
BlackZzzverrR [31]

Answer:

I think c 430 and if it is wrong I'm so sorry

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