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stich3 [128]
3 years ago
14

Pythagorean theorem help plz i really need help plz i dont really get it cuz i forgot

Mathematics
2 answers:
muminat3 years ago
7 0
7.62 a^2+b^2=c^2 have a great day
solmaris [256]3 years ago
7 0

Answer:

7.62 the answer is already rounded.

Step-by-step explanation:

C= the square root of A² + B²

c= square root of 7²+3²= 7.615773106=7.62

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What is one half added to seven tenths
Ivahew [28]
1. You need to change these so that they share a common denominator (bottom of the fraction)
2. Common denominator is 10 of these 2 fractions.
3. We don't need to change the 7/10 fraction because the denominator is already 10.
4. To make 1/2 have a denominator of 10, we need to times the top and bottom by the same number. To get 10 from 2, we have to times it by 5. Therefore, to change the top, we need to times it also by 5
5. 1 x 5 = 5
6. Add the two fractions together: 5/10 + 7/10 = 12/10
7. Simplify the expression. This is currently a top heavy fraction aka. improper fraction. To make it 'proper' we need to change it into a mixed number. To do this, we see how many of the bottom number goes into the top. In this case it is 1 with 2 left over. So the final fraction is...

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7 0
3 years ago
Read 2 more answers
Steven’s mother asked him to buy 750 grams of sugar. At the supermarket, he finds that the sugar packages give the weight in kil
9966 [12]
1, 000 grams = 1 kg
(divide both sides by 1, 000)
1 gram = 1/1, 000 kg
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Therefore, Steven needs to buy 0.75 kg of sugar!

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7 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}}

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3 years ago
What is 6.1 - 18.7 ?​
rodikova [14]

Answer:

-12.6

Step-by-step explanation:

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3 years ago
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I NEED HELP PLEASE!!!!!!!!!!
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