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3 years ago

What is the length of the leg on a right triangle with the other leg measuring 15 inches and the hypotenuse measuring 18 inches

Mathematics

1 answer:

Olegator [25]3 years ago

3 0

$\huge \cal \blue{ANSWER}$

In this triangle by applying Pythagoras theorem.

H²=B²+P²

18²=15²+P²

324=225+p²

324-225=P²

√99=p

9.94

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Which one is greater 10.05 or 10.005

Blababa [14]

10.05 = 10 5/100 = 10 1/20
10.005 = 10 5/1000 = 10 1/200

therefore, 10.05 is greater then 10.005

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3 years ago

Find the value of the given expression.<br> (see the image)

Andrews [41]

Answer:

d. 14

Step-by-step explanation:

3(2) = 3 × 2 = 6
Plug 6 in: 12 + {10 ÷ [11 - 6]}
11 - 6 = 5
Plug 5 in: 12 + {10 ÷ 5}
10 ÷ 5 = 2
Plug 2 in: 12 + 2
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I hope this helps!

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3 years ago

Goes through the 2 points (-3,5) (4,5). what's the equation?

Ierofanga [76]

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3 years ago

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2 years ago

A 500500-gallon tank initially contains 200200 gallons of brine containing 100100 pounds of dissolved salt. brine containing 11

mixer [17]

Let $A(t)$ denote the amount of salt in the tank at time $t$ . We're given that the tank initially holds $A(0)=100$ lbs of salt.

The rate at which salt flows in and out of the tank is given by the relation

$\dfrac{\mathrm dA}{\mathrm dt}=\underbrace{\dfrac{11\text{ lb}}{1\text{ gal}}\times\dfrac{44\text{ gal}}{1\text{ min}}}_{\text{rate in}}-\underbrace{\dfrac{A(t)}{200+(44-11)t}\times\dfrac{11\text{ gal}}{1\text{ min}}}_{\text{rate out}}$
$\implies A'(t)+\dfrac{11}{200+33t}A(t)=484$

Find the integrating factor:

$\mu(t)=\exp\left(\displaystyle\int\frac{11}{200+33t}\,\mathrm dt\right)=(200+33t)^{1/3}$

Distribute $\mu(t)$ along both sides of the ODE:

$(200+33t)^{1/3}A'(t)+11(200+33t)^{-2/3}A(t)=484(200+33t)^{-1/3}$
$\bigg((200+33t)^{1/3}A(t)\bigg)'=484(200+33t)^{-1/3}$
$A(t)=484\displaystyle\int(200+33t)^{-1/3}\,\mathrm dt$
$A(t)=22(200+33t)^{2/3}+C$

Since $A(0)=100$ , we get

$100=22(200)^{2/3}+C\implies C\approx-652.39$

so that the particular solution for $A(t)$ is

$A(t)=22(200+33t)^{2/3}-652.39$

The tank becomes full when the volume of solution in the tank at time $t$ is the same as the total volume of the tank:

$200+(44-11)t=500\implies 33t=300\implies t\approx9.09$

at which point the amount of salt in the solution would be

$A(9.09)\approx733.47\text{ lb}$

4 0

2 years ago