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4 years ago

13. If B is the midpoint of AC and AC = 8x -20, find BC. AB= 3x-1

Mathematics

1 answer:

maks197457 [2]4 years ago

3 0

Answer:

$\Huge \boxed{5x - 19}$

$\rule[225]{225}{2}$

Step-by-step explanation:

A ——— B ——— C

AC = 8x - 20

AB = 3x - 1

BC = AC - AB

BC = (8x - 20) - (3x - 1)

Distributing negative sign.

BC = 8x - 20 - 3x + 1

Combining like terms.

BC = 5x - 19

$\rule[225]{225}{2}$

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Use function notation to represent the area of a circle whose circumference is 136cm

Nimfa-mama [501]

Answer:

$A(\frac{68}{\pi})=\frac{4624}{\pi}$

Step-by-step explanation:

The given circle has circumference 136cm.

The formula for circumference is $C=2\pi r$

This implies that:

$136=2\pi r$

$\frac{136}{2\pi}=r$

$\frac{68}{\pi}=r$

The area of a circle is given as $A(r)=\pi r^2$

When $r=\frac{68}{\pi}$

$A(\frac{68}{\pi})=\pi*(\frac{68}{\pi})^2$

$A(\frac{68}{\pi})=\pi*\frac{(68)^2}{\pi^2}$

$A(\frac{68}{\pi})=\frac{4624}{\pi}$

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3 years ago

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Answer:

135

Step-by-step explanation:

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3 years ago

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3 years ago

A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flo

djverab [1.8K]

Answer:

a) $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) $31690.7 g/L$

Step-by-step explanation:

By definition, we have that the change rate of salt in the tank is $\frac{dy}{dt}=R_{i}-R_{o}$ , where $R_{i}$ is the rate of salt entering and $R_{o}$ is the rate of salt going outside.

Then we have, $R_{i}=80\frac{L}{min}*50\frac{g}{L}=4000\frac{g}{min}$ , and

$R_{o}=40\frac{L}{min}*\frac{y}{500} \frac{g}{L}=\frac{2y}{25}\frac{g}{min}$

So we obtain. $\frac{dy}{dt}=4000-\frac{2y}{25}$ , then

$\frac{dy}{dt}+\frac{2y}{25}=4000$ , and using the integrating factor $e^{\int {\frac{2}{25}} \, dt=e^{\frac{2t}{25}$ , therefore $(\frac{dy }{dt}+\frac{2y}{25}}=4000)e^{\frac{2t}{25}$ , we get $\frac{d}{dt}(y*e^{\frac{2t}{25}})= 4000 e^{\frac{2t}{25}$ , after integrating both sides $y*e^{\frac{2t}{25}}= 50000 e^{\frac{2t}{25}}+C$ , therefore $y(t)= 50000 +Ce^{\frac{-2t}{25}}$ , to find $C$ we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions $y(0)=10$ , so $10= 50000+Ce^{\frac{-0*2}{25}}$

$10=50000+C\\C=10-50000=-49990$

Finally we can write an expression for the amount of salt in the tank at any time t, it is $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) The tank will overflow due Rin>Rout, at a rate of $80 L/min-40L/min=40L/min$ , due we have 500 L to overflow $\frac{500L}{40L/min} =\frac{25}{2} min=t$ , so we can evualuate the expression of a) $y(25/2)=50000-49990e^{\frac{-2}{25}\frac{25}{2}}=50000-49990e^{-1}=31690.7$ , is the salt concentration when the tank overflows

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4 years ago

alvaro pesa 13 kg menos que javier y 6 kg mas que maria si alvaro pesa 38 kg cuanto pesan javier y maria

Nataliya [291]

<span>Maria tiene 32 kg y javier tiene 51 kg
</span>
(google translate, bad spanish)

5 0

3 years ago

13. If B is the midpoint of AC and AC = 8x -20, find BC. AB= 3x-1​

13. If B is the midpoint of AC and AC = 8x -20, find BC. AB= 3x-1