Are the triangles above similar?

Mathematics

2 answers:

Reptile [31]3 years ago

7 0

D, not similar. Hhhhhhhhhhhhhhhhhhh

Kamila [148]3 years ago

3 0

Answer:

Not similar

Hope this helps you

Can I have the brainliest please?

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-4 (2x + 13) + 3x = 80<br> How do you solve this out and get the answer

shusha [124]

Hey :)

x= -12

Use the distributive property to multiply −4 by 2x+13

−8x−52=3x+80
Subtract 3x from both sides
−8x−52−3x=80

Combine −8x and −3x to get −11x
−11x−52=80

Add 52 to both sides
−11x=80+52

Add 80 and 52 to get 132
−11x=132

Divide both sides by −11
x= 132/-11

Divide 132 by −11 to get −12
x=−12

Hope this helps :)

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3 years ago

What's the pattern for 1,3,5,7,5,3,1,3

natali 33 [55]

+2,+2,+2,-2,-2,-2,+2,+2,+2 etc

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4 years ago

A tank contains 240 liters of fluid in which 20 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pu

Novosadov [1.4K]

Answer:

$A(t)=240-220e^{-\frac{t}{40}}$

Step-by-step explanation:

A tank contains 240 liters of fluid in which 20 grams of salt is dissolved.

Volume of the tank = 240 liters
Initial Amount of Salt in the tank, A(0)=20 grams

Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 6 L/min

$R_{in}=$ (concentration of salt in inflow)(input rate of fluid)

$R_{in}=(1\frac{gram}{liter})( 6\frac{Liter}{min})=6\frac{gram}{min}$

$R_{out}=$ (concentration of salt in outflow)(output rate of fluid)

$R_{out}=(\frac{A(t)}{240})( 6\frac{Liter}{min})\\R_{out}=\frac{A}{40}$

Rate of change of the amount of salt in the tank:

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

$\dfrac{dA}{dt}=6-\dfrac{A}{40}$

We then solve the resulting differential equation by separation of variables.

$\dfrac{dA}{dt}+\dfrac{A}{40}=6\\$The integrating factor: e^{\int \frac{1}{40}dt} =e^{\frac{t}{40}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{40}}+\dfrac{A}{40}e^{\frac{t}{40}}=6e^{\frac{t}{40}}\\(Ae^{\frac{t}{40}})'=6e^{\frac{t}{40}}$

Taking the integral of both sides

$\int(Ae^{\frac{t}{40}})'=\int 6e^{\frac{t}{40}} dt\\Ae^{\frac{t}{40}}=6*40e^{\frac{t}{40}}+C, $(C a constant of integration)\\Ae^{\frac{t}{40}}=240e^{\frac{t}{40}}+C\\$Divide all through by e^{\frac{t}{40}}\\A(t)=240+Ce^{-\frac{t}{40}}$