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sammy [17]
3 years ago
9

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 :)
6 0
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}}

Recall that when t=0, A(t)=20 (our initial condition)

20=240+Ce^{-\frac{0}{40}}\\20-240=C\\C=-220\\$Therefore, the number A(t) of grams of salt in the tank at time t\\A(t)=240-220e^{-\frac{t}{40}}

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3 years ago
Help !!! multiply !!
Bess [88]

Answer:

-4x^2-x

Step-by-step explanation:

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If 8 key chains cost $9.20, then 16 key chains cost $ ___
mojhsa [17]
($9.20) 2 = $18.40 ...................
7 0
3 years ago
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