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3 years ago

Please Help! B belongs to AC D belongs to lineAF

Mathematics

1 answer:

Anvisha [2.4K]3 years ago

5 0

Answer:

x = 27

Step-by-step explanation:

∠ CAD and ∠ EDF are corresponding angles and are congruent, so

∠ CAD = 2x

The sum of the 3 angles in Δ ABD = 180°

sum the angles and equate to 180

72 + 2x + 2x = 180

72 + 4x = 180 ( subtract 72 from both sides )

4x = 108 ( divide both sides by 4 )

x = 27

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Find m∠U.<br> Write your answer as an integer or as a decimal rounded to the nearest tenth.

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Answer:

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$ut = \sqrt{ {6}^{2} + {5}^{2} } \\ \sqrt{36 + 25} \\ \sqrt{61 } \\ using \: cosine \: rule \\ cos \: u = \frac{ {5}^{2} + 61 - 36 }{2 \times 5 \times \sqrt{61} } \\ \\ = \frac{25 + 61 - 36}{78} \\ \\ = \frac{50}{78 } \\ \cos \: u = 0.64 \\ m \:u = 50.2$

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If one square yard costs 12 dollars how much is 180 square feet

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3 years ago

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Which expression is equivalent to log w (x2-6)4

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Log w (x^2-6)^4

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log w (x^2-6)^4 = log w + log (x^2-6)^4

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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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