All categories

3 years ago

Question on picture:

Mathematics

2 answers:

ValentinkaMS [17]3 years ago

6 0

Answer:

Formula for tan is opposite over adjacent, so do arctan of 150/100 to get 56.30993247 :))

Step-by-step explanation:

enyata [817]3 years ago

6 0

Answer:

B. 34

Step-by-step explanation:

sohcahtoa

tan x = 100/150

tan x = .667

x = 33.7

You might be interested in

24 is 44% of what number?

Zarrin [17]

Answer:

24 is 44% of 54.55

Step-by-step explanation:

24 ÷ 44% = y ÷ 100%

Solve for y - To do so, we will use cross multiplication.

44y = 24 × 100

44y = 2400

y = 2400 ÷ 100

y = 54.55

8 0

3 years ago

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}}$

4 0

3 years ago

Help me PLEASE! I’d really appreciate it!

quester [9]

Answer:

0, 2

1, -2

2, -6

3, -10

7 0

3 years ago

Write an equation and solve for x.

Sauron [17]

Answer:

120

Step-by-step explanation:

Angles opposite to equal sides are equal .

30 + 30 + x = 180 {Angle sum property of triangle}

60 + x = 180

x = 180 - 60

x = 120

8 0

3 years ago

A box with inside area of 1.1m2 is covered with an unknown 0.163m of insulation material. Determine the thermal conductivity for

kipiarov [429]

9514 1404 393

Answer:

0.40 W/(m·K)

Step-by-step explanation:

Units of thermal conductivity are ...

W/(m·K)

The length dimension in the denominator comes from the ratio of area to thickness. That is, we can compute the thermal conductivity from ...

(62 W)(0.163m)/(1.1 m²·23K) ≈ 0.40 W/(m·K)

5 0

3 years ago