How do I graph this ?

Any two rays will form angle. True False

mezya [45]

Answer:

True

Step-by-step explanation:

an angle is made up of two there could be 3 but then there would be 2 angles. Please give me the brailiest answer?

:) Hoped this helped!!! Have a good day!!! <3

3 0

4 years ago

Read 2 more answers

What is 41 divided by 4

geniusboy [140]

Answer:

10 with a remainder of 1

Step-by-step explanation:

and also, you should use a program like cymath for that. Only use brainly for problems that cant be answered by a website like that.

Good luck!

3 0

3 years ago

Whats the lcd of 1/7,14/17,12/13,5/6

gizmo_the_mogwai [7]

LCD (1/7, 14/7, 12/13, 5/6)

LCM = (7, 7, 13, 6)

= 2 * 3 * 7 * 13

= 546

1/7 = 78/546

14/7 = 1092/546

12/13 = 504/546

5/6 = 455/546

Calculation:

1/7 + 14/7

= 1 + 14/7

= 15/7

The common denominator you can calculate as the least common multiple of the both denominators: LCM (7, 7) = 7

Add:

15/7 + 12/13

= 15 . 13/7. 13 + 12 . 7/13 . 7

= 195/91 + 84/91

= 195 + 84/91

= 279/91

The common denominator you can calculate as the least common multiple of the both denominators: LCM (7, 13) = 91

Add:

279/91 + 5/6

= 279 . 6/91 . 6 + 5 . 91/6. 91

= 1674/546 + 455/546

= 1674 + 455/546

= 2129/546

The common denominator you can calculate as the least common multiple of the both denominators: LCM (91, 6) = 546

Hence, 546 is the LCM/LCD of (1/7, 14/17, 13/13, 5/6).

Hope that helps!!!!!!

3 0

3 years ago

Round 3:38 to the nearest ten

aev [14]

Answer:3

Step-by-step explanation:

Round 3.38 to the nearest ten

Rounding up to the nearest 10, we look at the digits on the left hand side of the decimal point. If the number on the right of the decimal point is less than 5, we ignore it.if it is greater than 5, we add one to the last digit on the right hand side.

In this case, the first digit on the left hand side is less than 5, so we ignore it. Rounding up to the nearest ten,the 0.38 is ignored.

Therefore, it becomes 3.0 = 3

To the nearest ten

7 0

3 years ago

A swimming pool with a volume of 30,000 liters originally contains water that is 0.01% chlorine (i.e. it contains 0.1 mL of chlo

SpyIntel [72]

Answer:

$R_{in}=0.2\dfrac{mL}{min}$

$C(t)=\dfrac{A(t)}{30000}$

$R_{out}= \dfrac{A(t)}{1500} \dfrac{mL}{min}$

$A(t)=300+2700e^{-\dfrac{t}{1500}},$ A(0)=3000$

Step-by-step explanation:

The volume of the swimming pool = 30,000 liters

(a) Amount of chlorine initially in the tank.

It originally contains water that is 0.01% chlorine.

0.01% of 30000=3000 mL of chlorine per liter

A(0)= 3000 mL of chlorine per liter

(b) Rate at which the chlorine is entering the pool.

City water containing 0.001%(0.01 mL of chlorine per liter) chlorine is pumped into the pool at a rate of 20 liters/min.

$R_{in}=$ (concentration of chlorine in inflow)(input rate of the water)

$=(0.01\dfrac{mL}{liter}) (20\dfrac{liter}{min})\\R_{in}=0.2\dfrac{mL}{min}$

(c) Concentration of chlorine in the pool at time t

Volume of the pool =30,000 Liter

$Concentration, C(t)= \dfrac{Amount}{Volume}\\C(t)=\dfrac{A(t)}{30000}$

(d) Rate at which the chlorine is leaving the pool

$R_{out}=$ (concentration of chlorine in outflow)(output rate of the water)

$= (\dfrac{A(t)}{30000})(20\dfrac{liter}{min})\\R_{out}= \dfrac{A(t)}{1500} \dfrac{mL}{min}$

(e) Differential equation representing the rate at which the amount of sugar in the tank is changing at time t.

$\dfrac{dA}{dt}=R_{in}-R_{out}\\\dfrac{dA}{dt}=0.2- \dfrac{A(t)}{1500}$

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

$\dfrac{dA}{dt}+\dfrac{A}{1500}=0.2\\$The integrating factor: e^{\int \frac{1}{1500}dt} =e^{\frac{t}{1500}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{1500}}+\dfrac{A}{1500}e^{\frac{t}{1500}}=0.2e^{\frac{t}{1500}}\\(Ae^{\frac{t}{1500}})'=0.2e^{\frac{t}{1500}}$

Taking the integral of both sides

$\int(Ae^{\frac{t}{1500}})'=\int 0.2e^{\frac{t}{1500}} dt\\Ae^{\frac{t}{1500}}=0.2*1500e^{\frac{t}{1500}}+C, $(C a constant of integration)\\Ae^{\frac{t}{1500}}=300e^{\frac{t}{1500}}+C\\$Divide all through by e^{\frac{t}{1500}}\\A(t)=300+Ce^{-\frac{t}{1500}}$

Recall that when t=0, A(t)=3000 (our initial condition)

$3000=300+Ce^{0}\\C=2700\\$Therefore:\\A(t)=300+2700e^{-\dfrac{t}{1500}}$

3 0

3 years ago