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alexandr1967 [171]
2 years ago
6

11k = -143 I need help with this question

Mathematics
2 answers:
garri49 [273]2 years ago
7 0

Answer:

k = -13

Step-by-step explanation:

-143/11 = -13

leva [86]2 years ago
6 0

Answer:

11k = -143

k=-143/11=-13 is a required answer

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butalik [34]
The answer would be 16 because 40 divided by 5 is 8 and 8 times 2 is sixteen so it would be 16
8 0
3 years ago
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An angle measures 111.4° more than the measure of its supplementary angle. What is the measure of each angle?
fomenos

The measure of the supplementary angles are 34.3 and 145.7 degrees.

<h3>What are supplementary angles?</h3>

Supplementary angles are those angles that sum up to 180 degrees.

In other words, two angles are Supplementary when they add up to 180 degrees.

Therefore, the angles measures 111.4° more than the measure of it's supplementary angle.

Hence,

let

x = measure of the other angle

x + x + 111.4 = 180

2x + 111.4 = 180

subtract 111.4 from both sides

2x + 111.4 - 111.4 = 180 - 111.4

2x = 68.6

divide both sides by 2

x = 68.6 / 2

x = 34.3

Other angle  = 34.3 + 111.4 = 145.7°

Therefore, the measure of the supplementary angles are 34.3 and 145.7 degrees.

learn more on supplementary angles here: brainly.com/question/15966137

#SPJ1

6 0
1 year ago
You pick a card at random. Without putting the first card back, you
Inga [223]

First, we need the probability of picking an odd number.

There are 5 cards in total, and 3 odd cards (3, 5, and 7).

That means that the probability that we'll draw an odd card would be \frac{3}{5}.

Then, we have 4 cards left, and 2 even cards (4 and 6), meaning that the probability that we draw an even card will be \frac{2}{4} or \frac{1}{2}.

To find the probability that these would happen in consecutive draws, we just multiply the probabilities together.

\frac{3}{5}*\frac{1}{2}=\frac{3}{10} or 0.3.

To convert this into a percentage, we multiply the decimal by 100.

0.3*100=30.

So the probability of picking an odd number and then picking an even number is 30%.

Hope this helps!

8 0
3 years ago
Can someone help me on this
tekilochka [14]

Answer:

the answers are

8) e. opposite sides

9) d. vertices

10) c. opposite angles

11) b. diagonals

12) a. consecutive sides

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
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