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almond37 [142]
2 years ago
15

Which one?????? i need help

Mathematics
1 answer:
borishaifa [10]2 years ago
5 0
<h3>Given</h3>

x + 2 ≥ 6

<h3>Find</h3>

the graph for the solution

<h3>Solution</h3>

We can solve for x by subtacting 2 from both sides of the inequality.

... x + 2 -2 ≥ 6 -2

... x ≥ 4 . . . . . . . . . . simplify

The answer is read, "x is greater than or equal to 4." This means the solution is those real numbers that are to the right of 4 on the number line, and including 4 on the number line. Numbers to the left of 4 are <em>not included</em>. The inclusion of the number 4 is shown by putting a solid dot there (not an open circle).

The appropriate choice is the 3rd one.

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Find the value of x that makes the equation true:<br><br>x + 3 = 4
Zolol [24]

Answer:

x = 1

Step-by-step explanation:

Subtract 3 from both sides

x+ 3 - 3 = 4 - 3

gives x=1

6 0
3 years ago
How much greater than m2+3mn−n2 is 4m2+5mn+6n2?<br> PLEASE HELP
tiny-mole [99]

Answer:

3m² + 2mn + 7n²

Step-by-step explanation:

Subtract m² + 3mn - n² from 4m² + 5mn + 6n², that is

4m² + 5mn + 6n² - (m² + 3mn - n²)

= 4m² + 5mn + 6n² - m² - 3mn + n² ← collect like terms

= 3m² + 2mn + 7n²


5 0
2 years ago
Help !!?!?! Emergency....A relative frequency table is made from data in a frequency table.
jeka57 [31]
33% Because Total for line A was 53%-20% From column C Equals 33%
7 0
2 years ago
Read 2 more answers
A tank contains 3,000 L of brine with 15 kg of dissolved salt. Pure water enters the tank at a rate of 30 L/min. The solution is
Shalnov [3]

Answer:

Step-by-step explanation:

Volume of tank is 3000L.

Mass of salt is 15kg

Input rate of water is 30L/min

dV/dt=30L/min

Let y(t) be the amount of salt at any time

Then,

dy/dt = input rate - output rate.

The input rate is zero since only water is added and not salt solution

Now, output rate.

Concentrate on of the salt in the tank at any time (t) is given as

Since it holds initially holds 3000L of brine then the mass to volume rate is y(t)/3000

dy/dt= dV/dt × dM/dV

dy/dt=30×y/3000

dy/dt=y/100

Applying variable separation to solve the ODE

1/y dy=0.01dt

Integrate both side

∫ 1/y dy = ∫ 0.01dt

In(y)= 0.01t + A, .A is constant

Take exponential of both side

y=exp(0.01t+A)

y=exp(0.01t)exp(A)

exp(A) is another constant let say C

y(t)=Cexp(0.01t)

The initial condition given

At t=0 y=15kg

15=Cexp(0)

Therefore, C=15

Then, the solution becomes

y(t) = 15exp(0.01t)

At any time that is the mass.

5 0
3 years ago
Which shape is a Parallelogram?
mariarad [96]
The top left is the correct answer, because the sides are parallel.

Hope this helps!
5 0
2 years ago
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