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Triss [41]
3 years ago
10

Solve p(x+q)=r for x

Mathematics
1 answer:
STatiana [176]3 years ago
3 0
(x+q) = r/p
x+q = r/p
x = (r/p) - q
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`y=2x+1` was a positive slope because...
Kobotan [32]

Step-by-step explanation:

.....because the coefficient of x is 2, a positive number.

6 0
2 years ago
Help answer asap pls ---------
Reika [66]

Answer:

d = 5

Step-by-step explanation:

We want to know what d would be when x = -1, so we can start off by plugging in -1 for x to get the following equation: \sqrt{8 - (-1)} = 2(-1)+d. On the left hand side, 2 negatives become a positive sign, so the -(-1) becomes a +1, and 8+1 would equal 9. On the right hand side, we just multiply 2 and -1 to get -2. So, the equation now looks like this: \sqrt{9} = d-2. We know that \sqrt{9} = 3, so we would get 3 = d-2. THen we would add 2 to both sides to get d = 5. Hope this helps!

8 0
2 years ago
The circumference is 12 mile. What is the diameter?
photoshop1234 [79]

To find the diameter of a circle, using only the circumference, you need to divide the circumference by π, or 3.14.

So in this case, you must divide 12 by 3.14, which gives you 3.82.

I even double checked this, and should be correct then.

8 0
2 years ago
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X/3-4=-7<br><br> please help me solve
Vanyuwa [196]
In order to solve this we need to get x by itself on one side.

We need o start by adding across the 4 so that we get:

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5 0
2 years ago
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Suppose brine containing 0.2 kg of salt per liter runs into a tank initially filled with 500 L of water containing 5 kg of salt.
Oliga [24]

Answer:

(a) 0.288 kg/liter

(b) 0.061408 kg/liter

Step-by-step explanation:

(a) The mass of salt entering the tank per minute, x = 0.2 kg/L × 5 L/minute = 1 kg/minute

The mass of salt exiting the tank per minute = 5 × (5 + x)/500

The increase per minute, Δ/dt, in the mass of salt in the tank is given as follows;

Δ/dt = x - 5 × (5 + x)/500

The increase, in mass, Δ, after an increase in time, dt, is therefore;

Δ = (x - 5 × (5 + x)/500)·dt

Integrating with a graphing calculator, with limits 0, 10, gives;

Δ = (99·x - 5)/10

Substituting x = 1 gives

(99 × 1 - 5)/10 = 9.4 kg

The concentration of the salt and water in the tank after 10 minutes = (Initial mass of salt in the tank + Increase in the mass of the salt in the tank)/(Volume of the tank)

∴ The concentration of the salt and water in the tank after 10 minutes =  (5 + 9.4)/500 = (14.4)/500 = 0.288

The concentration of the salt and water in the tank after 10 minutes = 0.288 kg/liter

(b) With the added leak, we now have;

Δ/dt = x - 6 × (14.4 + x)/500

Δ = x - 6 × (14.4 + x)/500·dt

Integrating with a graphing calculator, with limits 0, 20, gives;

Δ = 19.76·x -3.456 = 16.304

Where x = 1

The increase in mass after an increase in = 16.304 kg

The total mass = 16.304 + 14.4 = 30.704 kg

The concentration of the salt in the tank then becomes;

Concentration = 30.704/500 = 0.061408 kg/liter.

6 0
3 years ago
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