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Amanda [17]
2 years ago
15

Solve the Equation 3x + 5x - 5 = 19

Mathematics
1 answer:
Mila [183]2 years ago
8 0

Answer:

x=3

Step-by-step explanation:

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First you need to multiply 52 x 40 = 2080
then you need to divide 2080 from 60,000.
The answer will be 60000/2080 is approximately $28.85
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I think it’s 0.1

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A pair of dice is rolled. List the elements of sample space S. Let A denote the event “the sum is less than 5” and B the event “
Marina86 [1]
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5 0
3 years ago
I NEED HELP PLEASE!<br><br> If you can help me with this you amazing!<br><br> Thank you
krek1111 [17]

Answer:

a=-3

Step-by-step explanation:

we have

a-2=\frac{3+6a}{3}

Solve for a

That means ----> Isolate the variable a

Factor 3 in the numerator right side

a-2=\frac{3(1+2a)}{3}

Simplify right side

a-2=1+2a

subtract a both sides

a-2-a=1+2a-a

-2=1+a

subtract 1 both sides

-2-1=1+a-1

-3=a

Rewrite

a=-3

6 0
2 years ago
Water is leaking out of an inverted conical tank at a rate of 6800 cubic centimeters per min at the same time that water is bein
ivanzaharov [21]

Answer:

1508527.582 cm³/min

Step-by-step explanation:

The net rate of flow dV/dt = flow rate in - flow rate out

Let flow rate in = k. Since flow rate out = 6800 cm³/min,

dV/dt = k - 6800

Now, the volume of a cone V = πr²h/3 where r = radius of cone and h = height of cone

dV/dt = d(πr²h/3)/dt = (πr²dh/dt)/3 + 2πrhdr/dt (since dr/dt is not given we assume it is zero)

So, dV/dt = (πr²dh/dt)/3

Let h = height of tank = 12 m, r = radius of tank = diameter/2 = 3/2 = 1.5 m, h' = height when water level is rising at a rate of 21 cm/min = 3.5 m and r' = radius when water level is rising at a rate of 21 cm/min

Now, by similar triangles, h/r = h'/r'

r' = h'r/h = 3.5 m × 1.5 m/12 m = 5.25 m²/12 m = 2.625 m = 262.5 cm

Since the rate at which the water level is rising is dh/dt = 21 cm/min, and the radius at that point is r' = 262.5 cm.

The net rate of increase of water is dV/dt = (πr'²dh/dt)/3

dV/dt = (π(262.5 cm)² × 21 cm/min)/3

dV/dt = (π(68906.25 cm²) × 21 cm/min)/3

dV/dt = 1447031.25π/3 cm³

dV/dt = 4545982.745/3 cm³

dV/dt = 1515327.582 cm³/min

Since dV/dt = k - 6800 cm³/min

k = dV/dt - 6800 cm³/min

k = 1515327.582 cm³/min - 6800 cm³/min

k = 1508527.582 cm³/min

So, the rate at which water is pumped in is 1508527.582 cm³/min

5 0
3 years ago
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