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beks73 [17]
3 years ago
14

How can I answer this type of question?

Mathematics
1 answer:
Alex3 years ago
4 0
It would be: 0.38 / 0.25
Cancel both the decimals, 
= 38/25
Now, it becomes too easy can be simplified by division
= 38/25 = 1.52

In short, Your Answer would be 1.52

Hope this helps!
You might be interested in
The perimeters of the square and the triangle shown below are equal. What is the value of x?
Rudiy27
Perimeter is the sum of all the sides. So we can set up an equation:

\sf 3x+1+3x+1+3x+1+3x+1=2x+8+x+8+4x+2

Now solve for 'x', combine like terms:

When it comes to terms with variables it's just like normal addition but we keep the variable:

\sf 3x+3x+3x+3x=12x
\sf 2x+x+4x=7x

So we have:

\sf 12x+1+1+1+1=7x+8+8+2

Add:

\sf 12x+4=7x+18

Subtract 7x to both sides:

\sf 5x+4=18

Subtract 4 to both sides:

\sf 5x=14

Divide 5 to both sides:

\boxed{\sf x=2.8}
8 0
3 years ago
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
Help me please...!!!!!!
goblinko [34]
B would be the answer for sure
4 0
3 years ago
Read 2 more answers
Find the approximate area of the regions bounded by the curves y = x/(√x2+ 1) and y = x^4−x. (You may use the points of intersec
Finger [1]

The approximate area of the region bounded by the curves f(x) = x / √(x² + 1) and g(x) = x⁴ - x is approximately 0.806.

<h3>How to determine the approximate area of the regions bounded by the curves</h3>

In this problem we must use definite integrals to determine the area of the region bounded by the curves. Based on all the information given by the graph attached below, the area can be defined in accordance with this formula:

A = A₁ + A₂                                                                (1)

A₁ = ∫ [g(x) - f(x)] dx, for x ∈ [- 0.786, 0]                   (2)

A₂ = ∫ [f(x) - g(x)] dx, for x ∈ [0, 1.151]                       (3)

g(x) = x⁴ - x                                                               (4)

f(x) = x / √(x² + 1)                                                      (5)

Then, we proceed to find the integrals:

∫ g(x) dx = ∫ x⁴ dx - ∫ x dx = (1 / 5) · x⁵ - (1 / 2) · x²                          (6)

∫ f(x) dx = ∫ [x / √(x² + 1)] dx = (1 / 2) ∫ [2 · x / √(x² + 1)] dx = (1 / 2) ∫ [du / √u] = √u = √(x² + 1)                                                                                  (7)

And the complete expression for the integral is:

A = A₁ + A₂                                                                                      (1b)

A₁ = (1 / 5) · x⁵ - (1 / 2) · x² - √(x² + 1), for x ∈ [- 0.786, 0]               (2b)

A₂ = √(x² + 1) - (1 / 5) · x⁵ + (1 / 2) · x², for x ∈ [0, 1.151]                  (3b)

A₁ = 0.023

A₂ = 0.783

A = 0.023 + 0.783

A = 0.806

The approximate area of the region bounded by the curves f(x) = x / √(x² + 1) and g(x) = x⁴ - x is approximately 0.806.

To learn more on definite integral: brainly.com/question/14279102

#SPJ1

6 0
2 years ago
Compute the permutations and combinations. From a committee consisting of 5 men and 6 women, a sub-committee is formed consistin
kolbaska11 [484]

Answer:

360

Step-by-step explanation:

Multiply them all

3 0
3 years ago
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