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3 years ago

In the figure above abc is a square .. what is the value of x?

Mathematics

1 answer:

Len [333]3 years ago

4 0

If ABCD is a square, then m∠A = 90°.

Therefore we have the equation:

30° + 48° + x° = 90°

78° + x° = 90° |-78°

x° = 12°

Answer: 12

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How do you convert a mixed number into a improper fraction

Llana [10]

To convert a mixed number into an improper fraction, you can multiply the denominator of the fraction by the whole number. Then, you add the product with the numerator to get the new numerator. The denominator remains the same.

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3 years ago

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Which equation is the correct translation of the following statement?

leonid [27]

<h3>hello!</h3>

Let the number be x.

subtract 4:-

x-4 (4 less)

this expression equals 9:-

x-4=9 (Option C)

Hope everything is clear; if you need any explanation/clarification, kindly let me know, and I will comment and/or edit my answer :)

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3 years ago

Oxygen flows through one tube into a liter flask filled with air, and the mixture of oxygen and air (considered well stirred) es

yKpoI14uk [10]

The amount of oxygen in the flask through the intake tube illustrates proportions

The flask 99.35% will contain of oxygen after 5L have passed through the intake tube

<h3>How to calculate the percentage of oxygen?</h3>

To do this, we make use of the following representations

P for the proportion of oxygen
V for te volume of oxygen
x for the proportion of oxygen in the flask after 5 L passed through

So. we have the following derivative

dP = dV - P * dV

Divide through by dV

dP/dV = 1 - P

Next, we set up the integral

$\int\limits^x_{0.040} \frac {1}{1 - P}dP= \int\limits^5_0 {dV}$

Integrate both sides

$-\ln(|P - 1|)|\limits^x_{0.04} = 5$

Expand

-ln(|x - 1|) + ln(|0.04 - 1|) = 5

Evaluate the difference

-ln(|x - 1|) + ln(|-0.96|) = 5

Express as a fraction

ln(0.96/|x - 1|) = 5

Express both sides as exponents with a base of 2

e^ln(0.96/|x - 1|) = e^5

Solve for x

x = 1 - 0.96/e^5

Evaluate the difference

x = 0.9935

Express as percentage

x = 99.35%

Hence, the flask 99.35% will contain of oxygen after 5L have passed through the intake tube