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ohaa [14]
3 years ago
8

In ΔNLM, if m∠M=(4x-4)°, m∠L=(3x+12)°, and m∠N=(6x+3)°, what is x?

Mathematics
1 answer:
GarryVolchara [31]3 years ago
5 0

Answer:

  x = 13

Step-by-step explanation:

The sum of angle measures in a triangle is 180°.

  m∠M + m∠L + m∠N = 180°

  (4x-4)° + (3x+12)° + (6x+3)° = 180°

  13x +11 = 180 . . . . . . collect terms, divide by °

  13x = 169 . . . . . . . . . subtract 11

  x = 13 . . . . . . . . . . . . . divide by 13

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Step-by-step explanation:

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Percentage error=\frac{275-200}{200}\times 100=37.5\ \%

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3 years ago
How do I simplify 5 3/2
Volgvan

Note the improper fraction. Change it into a mixed fraction

3/2 = 2/2 + 1/2

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3 0
3 years ago
Find the area of the shaded region in the graph y=x^2-2
Artist 52 [7]

By applying the concept of <em>definite</em> integral and the given restrictions, we conclude that the area of the <em>shaded</em> region in the graph is - 11/12.

<h3>How to find the area of the shaded region in the graph y = x² - 2</h3>

The area below the curve can be found by using <em>definite</em> integrals, which is defined here in this form:

I = \int\limits^{1.5}_{0.5} {x^{2}-2} \, dx     (1)

Now we proceed to calculate the area:

I = \int\limits^{1.5}_{0.5} {x^{2}} \, dx - 2 \int\limits^{1.5}_{0.5}\, dx

I = \frac{x^{3}}{3}|_{0.5}^{1.5} - 2 \cdot x |_{0.5}^{1.5}

I = \frac{1.5^{3}-0.5^{3}}{3} - 2 \cdot (1.5 - 0.5)

I = -\frac{11}{12}

By applying the concept of <em>definite</em> integral and the given restrictions, we conclude that the area of the <em>shaded</em> region in the graph is - 11/12.

<h3>Remark</h3>

The statement is incomplete. Complete form is shown below:

Find the area of the shaded region in the graph y = x² - 2, between x = 0.5 and x = 1.5

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

#SPJ1

3 0
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