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Scilla [17]
2 years ago
11

Given that the two parallel lines are cut by a transversal, and the measure of angle 2 is 50 degrees. Find each of the missing a

ngles.

Mathematics
1 answer:
Mandarinka [93]2 years ago
7 0

Answer:

Step-by-step explanation:

Lines m and l are the parallel lines and a line 'n' is a transverse intersecting these lines.

m∠2 = 50°

m∠1 + m∠2 = 180° [Linear pair of angles]

m∠1 = 180° - 50°

m∠1 = 130°

m∠3 = m∠1 = 130° [Vertically opposite angles]

m∠3 + m∠5 = 180° [Consecutive interior angles]

m∠5 = 180° - m∠3

        = 180° - 130°

        = 50°

m∠6 + m∠5 = 180° [Linear pair of angles]

m∠6 = 180° - 50° = 130°

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I think it's answer D. (141).  :D

Step-by-step explanation:

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Verify the given linear approximation at a = 0. Then determine the values of x for which the linear approximation is accurate to
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Answer:

Part 1)

See Below.

Part 2)

\displaystyle (-0.179, -0.178) \cup (-0.010, 0.012)

Step-by-step explanation:

Part 1)

The linear approximation <em>L</em> for a function <em>f</em> at the point <em>x</em> = <em>a</em> is given by:

\displaystyle L \approx f'(a)(x-a) + f(a)

We want to verify that the expression:

1-36x

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\displaystyle f(x) = \frac{1}{(1+9x)^4}

At <em>x</em> = 0.

So, find f'(x). We can use the chain rule:

\displaystyle f'(x) = -4(1+9x)^{-4-1}\cdot (9)

Simplify. Hence:

\displaystyle f'(x) = -\frac{36}{(1+9x)^{5}}

Then the slope of the linear approximation at <em>x</em> = 0 will be:

\displaystyle f'(1) = -\frac{36}{(1+9(0))^5} = -36

And the value of the function at <em>x</em> = 0 is:

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Hence verified.

Part B)

We want to determine the values of <em>x</em> for which the linear approximation <em>L</em> is accurate to within 0.1.

In other words:

\displaystyle \left| f(x) - L(x) \right | \leq 0.1

By definition:

\displaystyle -0.1\leq f(x) - L(x) \leq 0.1

Therefore:

\displaystyle -0.1 \leq \left(\frac{1}{(1+9x)^4} \right) - (1-36x) \leq 0.1

We can solve this by using a graphing calculator. Please refer to the graph shown below.

We can see that the inequality is true (i.e. the graph is between <em>y</em> = 0.1 and <em>y</em> = -0.1) for <em>x</em> values between -0.179 and -0.178 as well as -0.010 and 0.012.

In interval notation:

\displaystyle (-0.179, -0.178) \cup (-0.010, 0.012)

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