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Novosadov [1.4K]
3 years ago
9

Match the reasons with the statements in the proof if the last line of the proof would be 6. ∠1 and ∠7 are supplementary by defi

nition. Given: s || t Prove: 1, 7 are supplementary 1. s||t Given 2. ∠5 and ∠7 are supplementary. Exterior sides in opposite rays. 3. m∠5 + m∠7 = 180° Definition of supplementary angles. 4. m∠1 = m∠5 Substitution 5. m∠1 + m∠7 = 180° If lines are ||, corresponding angles are equal.
Mathematics
2 answers:
nataly862011 [7]3 years ago
7 0

Answer:

Step-by-step explanation:

Given: s II t

Substitution: m<1 + m<7 = 180*

If lines are II, corresponding angles are equal: m<1 = m<5

<5 and <7 are supplementary: definition of supplementary angles

m<5 + m<7 = 180* : exterior sides in opposite rays

Alecsey [184]3 years ago
4 0

Answer:

Given : s||t

Exterior sides in opposite rays : ∠5 and ∠7 are supplementary

Definition of supplementary angles : m∠5 + m∠7 = 180°

If lines are ||, corresponding angles are equal : m∠1 = m∠5

Substitution : m∠1 + m∠7 = 180°

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If f(x) is = to 0 then that means that you would have to plug the 0 in for all of the x that are in the function and the solve it.
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0        0        0 -1
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Are rational numbers closed under squaring?
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3 years ago
Evaluate Dx / ^ 9-8x - x2^
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Since you use parentheses appropriately in the answer choices, I'm going to go out on a limb here and assume something like "^x^" stands for \sqrt x.

In that case, you want to find the antiderivative,

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Complete the square in the denominator:

9-8x-x^2=25-(16+8x+x^2)=5^2-(x+4)^2

Now substitute x+4=5\sin y, so that \mathrm dx=5\cos y\,\mathrm dy. Then

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\int\frac{5\cos y}{\sqrt{5^2-(5\sin y)^2}}\,\mathrm dy

which simplifies to

\displaystyle\int\frac{5\cos &#10;y}{5\sqrt{1-\sin^2y}}\,\mathrm dy=\int\frac{\cos y}{\sqrt{\cos^2y}}\,\mathrm dy

Now, recall that \sqrt{x^2}=|x|. But we want the substitution we made to be reversible, so that

x+4=5\sin y\iff y=\sin^{-1}\left(\dfrac{x+4}5\right)

which implies that -\dfrac\pi2\le y\le\dfrac\pi2. (This is the range of the inverse sine function.)

Under these conditions, we have \cos y\ge0, which lets us reduce \sqrt{\cos^2y}=|\cos y|=\cos y. Finally,

\displaystyle\int\frac{\cos y}{\cos y}\,\mathrm dy=\int\mathrm dy=y+C

and back-substituting to get this in terms of x yields

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\sin^{-1}\left(\frac{x+4}5\right)+C
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