1 year ago

In exercises 1 and 2, find m<1 and m<2. Tell which theorem you used in each case.

Mathematics

1 answer:

solniwko [45]1 year ago

8 0

Answer:

m∠2 = 78°

Step-by-step explanation:

From the given picture,

lines 'm' and 'n' are the parallel lines and line 't' is the transverse line.

∠1 and angle with the measure of 78° are corresponding angles.

Therefore, m∠1 = 78°

m∠1 = m∠2 [Vertical angles are equal]

m∠2 = 78°

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Consider the function represented by the equation y-6x-9=0. Which answer shows the equation written in function notation with x

spin [16.1K]

we have

$y-6x-9=0$

Step $1$

Clear the variable y

$y-6x-9=0$

Adds $(6x+9)$ both sides

$y-6x-9+(6x+9)=0+(6x+9)$

$y=(6x+9)$

Step $2$

Convert in function notation

Let

$f(x)=y$

$f(x)=(6x+9)$

therefore

the answer is

$f(x)=(6x+9)$

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2 years ago

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kati45 [8]

Answer:

B is 50° and C is 130°

Step-by-step explanation:

complementary angle

x+x=90°

40°+x=90°

x=90°-40°

x=50°

supplementary angles

x+x=180

50+x=180

x=180-50

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2 years ago

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Complete the following proof. Given: Points R, S, T, Q on circle O Prove: m\overarc RS + m\overarc ST + m\overarc TQ + m\overarc

LekaFEV [45]

Answer:

360°

Step-by-step explanation:

From the figure attached,

R, S, T and Q are the points on a circle O.

Since, "measure of an arc of a circle is equal to the measure of the angle subtended by the arc at the center."

By this statement,

$m(\widehat{RS})=m(\angle ROS)$

$m(\widehat{ST})=m(\angle SOT)$

$m(\widehat{TQ})=m(\angle TOQ)$

m(major arc RQ) = m(∠QOR)

Now $m(\widehat {RS})+m(\widehat{ST})+m(\widehat{TQ})$ + m(major arc RQ) = $m(\angle ROS)+m(\angle SOT)+m(\angle TOQ)+m(\angle QOR)$

Since sum of all angles at a point = 360°

$m(\widehat {RS})+m(\widehat{ST})+m(\widehat{TQ})$ + m(major arc RQ) = 360°

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Answer:

y= -intercept = 10

Step-by-step explanation:

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