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2 years ago

Explain how rays AB and AC form both a line and an angle.

Mathematics

2 answers:

konstantin123 [22]2 years ago

5 0

Answer:

An angle is defined as two rays with a common endpoint, so CAB (or BAC) is an angle. A line is described as an infinite set of points that extend forever in either direction, which these rays also do.

Step-by-step explanation:

GuDViN [60]2 years ago

4 0

A straight line (180o) is still an angle. Would you be wondering if the question said "Can 179o be a real angle?" No you would simply draw it. 1 degree more and you have 180o and it is an angle as well.

So you have a line that is both a line and an angle with A in the middle.

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Do you know the answer toPart A: 25% of 50 = Part B: 25% of 60 = Part C: 50% of 60 = Part D: 75% of 60 = Part E: 75% of 30 = Par

Verdich [7]

Answer:

Part A: 25% of 50 = 12.5

Part B: 25% of 60 = 15

Part C: 50% of 60 = 30

Part D: 75% of 60 = 45

Part E: 75% of 30 = 22.5

Part F: 100% of 22.5 = 22.5

Part G: 10% of 22.5 = 2.25

Part H: 50% of 45.7 = 22.85

Step-by-step explanation:

To find - Do you know the answer to

Part A: 25% of 50 =

Part B: 25% of 60 =

Part C: 50% of 60 =

Part D: 75% of 60 =

Part E: 75% of 30 =

Part F: 100% of 22.5 =

Part G: 10% of 22.5 =

Part H: 50% of 45.7 =

Proof -

Part A :

25% of 50 = $\frac{25}{100} *50$ = $\frac{1}{4}*50$ = 12.5

⇒25% of 50 = 12.5

Part B :

25% of 60 = $\frac{25}{100} *60$ = $\frac{1}{4}*60$ = 15

⇒25% of 60 = 15

Part C :

50% of 60 = $\frac{50}{100} *60$ = $\frac{1}{2}*60$ = 30

⇒50% of 60 = 30

Part D :

75% of 60 = $\frac{75}{100} *60$ = $\frac{75}{10}*6$ = 45

⇒75% of 60 = 45

Part E :

75% of 30 = $\frac{75}{100} *30$ = $\frac{75}{10}*3$ = 22.5

⇒75% of 30 = 22.5

Part F :

100% of 22.5 = $\frac{100}{100} *22.5$ = 22.5

⇒ 100% of 22.5 = 22.5

Part G :

10% of 22.5 = $\frac{10}{100} *22.5$ = 2.25

⇒ 10% of 22.5 = 2.25

Part H :

50% of 45.7 = $\frac{50}{100} *45.7$ = $\frac{1}{2}*45.7$ = 22.85

⇒50% of 45.7 = 22.85

6 0

2 years ago

Can someone help me with this problem

balandron [24]

Hi! Your answer should be, 1,960.89

Hope this helps you!

3 0

2 years ago

it took jake 28 minutes to run 5km last months. this month he ran 5km in 21 minutes. by what percentage had jake improve his run

mars1129 [50]

Answer:

25%

Step-by-step explanation:

28-21=7

7/28=×/100

7×100=700

700÷28=25

6 0

2 years ago

Read 2 more answers

Two similar trapezoids have areas

puteri [66]

It d i hope this helps

3 0

2 years ago

Read 2 more answers

1) find the equation of the line parallel to

In-s [12.5K]

Answer:

1) The equation of the line parallel to x-5y=6 and through (4,-2) is 5y = x -14

2) The equation of the line perpendicular to y= -2/5x + 3 and through (2,-1) is 2y = 5x -12

Solution:

1) find the equation of the line parallel to x-5y=6 and through (4,-2).

Given, line equation is x – 5y = 6

We have to find the line equation that is parallel to given line and passing through the point (4, -2)

Now, let us find slope of the given line.

$\text { Slope }=\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-1}{-5}=\frac{1}{5}$

Now, we know that, slope of parallel lines are equal.

So, slope of required line is 1/5 and it passes through (4, -2)

Now, using point slope form

$y-y_{1}=m\left(x-x_{1}\right) \text { where } m \text { is slope and }\left(x_{1}, y_{1}\right) \text { is point on line. }$

$y-(-2)=\frac{1}{5}(x-4) \rightarrow 5(y+2)=x-4 \rightarrow 5 y+10=x-4 \rightarrow x-5 y=14$

Hence, the line equation is 5y = x -14

2) find the equation of the line perpendicular to y= -2/5x + 3 and through (2,-1)

$\text { Given, line equation is } y=-\frac{2}{5} x+3 \rightarrow 5 y=-2 x+15 \rightarrow 2 x+5 y=15$

We have to find the line equation that is perpendicular to given line and passing through the point (2, -1)

Now, let us find slope of the given line.

$\text { Slope }=\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-2}{5}=-\frac{2}{5}$

Now, we know that, product of slopes of perpendicular lines equals to -1.

So, slope of required line $\times$ slope of given line = -1

slope of required line = $-1 \times \frac{5}{-2}=\frac{5}{2}$

And it passes through (2, -1)

Now, using point slope form

$\begin{array}{l}{\text { Line equation is } y-(-1)=\frac{5}{2}(x-2) \rightarrow 2(y+1)=5(x-2)} \\\\ {\rightarrow 2 y+2=5 x-10 \rightarrow 5 x-2 y=12}\end{array}$

Hence, the line equation is 2y = 5x -12

4 0

2 years ago