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sergey [27]
2 years ago
15

4. Two students created a list of steps for the following construction. Which student has steps in the correct order, and which

does not? Explain.
You are given and point C. Construct a line parallel to that passes through point C.

Student A Steps:
Student B Steps:
Draw a line that intersects points B and C.
Draw a line through point C and point G.
Keep the compass at the same width, and place it on point C.
Keeping the compass at the same width, place it on point F.
Mark the intersection of the two arcs as point G.
Open the compass to the width between points D and E.
Place the compass on point B, and swing an arc that crosses line AB and line BC. Label the points D and E.
Swing an arc that crosses line BC, and label the point F.
Swing an arc that intersects the arc created from line BC at point C.
Draw a line that intersects points B and C.
Place the compass on point B, and swing an arc that crosses line AB and line BC. Label the points D and E.
Keep the compass at the same width, and place it on point C.
Swing an arc that crosses line BC, and label the point F.
Open the compass to the width between points D and E.
Keeping the compass at the same width, place it on point F.
Swing an arc that intersects the arc created from line BC at point C.
Mark the intersection of the two arcs as point G.
Draw a line through point C and point G.
Mathematics
1 answer:
shtirl [24]2 years ago
7 0

Answer:

Student B

Step-by-step explanation:

student A give to little step the beging to get pointe G Student A is wrong

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i f N(-7,-1) is a point on the terminal side of ∅ in standard form, find the exact values of the trigonometric functions of ∅.
Natali [406]

Answer:

sin\ \varnothing = \frac{-1}{10}\sqrt 2

cos\ \varnothing = \frac{-7}{10}\sqrt 2

tan\ \varnothing = \frac{1}{7}

cot\ \varnothing = 7

sec\ \varnothing = \frac{-5}{7}\sqrt 2

csc\ \varnothing = -5\sqrt 2

Step-by-step explanation:

Given

N = (-7,-1) --- terminal side of \varnothing

Required

Determine the values of trigonometric functions of \varnothing.

For \varnothing, the trigonometry ratios are:

sin\ \varnothing = \frac{y}{r}       cos\ \varnothing = \frac{x}{r}       tan\ \varnothing = \frac{y}{x}

cot\ \varnothing = \frac{x}{y}       sec\ \varnothing = \frac{r}{x}       csc\ \varnothing = \frac{r}{y}

Where:

r^2 = x^2 + y^2

r = \sqrt{x^2 + y^2

In N = (-7,-1)

x = -7 and y = -1

So:

r = \sqrt{(-7)^2 + (-1)^2

r = \sqrt{50

r = \sqrt{25 * 2

r = \sqrt{25} * \sqrt 2

r = 5 * \sqrt 2

r = 5 \sqrt 2

<u>Solving the trigonometry functions</u>

sin\ \varnothing = \frac{y}{r}

sin\ \varnothing = \frac{-1}{5\sqrt 2}

Rationalize:

sin\ \varnothing = \frac{-1}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}

sin\ \varnothing = \frac{-\sqrt 2}{5*2}

sin\ \varnothing = \frac{-\sqrt 2}{10}

sin\ \varnothing = \frac{-1}{10}\sqrt 2

cos\ \varnothing = \frac{x}{r}

cos\ \varnothing = \frac{-7}{5\sqrt 2}

Rationalize

cos\ \varnothing = \frac{-7}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}

cos\ \varnothing = \frac{-7*\sqrt 2}{5*2}

cos\ \varnothing = \frac{-7\sqrt 2}{10}

cos\ \varnothing = \frac{-7}{10}\sqrt 2

tan\ \varnothing = \frac{y}{x}

tan\ \varnothing = \frac{-1}{-7}

tan\ \varnothing = \frac{1}{7}

cot\ \varnothing = \frac{x}{y}

cot\ \varnothing = \frac{-7}{-1}

cot\ \varnothing = 7

sec\ \varnothing = \frac{r}{x}

sec\ \varnothing = \frac{5\sqrt 2}{-7}

sec\ \varnothing = \frac{-5}{7}\sqrt 2

csc\ \varnothing = \frac{r}{y}

csc\ \varnothing = \frac{5\sqrt 2}{-1}

csc\ \varnothing = -5\sqrt 2

3 0
2 years ago
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