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2 years ago

Question 2 (Multiple Choice Worth 1 points)

1 answer:

3 0

A vertical line can only pass through the x axsis not the y unless it was at 0 then it would pass through all the y points. A diagonal line can pass through (-5,1) but not a vertical choose x=-5 if you want it to be a vertical line going through (-5,0)

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Given 4 coordinate points â€” A(2, 0), B(4, 0), C(5, â€“2), D(1, â€“2) â€" identify the figure showing a

babymother [125]

Answer:

Option B

Step-by-step explanation:

First we would identify the coordinates of the points before they are rotated.

A(2, 0), B(4, 0), C(5, -2), D(1, -2)

Plot a graph with the coordinates.

We find that the figure appears in the fourth quadrant.

Find attached the diagram.

To rotate the figure obtained by 180° about the origin zero, we would change the signs of the coordinates to become opposite.

That is positive sign becomes negative and vice versa:

A(-2, 0), B(-4, 0), C(-5, 2), D(-1, 2)

This implies that we rotated the figure 90° twice. When this is done, the figure would appear in second quadrant.

Plotting the new coordinates on a graph would give option B.

8 0

2 years ago

Can anybody show me how to do this half angle identity problem STEP BY STEP? I always seem to be missing something

klio [65]

Answer: $\bold{-\dfrac{7\sqrt2}{10}}$

<u>Step-by-step explanation:</u>

It is given that θ is between 270° and 360°, which means that θ is located in Quadrant IV ⇒ (x > 0, y < 0). Furthermore, the half-angle will be between 135° and 180°, which means the half-angle is in Quadrant II ⇒ $cos\ \dfrac{\theta}{2}$

It is given that sin θ = $-\dfrac{7}{25}$ ⇒ y = -7 & hyp = 25

Use Pythagorean Theorem to find "x":

x² + y² = hyp²

x² + (-7)² = 25²

x² + 49 = 625

x² = 576

x = 24

Use the "x" and "hyp" values to find cos θ:

$cos\ \theta=\dfrac{x}{hyp}=\dfrac{24}{25}$

Lastly, input cos θ into the half angle formula:

$cos\bigg(\dfrac{\theta}{2}\bigg)=\pm \sqrt{\dfrac{1+cos\ \theta}{2}}\\\\\\.\qquad \quad =\pm \sqrt{\dfrac{1+\dfrac{24}{25}}{2}}\\\\\\.\qquad \quad =\pm \sqrt{\dfrac{\dfrac{25}{25}+\dfrac{24}{25}}{2}}\\\\\\.\qquad \quad =\pm \sqrt{\dfrac{\dfrac{49}{25}}{2}}\\\\\\.\qquad \quad =\pm \sqrt{\dfrac{49}{50}}\\\\\\.\qquad \quad =\pm \dfrac{7}{5\sqrt2}}\\\\\\.\qquad \quad =\pm \dfrac{7}{5\sqrt2}}\bigg(\dfrac{\sqrt2}{\sqrt2}\bigg)\\\\\\.\qquad \quad =\pm \dfrac{7\sqrt2}{10}$

Reminder: We previously determined that the half-angle will be negative.

6 0

2 years ago

Quadrilateral ABCD is inscribed in this circle. What is the measure of angle C?

dlinn [17]

Hope it would help!!!

6 0

3 years ago

Solve the inequality 2x>30+5/4x

insens350 [35]

Answer:

Step-by-step explanation:

$2x > 30+\frac{5}{4x} \\2x-\frac{5}{4x} > 30\\\frac{8x^2-5}{4x} > 30\\case~1\\if~x > 0\\8x^2-5 > 120x\\8x^2-120x > 5\\x^2-15x > \frac{5}{8} \\adding~(-\frac{15}{2} )^2~to~both~sides\\(x-\frac{15}{2} )^2 > \frac{5}{8}+\frac{225}{4} \\(x-\frac{15}{2} )^2 > \frac{455}{8} \\x-\frac{15}{2} < -\sqrt{\frac{455}{8} } \\x < \frac{15}{2}-\sqrt{\frac{455}{8} } \\or~x < 0\\rejected~as~x > 0$

$x-\frac{15}{2} > \sqrt{\frac{455}{8} } \\x > \frac{15}{2} +\sqrt{\frac{455}{8} }$

case~2

$if~x < 0\\8x^2-5 < 120x\\8x^2-120x < 5\\x^2-15x < \frac{5}{8} \\adding~(-\frac{15}{2} )^2\\(x-\frac{15}{2} )^2 < \frac{5}{8} +(-\frac{15}{2} )^2\\|x-\frac{15}{2} | < \frac{5+450}{8} \\-\sqrt{\frac{455}{8} } < x-\frac{15}{2} < \sqrt{\frac{455}{8} } \\\frac{15}{2} -\sqrt{\frac{455}{8} } < x < \frac{15}{2} +\sqrt{\frac{455}{8} } \\but~x < 0\\7.5-\sqrt{\frac{455}{8} } < x < 0$

8 0

1 year ago

Question 6(Multiple Choice Worth 5 points)

9966 [12]

Question 6: $38.82
question 7: t=5w
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3 0

2 years ago

Read 2 more answers