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Kay [80]
3 years ago
14

In the figure below, \overline{AD} AD start overline, A, D, end overline and \overline{BE} BE start overline, B, E, end overline

are diameters of circle PPP. What is the arc measure of \stackrel{\LARGE{\frown}}{CD} CD ⌢ C, D, start superscript, \frown, end superscript in degrees? ^\circ ∘ degrees

Mathematics
1 answer:
VLD [36.1K]3 years ago
6 0

Answer:

<u>The measure of the arc CD = 64°</u>

Step-by-step explanation:

It is required to find the measure of the arc CD in degrees.

So, as shown at the graph

BE and AD are are diameters of circle P

And ∠APE is a right angle ⇒ ∠APE = 90°

So, BE⊥AD

And so, ∠BPE = 90° ⇒(1)

But it is given: ∠BPE = (33k-9)° ⇒(2)

From (1) and (2)

∴ 33k - 9 = 90

∴ 33k = 90 + 9 = 99

∴ k = 99/33 = 3

The measure of the arc CD = ∠CPD = 20k + 4

By substitution with k

<u>∴ The measure of the arc CD = 20*3 + 4 = 60 + 4 = 64°</u>

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X + 2y = 5<br> x - 3y = 7 <br> What is the value of the y-determinant? <br> -2<br> -1 <br> 2
grandymaker [24]

Answer:

y-determinant = 2

Step-by-step explanation:

Given the following system of equation:

  • x + 2y = 5
  • x - 3y = 7

Let's represent it using a matrix:

\left[\begin{array}{ccc}1&2\\1&-3\end{array}\right] = \left[\begin{array}{ccc}5\\7\end{array}\right]

The y‐numerator determinant is formed by taking the constant terms from the system and placing them in the y‐coefficient positions and retaining the x‐coefficients. Then:

\left[\begin{array}{ccc}1&5\\1&7\end{array}\right]

y-determinant = (1)(7) - (5)(1) = 2.

Therefore, the y-determinant = 2

5 0
3 years ago
What does subsequent mean
Olenka [21]
Subsequent means following (in time, order or place f.e.)
6 0
3 years ago
Please Help Brain list and 25 Points whoever gets it right. You have to have a actual explanation and a real answer. Not just B,
den301095 [7]

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3 0
3 years ago
PLEASE HELP! 20 POINTS 1) A ball is thrown starting at a time of 0 and a height of 2 meters. The height of the ball follows the
drek231 [11]

Answer:

1) The height of the ball from 0 to 5  seconds are;

At t = 0 second, height = 2

At t = 1 second, height h = 22.1

At t = 2 seconds, height h = 32.4

At t = 3 seconds, height h = 32.9

At t = 4 seconds, height h = 23.6

At t = 5 seconds, height h = 4.5

2)  The correct option is;

D. -16·t² + 25·t + 1

Step-by-step explanation:

1) The equation of motion of the ball is given as follows;

H(t) = -4.9·t² + 25·t + 2

The height of the ball from 0 to 5 seconds are;

H(0) = -4.9×(0)² + 25×(0) + 2 = 2

H(1) = -4.9×(1)² + 25×(1) + 2 = 22.1

H(2) = -4.9×(2)² + 25×(2) + 2 = 32.4

H(3) = -4.9×(3)² + 25×(3) + 2 = 32.9

H(4) = -4.9×(4)² + 25×(4) + 2 = 23.6

H(5) = -4.9×(5)² + 25×(5) + 2 = 4.5

Therefore, we have;

The height of the ball are

At t = 0 second, height = 2

At t = 1 second, height h = 22.1

At t = 2 seconds, height h = 32.4

At t = 3 seconds, height h = 32.9

At t = 4 seconds, height h = 23.6

At t = 5 seconds, height h = 4.5

2) Given that the equation of the ball is that of a projectile motion, such as follows;

h = h₀ + v₀·sin(θ₀)·t - 1/2·g·t² which is equivalent to h = -1/2·g·t²+ h₀+v₀·sin(θ₀)·t

it is best represented by the quadratic equation of an upside down parabola which is option D. -16·t² + 25·t + 1

6 0
3 years ago
How would you determine the axis of symmetry in order to graph x^2 = 8y^2.
Tanya [424]
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Finding the equation of a parabola is quite difficult but under certain cicumstances we may easily find an equation. Let's place the focus and vertex along the y axis with the vertex at the origin. Suppose the focus is at (0,p). Then the directrix, being perpendicular to the axis, is a horizontal line and it must be p units away from V. The directrix then is the line y=-p. Consider a point P with coordinates (x,y) on the parabola and let Q be the point on the directrix such that the line through PQ is perpendicular to the directrix. The distance PF is equal to the distance PQ. Rather than use the distance formula (which involves square roots) we use the square of the distance formula since it is also true that PF2 = PQ2. We get

<span>(x-0)2+(y-p)2 = (y+p)2+(x-x)2. 
x2+(y-p)2 = (y+p)2.</span>If we expand all the terms and simplify, we obtain<span>x2 = 4py.</span>

Although we implied that p was positive in deriving the formula, things work exactly the same if p were negative. That is if the focus lies on the negative y axis and the directrix lies above the x axis the equation of the parabola is

<span>x2 = 4py.</span><span>The graph of the parabola would be the reflection, across the </span>x<span> axis of the parabola in the picture above. A way to describe this is if p > 0, the parabola "opens up" and if p < 0 the parabola "opens down".</span>

Another situation in which it is easy to find the equation of a parabola is when we place the focus on the x axis, the vertex at the origin and the directrix a vertical line parallel to the y axis. In this case, the equation of the parabola comes out to be

<span>y2 = 4px</span><span>where the directrix is the verical line </span>x=-p and the focus is at (p,0). If p > 0, the parabola "opens to the right" and if p < 0 the parabola "opens to the left". The equations we have just established are known as the standard equations of a parabola. A standard equation always implies the vertex is at the origin and the focus is on one of the axes. We refer to such a parabola as a parabola in standard position.

Parabolas in standard positionIn this demonstration we show how changing the value of p changes the shape of the parabola. We also show the focus and the directrix. Initially, we have put the focus on the y axis. You can select on which axis the focus should lie. Also you may select positive or negative values of p. Initially, the values of p are positive. 

5 0
4 years ago
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