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arlik [135]
2 years ago
6

Find mPRT. P 649U 42° R s

Mathematics
1 answer:
Mkey [24]2 years ago
3 0

Answer: 296 degrees

Step-by-step explanation:

Arc PRT is just the entire circumference - arc PT.

Since the 64 degree angle is a central angle, arc PT measures 64 degrees.

So arc PRT = 360 degree - 64 degree = 296 degrees.

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Answer: Using context clues

Step-by-step explanation:

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3 years ago
The reflection pool is 1 meter deep, 10 meters wide, and 20.5 meters long. What is the volume of the pool
gtnhenbr [62]
Volume = length x width x height
Volume = 1 x 10 x 20.5
Volume = 205 m^3
3 0
3 years ago
13is 26% of what number
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x-\ number\\\\
26\%*x=13\\\\
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4 0
4 years ago
Write an equation that represents a vertical translation 7 units down of the graph of g(x) = 21.
xenn [34]

Answer:

<h2>h(x) = 14</h2>

Step-by-step explanation:

f(x) + n - translation n units up

f(x) - n - translation n units down

f(x + n) - translation n units to the left

f(x - n) - translation n units to the right

===========================================

g(x) = 21

translation 7 units down: g(x) - 7 = 21 - 7 = 14

3 0
3 years ago
A circle is translated 4 units to the right and then reflected over the x-axis. Complete the statement so that it will always be
irga5000 [103]

Answer:

The statement is now presented as:

\exists\, (h,k)\in \mathbb{R}^{2} /f: (x-h^{2})+(y-k)^{2}=r^{2}\implies f': [x-(h+4)]^{2}+[y-(-k)]^{2} = r^{2}

In other words, this mathematical statement can be translated as:

<em>There is a point (h, k) in the set of real ordered pairs so that a circumference centered at (h,k) and with a radius r implies a equivalent circumference centered at (h+4,-k) and with a radius r. </em>

Step-by-step explanation:

Let C = (h,k) the coordinates of the center of the circle, which must be transformed into C'=(h', k') by operations of translation and reflection. From Analytical Geometry we understand that circles are represented by the following equation:

(x-h)^{2}+(y-k)^{2} = r^{2}

Where r is the radius of the circle, which remains unchanged in every operation.

Now we proceed to describe the series of operations:

1) <em>Center of the circle is translated 4 units to the right</em> (+x direction):

C''(x,y) = C(x, y) + U(x,y) (Eq. 1)

Where U(x,y) is the translation vector, dimensionless.

If we know that C(x, y) = (h,k) and U(x,y) = (4, 0), then:

C''(x,y) = (h,k)+(4,0)

C''(x,y) =(h+4,k)

2) <em>Reflection over the x-axis</em>:

C'(x,y) = O(x,y) - [C''(x,y)-O(x,y)] (Eq. 2)

Where O(x,y) is the reflection point, dimensionless.

If we know that O(x,y) = (h+4,0) and C''(x,y) =(h+4,k), the new point is:

C'(x,y) = (h+4,0)-[(h+4,k)-(h+4,0)]

C'(x,y) = (h+4, 0)-(0,k)

C'(x,y) = (h+4, -k)

And thus, h' = h+4 and k' = -k. The statement is now presented as:

\exists\, (h,k)\in \mathbb{R}^{2} /f: (x-h^{2})+(y-k)^{2}=r^{2}\implies f': [x-(h+4)]^{2}+[y-(-k)]^{2} = r^{2}

In other words, this mathematical statement can be translated as:

<em>There is a point (h, k) in the set of real ordered pairs so that a circumference centered at (h,k) and with a radius r implies a equivalent circumference centered at (h+4,-k) and with a radius r. </em>

<em />

4 0
3 years ago
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