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Anarel [89]
3 years ago
11

The question is in the attachment. Please answer

Mathematics
1 answer:
MAVERICK [17]3 years ago
6 0

Answer:

106 = arc AC

Step-by-step explanation:

Inscribed Angle = 1/2Intercepted Arc

53 = 1/2 arc AC

Multiply each side by 2

2*53 = arc AC

106 = arc AC

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Jill says he did it wrong. He should have multiplied the bottom first (3 x 5) and then multiplied by the height. Explain to Jill
inna [77]

Answer:

Step-by-step explanation:

Jack's method is equivalent because of the "law of commutative property", which means that multiplication problems have the same result no matter the order.

2x5 is the same as 5x2

8 0
3 years ago
What is radians converted to degrees? If necessary, round your answer to the nearest degree.
Tomtit [17]

To convert radians to degrees, make use of the fact that pi<span> radians equals one half circle, or 180º.</span>

<span>This means that if you divide radians by </span>pi, the answer is the number of half circles. Multiplying this by 180º will tell you the answer in degrees.

<span>So, to convert radians to degrees, multiply by <span>180/</span></span>pi, like this:

Degrees= radians X 180/pi

<span>                                    </span>

3 0
3 years ago
Solve for x. 15x + 6 = 10x + 21 x = -5 x = 5 x = 3 x = 5
klio [65]
Move all the x terms to one side and solve for x. 15x-10x=21-6 : 5x/5=15/5 : x=3
4 0
3 years ago
Does anyone know how to Graph f(x)=51(2)^x
3241004551 [841]
This is an exponential function.   

If x = 0, 2^x = 2^0 = 1.  The beginning value of 2^x is 1 and the beginning value of 51*2^x is 51.

Make a table and graph the points:

x        y=51*2^x                                point (x,y)
--       ---------------                            ---------------
0             51                                       (0,51)
2            51*2^2 = 51(4) = 204           (2,204)             and so on.

The graph shows up in both Quadrants I and II.  Its y-intercept is (0,51).  Its slope is always positive.

5 0
3 years ago
Read 2 more answers
What is the area of the region bounded between the curves y=x and y=sqrt(x)?
erastova [34]

Answer:

\frac{1}{6}

Step-by-step explanation:

y = x     .....(1)

y=\sqrt{x}     .... (2)

By solving equation (1) and equation (2)

x = \sqrt{x}

\sqrt{x}\left ( \sqrt{x}-1 \right )=0

\sqrt{x}=0 or \left ( \sqrt{x}-1 \right )=0

x = 0, x = 1

y = 0, y = 1

A = \int_{0}^{1}ydx(curve)-\int_{0}^{1}ydx(line)

A = \int_{0}^{1}\sqrt{x}dx-\int_{0}^{1}xdx

A = \frac{2}{3}[x^\frac{3}{2}]_{0}^{1}-\frac{1}{2}[x^2]_{0}^{1}

A = \frac{2}{3}-\frac{1}{2}

A = \frac{1}{6}

8 0
3 years ago
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