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jeka57 [31]
3 years ago
15

If circle B has a radius of 4 and mAC =36, what is the area of the sector ABC?

Mathematics
2 answers:
Dennis_Churaev [7]3 years ago
8 0
Area of circle B = 16pi
16*36/360 = 1.6 pi ~ 5.024 units^2
Svetach [21]3 years ago
7 0
8π/5 or 5.026 :)) 

36/360 --> 1/10 x π(4)²

1/10*16π


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Read the word problem below.
salantis [7]

Answer:

1 3/5 miles/hr

Step-by-step explanation:

Let C be the speed of the current

Let D be the distance between Chan's house and the park

We know that Distance, D, is = Speed x Time

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Chan's speed going upstream is (8 - C)mph.

That gives us:

D = (8-C)(3 hr)

Chan's speed going downstream is (8+C)mph

So we have:

D = (8+C)(2 hr)

We know that the d is the same for these two equations, so:

(8-C)(3 hr)  = (8+C)(2 hr)

24 - 3C = 16 + 2C

5C = 8

C = (8/5) or 1 3/5 mph

5 0
2 years ago
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Writing an equation of a probably given the vertex and focus
White raven [17]

The equation of the vertical parabola in vertex form is written as

y=\frac{1}{4p}(x-h)^2+k

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p=1-6=-5

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6 0
9 months ago
Find the measure of the unknown angle.
Tomtit [17]

Answer:

131° Degrees

Step-by-step explanation:

It's the same just put in different places

4 0
2 years ago
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Find two unit vectos that are orthogonal to both [0,1,2] and [1,-2,3]
alekssr [168]

Answer:

Let the vectors be

a = [0, 1, 2] and

b = [1, -2, 3]

( 1 ) The cross product of a and b (a x b) is the vector that is perpendicular (orthogonal) to a and b.

Let the cross product be another vector c.

To find the cross product (c) of a and b, we have

\left[\begin{array}{ccc}i&j&k\\0&1&2\\1&-2&3\end{array}\right]

c = i(3 + 4) - j(0 - 2) + k(0 - 1)

c = 7i + 2j - k

c = [7, 2, -1]

( 2 ) Convert the orthogonal vector (c) to a unit vector using the formula:

c / | c |

Where | c | = √ (7)² + (2)² + (-1)²  = 3√6

Therefore, the unit vector is

\frac{[7,2,-1]}{3\sqrt{6} }

or

[ \frac{7}{3\sqrt{6} } , \frac{2}{3\sqrt{6} } , \frac{-1}{3\sqrt{6} } ]

The other unit vector which is also orthogonal to a and b is calculated by multiplying the first unit vector by -1. The result is as follows:

[ \frac{-7}{3\sqrt{6} } , \frac{-2}{3\sqrt{6} } , \frac{1}{3\sqrt{6} } ]

In conclusion, the two unit vectors are;

[ \frac{7}{3\sqrt{6} } , \frac{2}{3\sqrt{6} } , \frac{-1}{3\sqrt{6} } ]

and

[ \frac{-7}{3\sqrt{6} } , \frac{-2}{3\sqrt{6} } , \frac{1}{3\sqrt{6} } ]

<em>Hope this helps!</em>

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Answer:

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Step-by-step explanation:

10^2+(3+5)^2-5

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-> 100 + 8^2 - 5

-> 100 + 64 - 5

-> 159

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