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Novosadov [1.4K]
3 years ago
7

What is the GCF of 56 and 27

Mathematics
2 answers:
almond37 [142]3 years ago
6 0
I think it may be 3))))
valina [46]3 years ago
5 0
The GCF for 56 and 27 is 1
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An 8-inch dinner knife is sitting on a circular plate so that its ends are on the edge of the plate. If the minor arc that is in
ankoles [38]

The diameter of the plate is 9.24 inches

Step-by-step explanation:

Let us consider the radius to be 'r'

Angle = 120°

Chord length = 8 inches

Chord length = 2rsin(θ/2)

8 = 2r sin(120/2)

8 = 2r sin60°

8/2 = r (√3/2)

4/(√3/2) = r

r = 8/√3

r = 8/1.73

r = 4.62 inches

Diameter of the plate = 2(4.62)

= 9.24 inches

The diameter of the plate is 9.24 inches

7 0
3 years ago
What is the value of the x in the figure? Plz help ASAP
zalisa [80]

Answer:

2x+15=145........ vertically Opposite

2x=145-15

2x=130

x=130/2

65

x=65

4 0
2 years ago
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In the same diagram of the staircase railing, what is the value of n? (Picture included)
BaLLatris [955]
Hello,

Angle BCH and CHJ are alternate-interior ( and have the same measure)

82+30=3n+7
==>3n=105
==>n=35
8 0
3 years ago
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Find the equation for parabola<br> Focus(3,4)<br> Directrix y=2
sveta [45]

Answer:

Step-by-step explanation:

we know that distance d from the focus to P should be the same to the distance from P to the directrix

(x-h)^2=4p(y-k)

we need to find the y coordinate,

x is the same from focus, 3

y=(3, (4+2)/2)=(3,3)

we find p now by subtracting the y from the focus from the y that we just found

p=4-3=1

again (x-h)^2=4p(y-k), p=1

(x-3)^2=4(1)(y-3)

(x-3)^2=4(y-3),  (x-3)^2=4y-12

simplify

4y=(x-3)^2+12

y=((x-3)^2)/4 + 3

4 0
3 years ago
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Let v⃗ 1=⎡⎣⎢033⎤⎦⎥,v⃗ 2=⎡⎣⎢1−10⎤⎦⎥,v⃗ 3=⎡⎣⎢30−3⎤⎦⎥ be eigenvectors of the matrix A which correspond to the eigenvalues λ1=−1, λ2
kaheart [24]

Answer:

- x as a linear combination :

x = -1 v1+ 0 v2+ 1 v3.

- Transpose Ax = (12, -6, -6)

Step-by-step explanation:

Given v1 = (0 3 3),v2 = (1 −1 0), v3 = (3 0 −3) be eigenvectors of the matrix A which correspond to the eigenvalues λ1 = −1, λ2 = 0, and λ3 = 1, respectively, and let x = (−2 −4 0). Express x as a linear combination of v1, v2, and v3, and find Ax .

To write x as a linear combination of v1, v2, and v3

x = -1 v1+ 0 v2+ 1 v3.

To find Ax

Write A = (0 ......3 ......3 )

...................(1 ......-1 ......0)

...................(3 ......0......-3)

Since transpose x = (-2, 4, 0)

Ax =......... (0 ......3......3 )(-2)

...................(1 ......-1 ......0)(4)

...................(3 ......0......-3)(0)

= (0×-2 + 3×4 + 3×0)

...(1×-2 + -1×4 + 0×0)

.. (3×-2 + 0×4 + -3×0)

As = (12)

....(-6)

....(-6)

Transpose Ax = (12, -6, -6)

7 0
3 years ago
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