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liraira [26]
3 years ago
15

What is the range of the function y = -3cosx - 1?

Mathematics
1 answer:
Bingel [31]3 years ago
8 0
Range of y = -3 cosx -1 is -4 < y < 2
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luda_lava [24]

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Im stuck on this question plz help
astra-53 [7]

Answer:

C) x = 20.8

Step-by-step explanation:

Since the line with the length equal to 6 forms a right angle on line x then we know it is perpendicular and bisects it.

Plug in the values: a^2 + b^2 = c^2

a^2 + 6^2 = 12^2

a^2 + 36 = 144

a^2 + 36 - 36 = 144 - 36

a^2 = 108

√a^2 = √108

a = √108

a = 10.3923048454

a = 1/2 x

2a = x

2(10.3923048454) = x

x = 20.7846096908

x = 20.8

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3 years ago
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A hotel offers two activity packages. One costs $192 and includes 3 h
poizon [28]

Let x = Horseback riding and y = Parasailing

Eq. 1

3x + 2y = 192

Eq. 2

2x + 3y = 213

By elimination:

<span>     2(3x + 2y = 192)</span>

+ -3(2x + 3y = 213)

<span>              -5y = - 255</span>

y = $ 51

Substitute y :

3x + 2y = 192

3x + 2(51) = 192

<span>x = $ 30
</span>


<span>Hope this answer will be a good help for you.</span>

3 0
3 years ago
Help me. i swear... if u dont. u are an uncultured swine
igor_vitrenko [27]

Answer:

a

Step-by-step explanation:

6 0
2 years ago
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In P2, find the change-of-coordinates matrix from the basis B = {1 − 3t 2 , 2 + t − 5t 2 , 1 + 2t} to the standard basis of P2.
Serjik [45]

Complete Question:

In P2, find the change-of-coordinates matrix from the basis B = {1 − 3t² , 2+t− 5t² , 1 + 2t} to the standard basis C = {1, t, t²}. Then, write t² as a linear combination of the polynomials in B.

Answer:

The change of coordinate matrix is :

M = \left[\begin{array}{ccc}1&2&1\\0&1&2\\-3&-5&0\end{array}\right]

U = t² = 3 [1 − 3t²] - 2 [2+t− 5t²] + [1 + 2t]

Step-by-step explanation:

Let U =  {D, E, F} be any vector with respect to Basis B

U = D [1 − 3t²] + E [2+t− 5t²] + F[1 + 2t]..............(*)

U = [D+2E+F]+ t[E+2F] + t²[-3D-5E]...................(**)

In Matrix form;

\left[\begin{array}{ccc}1&2&1\\0&1&2\\-3&-5&0\end{array}\right] \left[\begin{array}{ccc}D\\E\\F\end{array}\right] = \left[\begin{array}{ccc}D+2E+F\\E+2F\\-3D-5E\end{array}\right]

The change of coordinate matrix is therefore,

M = \left[\begin{array}{ccc}1&2&1\\0&1&2\\-3&-5&0\end{array}\right]

To find D, E, F in (**) such that U = t²

D + 2E + F = 0.................(1)

E + 2F = 0.........................(2)

-3D -5E = 1........................(3)

Substituting eqn (2) into eqn (1 )

D=3F...................................(4)

Substituting equations (2) and (4) into eqn (3)

-9F+10F=1

F = 1

Put the value of F into equations (2) and (4)

E = -2(1) = -2

D = 3(1) = 3

Substituting the values of D, E, and F into (*)

U = t² = 3 [1 − 3t²] - 2 [2+t− 5t²] + [1 + 2t]

3 0
2 years ago
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