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LenaWriter [7]
3 years ago
15

The lifespan of a guinea pig is 6 years less than that of a giraffe. The lifespan of a tiger is 4 times that of the guinea pig.

If the total lifespan of the animals is 30 years, calculate the longevity for the giraffe.
Mathematics
1 answer:
Verdich [7]3 years ago
4 0

Answer

calculate the longevity for the giraffe.

To prove

Let us assume lifespan of a guinea pig = u

Let us assume lifespan of a giraffe = v

Let us assume lifespan of a tiger = w

As given

The lifespan of a guinea pig is 6 years less than that of a giraffe.

The lifespan of a tiger is 4 times that of the guinea pig.

Than

u = v - 6

w = 4u

As given

total lifespan of the animals is 30 years.

than the equation becomes

u + v + w = 30

put the u = v - 6 ,w = 4u in the above equation

v - 6 + v + 4u = 30

put u = v - 6 in the above equation

v-6 + v + 4 × (v - 6) = 30

v - 6 + v + 4v - 24 = 30

6v = 30 + 24 + 6

6v = 60

v = \frac{60}{6}

v = 10 years .

Therefore the  longevity for the giraffe is 10 years .


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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

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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]

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X + 2y =6<br> X- y =3<br> Solution?
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Since one equation has a negative y and the other has a positive y, I'm going to use those since they cancel each other out. Before that, the two y's need to be equal to each other.

x+2y=6
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Multiply the bottom equation by two so then you have:

x+2y=6
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The y's now cancel out:

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Add them together

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Divide

x=4.

To find y, plug x into either equation (*don't have to do both, but I will)

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The answer is:
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