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3 years ago

A map measuring 60 cm by 25 cm is reduced twice in the ratio 3:5. Calculate the final dimensions of the map.

1 answer:

8 0

60 / 3 = 20. 25 / 5 = 5. This means that the you get 20cm by 5 cm. However, you must do the sum again with the new dimensions. 20 / 3 = 6.6666666. 5 / 5 = 1. The final dimensions of the map are 6.66666cm (6 1/3 cm) by 1cm.

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Consider the linear transformation T from V = P2 to W = P2 given by T(a0 + a1t + a2t2) = (2a0 + 3a1 + 3a2) + (6a0 + 4a1 + 4a2)t

Svet_ta [14]

Answer:

$[T]EE=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]$

Step-by-step explanation:

First we start by finding the dimension of the matrix [T]EE

The dimension is : Dim (W) x Dim (V) = 3 x 3

Because the dimension of P2 is the number of vectors in any basis of P2 and that number is 3

Then, we are looking for a 3 x 3 matrix.

To find [T]EE we must transform the vectors of the basis E and then that result express it in terms of basis E using coordinates and putting them into columns. The order in which we transform the vectors of basis E is very important.

The first vector of basis E is e1(t) = 1

We calculate T[e1(t)] = T(1)

In the equation : 1 = a0

$T(1)=(2.1+3.0+3.0)+(6.1+4.0+4.0)t+(-2.1+3.0+4.0)t^{2}=2+6t-2t^{2}$

$[T(e1)]E=\left[\begin{array}{c}2&6&-2\\\end{array}\right]$

And that is the first column of [T]EE

The second vector of basis E is e2(t) = t

We calculate T[e2(t)] = T(t)

in the equation : 1 = a1

$T(t)=(2.0+3.1+3.0)+(6.0+4.1+4.0)t+(-2.0+3.1+4.0)t^{2}=3+4t+3t^{2}$

$[T(e2)]E=\left[\begin{array}{c}3&4&3\\\end{array}\right]$

Finally, the third vector of basis E is $e3(t)=t^{2}$

$T[e3(t)]=T(t^{2})$

in the equation : a2 = 1

$T(t^{2})=(2.0+3.0+3.1)+(6.0+4.0+4.1)t+(-2.0+3.0+4.1)t^{2}=3+4t+4t^{2}$

Then

$[T(t^{2})]E=\left[\begin{array}{c}3&4&4\\\end{array}\right]$

And that is the third column of [T]EE

Let's write our matrix

$[T]EE=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]$

T(X) = AX

Where T(X) is to apply the transformation T to a vector of P2,A is the matrix [T]EE and X is the vector of coordinates in basis E of a vector from P2

For example, if X is the vector of coordinates from e1(t) = 1

$X=\left[\begin{array}{c}1&0&0\\\end{array}\right]$

$AX=\left[\begin{array}{ccc}2&3&3\\6&4&4\\-2&3&4\end{array}\right]\left[\begin{array}{c}1&0&0\\\end{array}\right]=\left[\begin{array}{c}2&6&-2\\\end{array}\right]$

Applying the coordinates 2,6 and -2 to the basis E we obtain

$2+6t-2t^{2}$

That was the original result of T[e1(t)]

8 0

3 years ago

Additive inverse of 2/8

balu736 [363]

Answer:

-2/8is it's answer

Step-by-step explanation:

mark as Brainlist

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3 years ago

Neeeed help plssss solve just one to see how the way to solve others because I don’t understand the question ❤️ pls pls pleaaaas

spin [16.1K]

The descriminant is b^2-4ac so for 26 it would be -5^2 - 4 (1) (7). The quadratic formula is -B plus or minus the square root of b squared minus 4ac over 2a

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2 years ago

What is 10^-10 in decimal form?

S_A_V [24]

10^-10 in decimal form is 0.000000001. Because the exponent is negative, you move the decimal point ( decimal point is at the end of 0 of the base number 10) 10 place to the right of 10 and u get the answer above. I discarded the 0 later in the answer because now that you've moved the decimal, that 0 have no value. But when forming it in decimal you have to use that 0 or you will get a wrong answer, instead of eight 0's, you will end up with nine 0's.

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3 years ago

The diameter of a circle is equivalent to the

jasenka [17]

Answer:

Step-by-step explanation:

The diameter of a circle is twice the radius, so the statement is false.

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2 years ago

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