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3 years ago

13

My teacher is making us do online and my problems is to find 62×1000 annex Then there's a blank zeros to Then another blank to f

orm the product So can you help

1 answer:

Readme [11.4K]3 years ago

4 0

Answer:

Step-by-step explanation:

We are to find 62×1000

We can write in standard form

62×10³

6.2×10×10³

Using indices

a^m × a^n = a^(m+n)

Therefore,

6.2×10¹+³

6.2×10⁴

Using the normal multiplication

................1000

...............×. .62

...…......-------------

....., ... ..2 0 0 0

.......+.6 0 0 0

-------------------

..........6 2 0 0 0

----------------------

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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

4 times x with an exponent of 2 plus 2 times y with an exponent of 4. X=-3 y=2

Digiron [165]

Answer:

Step-by-step explanation:

Your equation is: $4*x^{2} +2 * y^{4}$

You also know that x = -3 and y = 2. You can rewrite the equation by substituting x and y with their actual values:

$4*(-3)^{2} +2 * 2^{4}$

First, you want to do the exponentiation (meaning you want to calculate $(-3)^{2}$ and $2^{4}$ parts). $(-3)^{2}$ equals 9 and $2^{4}$ equals 16.

So the equation now is:

$4*9+2*16$

Now you want to do multiplication. $4*9$ equals 36 and $2*16$ equals 32. So you're left with:

$36+32$

Which equals 68.

8 0

3 years ago

What is 30+234567654567

prohojiy [21]

The answer would be 234,567,654,597

5 0

3 years ago

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Which expression best estimates 6 and three-fourths divided by 1 and two-thirds?

bixtya [17]

Answer: The expression is 7/2.

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2 years ago

Georgia ran a 5-kilometer race. How many meters did she run?

kumpel [21]

Answer:

Georgia ran 5000 meters

Step-by-step explanation:

1 kilometer = 1000 meter

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3 years ago

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