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3 years ago

What are the steps to converting 720 seconds into hours?

2 answers:

7 0

First you have to know how many minutes are in 720 seconds. A minute is 60 seconds, so you can divide 720 by 60 and get 12 minutes. Then, since there are 60 minutes in an hour, you divide 12 by 60 which gets you two-tenths of an hour, or .2 of an hour, or 20% of an hour.

Nataly [62]3 years ago

3 0

This conversion of 720 seconds to hours has been calculated by multiplying 720 seconds by 0.0002 and the result is 0.2 hours.

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GUYS CAN YOU PLEASE HELP ME? Write an equation for each word problem.

svp [43]

1. 584x3=1752
2. 90/2=45. 45-3=42. 45+3=48. Madison has $48 and Saul has $42.

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3 years ago

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A shoe company sends out 7 trucks loaded with nursing shoes that cost the company $ 1,572per truck if they send out 750 boxes pe

weqwewe [10]

1572*7=11004

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3 years ago

What is the value of t in the equation t+12=30 ?

masha68 [24]

Simplifying X + 12 = 30
Reorder the terms: 12 + X = 30
Solving 12 + X = 30
Solving for variable 'X'.
Move all terms containing X to the left, all other terms to the right.
Add '-12' to each side of the equation. 12 + -12 + X = 30 + -12
Combine like terms: 12 + -12 = 0 0 + X = 30 + -12 X = 30 + -12
Combine like terms: 30 + -12 = 18 X = 18
Simplifying X = 18

6 0

2 years ago

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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)]

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3 years ago

What is the standard form of the equation of the circle?

lisabon 2012 [21]

The standard form of the equation of a circle is:
$(x - a) {}^{2} + (y - b) {}^{2} = r {}^{2}$
Where the point (a,b) represents the centre and r is the radius. So, given your information, the equation will be:
${(x + 4)}^{2} + {(y + 3)}^{2} = 25$

4 0

3 years ago

Read 2 more answers