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sineoko [7]
3 years ago
6

Solve each eqaution justify each epaution 20-3h=2

Mathematics
1 answer:
OleMash [197]3 years ago
5 0
20-3h=2=>3h=20-2=18=>h=18/3=6
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Which mathematical property is demonstrated? 7+9+3=7+3+9
ipn [44]
The property that allows 7 + 9 + 3 = 7 + 3 + 9 is called the Commutative Property of Addition and Multiplication. This means that as long as the sign is either a multiplication symbol or an addition symbol, we can move (or "commute") the values wherever we want and the expression will still equal the same amount.
Hope that helped =)
6 0
3 years ago
Read 2 more answers
Volume of cylinder with a base radius of 5 and height of 3
Anna007 [38]

Answer:

235.5.

Step-by-step explanation:

<u>VOLUME: THE AMOUNT OF SPACE OCCUPIED BY A 3D OBJECT.</u>

The formula for volume of a cylinder is:

pi*r^2*h (3.14 * radius *radius * height)

The radius of this cylinder is 5 and the height is 3.

Now, we can plug the numbers in the formula.

<u>SOLVE:</u>

<u />pi *5^2*3(3.14*5*5*3)=235.5.<u />

The product of the equation is 235.5.

I, therefore, believe the volume of this cylinder is 235.5.

6 0
3 years ago
5,3,7,9, 5, 19,4
Helga [31]
3) median and mode!
Explanation:
Median- 3,4,5,(5),7,9,19
Mode- 5 is repeated.
4 0
2 years ago
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
3/4 + 3/4 = 3/11 + 6/11= 3/10 +
valentina_108 [34]
+ what??????????????
3 0
3 years ago
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