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2 years ago

Solve for x. 3/8= 15x

Mathematics

2 answers:

Colt1911 [192]2 years ago

7 0

Answer:

x=1/40

Step-by-step explanation:

The decimal form is x=0.025

Mariulka [41]2 years ago

3 0

^^^^^^^ agreed ^^^^^^

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3(4k+n) = 12k +n or 12k+3n?

blagie [28]

Answer:

12k+3n

Step-by-step explanation:

5 0

2 years ago

PLS HELP IF YOU CAN!THIS IS DUE IN 30 MINUTES :(

ExtremeBDS [4]

Answer:

Length: 20 or 40 feet

Width: 40 or 20 feet

Step-by-step explanation:

Area = 800 = length*width = x*(60-x)= 60x - x^2

- x^2 + 60x - 800 = 0

x = 20

Or x = 40

So length can be either 20 feet or 40 feet and width can be 40 (60-20) or 20 (60-40)

6 0

2 years ago

3, 12, 48, 192, 768, . . .

Sav [38]

The answer would be the sequence has a common ratio of 4.
The explanation for this would be:
To get the common ratio for get the simplest form of the two numbers, so 12/3 = 4 and 48/12 = 4So we know that their common ratio is 4.
To check:
3 x 4 = 12
12 x 4 = 48
48 x 4 = 192
192 x 4 = 768

7 0

3 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

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

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