3/1/2+1/5/6÷1/3/5 <br>hello good afternoon can u please solve this question

8 times the difference between a number and 6 is equal to 4 times the number...

astraxan [27]

Answer:

(A) 12

Step-by-step explanation:

First, lets define the number as x.

"8 times the difference between a number and 6 is equal to 4 times the number" is translated into "8(x - 6) = 4x"

8(x - 6) = 4x

8x - 48 = 4x

4x = 48

x = 12

6 0

2 years ago

Read 2 more answers

45 miles in 40 minutes

charle [14.2K]

Hi there!

45 mi to 40 min = 1.125 mi per min

45 ÷ 40 = 1.124

Hope this helps!

8 0

3 years ago

Tamara makes 12 identical flower arrangements, each with 3 roses and some carnations. She uses a total of 96 flowers.

storchak [24]

Answer:

5 carnations in each arrangement

Total of 60 carnations used

Step-by-step explanation:

First calculate how many flowers are in each arrangement by dividing the total number of flowers by the number of arrangements:

96 ÷ 12 = 8

Given each arrangement has 3 roses, subtract 3 from 8 to find the number of carnations in each arrangement:

8 - 3 = 5

Therefore, there are 5 carnations in each arrangement

Total number of carnations used = 5 x 12 = 60

5 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

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

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