All categories

4 years ago

The slope of the graph of y = x is

2 answers:

6 0

Step One: Identify two points on the line.Step Two: Select one to be (x1, y1) and the other to be (x2, y2).Step Three: Use the slope equation to calculate slope. hope this helps

Oliga [24]4 years ago

5 0

Y=X, X=1 therefore y=1. Just like you wouldn't say 1X you say X.

You might be interested in

3(x - 5) = 30 I dont know how to solve this

Anastaziya [24]

Answer:

x=15

Step-by-step explanation:

3(x-5)=30

First, you would distribute the three to the x and the negative five.

3x-15=30

Then you would add 15 to the 15 to cancel it and to the 30 to make it even.

3x-15=30

+15=+15=

3x=45

Finally, you would divide 3 into both sides, cancelling out the three and fiving you,

x=15

Hope this helped!

3 0

3 years ago

Read 2 more answers

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

How tall is the flag pole (please give explanation too)

Tpy6a [65]

Answer:

22 3/4 ft

Step-by-step explanation:

The diagram seems to show a 6 1/2 ft person casting a 9 ft shadow at the same time a flagpole is casting a 31 1/2 ft shadow. You want the height of the flagpole.

<h3>Proportion</h3>

Shadow lengths are presumed proportional to the height of the object making them. This means ...

h/(31 1/2) = (6 1/2)/9 . . . . . all lengths in feet

Multiplying by 31 1/2, we get ...

h = (31 1/2)(6 1/2)/9 = (819/4)/9 = 91/4

h = 22 3/4 . . . . ft

The flagpole is 22 3/4 feet tall.

3 0

2 years ago

You want to walk from home to a clothing store that is 3/8 miles away. You stop for a rest after 1/4 miles. How much farther do

choli [55]

3/8-1/4

3/8-2/8

1/8 miles left

3 0

3 years ago

Read 2 more answers

Help me convert 3/38 into a decimal!

Leya [2.2K]

Answer:

0.07894. . .

Step-by-step explanation:

3 0

3 years ago