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2 years ago

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bridget was thinking of a number bridget subtracts 7.7 from the number and gets an answer of 68 form an equation with x from the

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1 answer:

bekas [8.4K]2 years ago

8 0

Answer:

68 + 7.7 = x

Step-by-step explanation:

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mart [117]

Answer:

the common ratio is

1/3

because

9*1/3=3

3*1/3=1

1*1/3=1/3

...

8 0

2 years ago

The points (3,10) and (5,t) fall on a line with a slope of -4 what t value

Elenna [48]

Answer:

t=2

Step-by-step explanation:

The slope is given by

m = (y2-y1)/(x2-x1)

We have points (3,10) and (5,t) and the slope is -4

Substituting these in

-4 = (t-10)/ (5-3)

Simplifying

-4 = (t-10) / 2

Multiply by 2 on both sides

-4 = (t-10) / 2 *2

-8 = t-10

Add 10 on both sides

-8+10 = t-10+10

2 = t

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3 years ago

Solve for x and y<br> y+2x=2<br> y=-4x+10

Katena32 [7]

Answer:

x=4 and y=(-6)

Step-by-step explanation:

Solve for the first variable in one of the equations, then substitute the result into the other equation.

7 0

2 years ago

I have 2 questions 1.Matt needs to buy 13 cases of oil. Each case contains 24 bottles. How many bottles of oil will Matt have?

hammer [34]

Answer:

cool

Step-by-step explanation:

8 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