Jerry's return trip took 2/5 hour longer. How long did his return trip take?

Resolver la siguiente ecuación por eliminación o sustitución: a-8b= -9 a-2b= -7

Crazy boy [7]

Answer: Solution: $a=\dfrac{-19}{3}$ and $b=\dfrac{1}{3}$ .

Step-by-step explanation:

The given pair of equations:

$a-8b= -9\ ...(i)\\\\ a-2b= -7\ ...(ii)$

Eliminate equation (i) from (ii) , we get

$-2b-(-8b)=-7-(-9)\\\\\Rightarrow\ -2b+8b=-7+9\\\\\Rightarrow\ 6b=2\\\\\Rightarrow\ b=\dfrac{1}{3}$

Put value of b in (ii) , we get

$a-2\times\dfrac{1}{3}=-7\\\\\Rightarrow\ a=-7+\dfrac{2}{3}\\\\\Rightarrow\ a=\dfrac{-21+2}{3}\\\\\Rightarrow\ a=\dfrac{-19}{3}$

Solution: $a=\dfrac{-19}{3}$ and $b=\dfrac{1}{3}$ .

4 0

3 years ago

I will give brainliest, you have to put these in the correct order (24 points)

Vlada [557]

Answer:

First i do not know sorry please help me

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

a line passes through the point (6, -9) and has a slope of 3/2. write an equation in point-slope form for this line

Stella [2.4K]

Y=3/2x is the answer

3 0

3 years ago

Read 2 more answers

HELP! WILL GIVE BRANLIEST!

avanturin [10]

Answer: 5 Hope this helped-

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

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

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

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