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vlada-n [284]
3 years ago
10

What is the ordered pair of y=4.5x

Mathematics
1 answer:
garri49 [273]3 years ago
5 0

Answer:(1,1)

Step-by-step explanation:

y=4.5x

x=y/4.5

put value of x

y=4.5(y/4.5)

y=y

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I don’t need you to explain just answer.
Bumek [7]

Answer:

Simplify the expression.

i could not type in exponent so just open the screen shot

Step-by-step explanation:

i Hope it works

8 0
3 years ago
The seventh grade students are having an end-of-year party in the school
Ahat [919]

The complete statement is:

  • Determine the number of tables by dividing 754 by 6
  • If there is a remainder, the answer will need to be rounded.
  • There is a remainder, so 126 tables are needed.

<h3>How to determine the number of tables?</h3>

The given parameters are:

Students = 754

Student per table = 6

The number of tables is calculated as:

Table = Students / Student per table

This gives

Table = 754/6

Evaluate the quotient

Table = 125.7

Approximate

Table = 126

Hence, the complete statement is:

Determine the number of tables by dividing 754 by 6

If there is a remainder, the answer will need to be rounded.

There is a remainder, so 126 tables are needed.

Read more about quotient at:

brainly.com/question/629998

#SPJ1

6 0
1 year ago
A 13.5-meter ladder leans against a brick wall and the ladder makes a 75∘ with the wall. The distance d the ladder goes up the w
Alex73 [517]

Answer:

3.5 m

Step-by-step explanation:

Use Pythagorean theorem to find the distance (d) the distance goes up the wall = adjacent to the angle 75° that is formed with the wall.

Length of ladder = the hypothenuse = 13.5 m

Thus, we would use:

Cos75 = adjacent/hypothenuse

Cos 75 = d/13.5

Multiply both sides by 13.5

13.5*cos(75) = d

13.5*0.2588 = d

3.4938 = d

d ≈ 3.5 m (nearest tenth)

3 0
3 years ago
Factor Completely.<br> 12k^2-16k-60
faltersainse [42]

Answer:

4(k - 3)(3k + 5)

Step-by-step explanation:

Given

12k² - 16k - 60 ← factor out 4 from each term

= 4(3k² - 4k - 15) ← factor the quadratic

Consider the factors of the product of the coefficient of the k² term and the constant term which sum to give the coefficient of the k term

product = 3 × - 15 = - 45 , sum = - 4

Factors are - 9 and + 5

Use these factors to split the middle term

3k² - 9k + 5k - 15 → ( factor the first/second and third/fourth terms

= 3k(k - 3) + 5(k - 3) ← factor out (k - 3)

= (k - 3)(3k + 5)

Hence

12k² - 16k - 60 = 4(k - 3)(3k + 5) ← in factored form

8 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
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