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2 years ago

What's a mural with an area of 18m2

Mathematics

1 answer:

Sphinxa [80]2 years ago

6 0

Assuming that the mural is a perfect square, therefore the formula for area is simply:

A = s^2

where A is area and s is the measure of the sides

Therefore from the formula, we can calculate the side lengths of the mural, s:

18 m^2 = s^2

s = sqrt (18 m^2)

s = 4.24 m

<span>So the mural has a dimensions of 4.24 m by 4.24 m</span>

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Each of the 10 students in the baking club made 2 chocolate cakes for a fundraiser. They all used the same recipe, using C cups

8090 [49]

Answer:

The expression is C/20 cups of flour in one cake

Step-by-step explanation:

* Lets explain how to solve the problem

- There are 10 students

- Each one made 2 chocolate cakes

- They all used the same recipe

- The total amount of flour is C cups

* Lets change these information to equation

∵ There are 10 students

∵ Each student made 2 cakes

∴ The total number of cakes = 10 × 2 = 20 cakes

∵ The total amount of flour for the twenty cakes is C cups

- To find the amount of flour in each cake divide the total amount

of flour by the total number of the cakes

∴ The amount of flour in each cake = C/20

* The expression is C/20 cups of flour in one cake

7 0

2 years ago

Let e1= 1 0 and e2= 0 1 , y1= 4 5 , and y2= −2 7 , and let T: ℝ2→ℝ2 be a linear transformation that maps e1 into y1 and maps

Furkat [3]

Answer:

The image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is $\left[\begin{array}{c}24&-8\end{array}\right]$

Step-by-step explanation:

We know that $T:$ $IR^{2}$ → $IR^{2}$ is a linear transformation that maps $e_{1}$ into $y_{1}$ ⇒

$T(e_{1})=y_{1}$

And also maps $e_{2}$ into $y_{2}$ ⇒

$T(e_{2})=y_{2}$

We need to find the image of the vector $\left[\begin{array}{c}4&-4\end{array}\right]$

We know that exists a matrix A from $IR^{2x2}$ (because of how T was defined) such that :

$T(x)=Ax$ for all x ∈ $IR^{2}$

We can find the matrix A by applying T to a base of the domain ( $IR^{2}$ ).

Notice that we have that data :

$B_{IR^{2}}=$ { $e_{1},e_{2}$ }

Being $B_{IR^{2}}$ the cannonic base of $IR^{2}$

The following step is to put the images from the vectors of the base into the columns of the new matrix A :

$T(\left[\begin{array}{c}1&0\end{array}\right])=\left[\begin{array}{c}4&5\end{array}\right]$ (Data of the problem)

$T(\left[\begin{array}{c}0&1\end{array}\right])=\left[\begin{array}{c}-2&7\end{array}\right]$ (Data of the problem)

Writing the matrix A :

$A=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]$

Now with the matrix A we can find the image of $\left[\begin{array}{c}4&-4\\\end{array}\right]$ such as :

$T(x)=Ax$ ⇒

$T(\left[\begin{array}{c}4&-4\end{array}\right])=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]\left[\begin{array}{c}4&-4\end{array}\right]=\left[\begin{array}{c}24&-8\end{array}\right]$

We found out that the image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is the vector $\left[\begin{array}{c}24&-8\end{array}\right]$

3 0

2 years ago

Alicia records the total number of points scored in two games by 10 players on her basketball team. 28, 13, 4, 8, 15, 10, 22, 7,

madam [21]

Answer:

Choices are missing here.

However, we can find the plot that represents the data.

To draw a scatterplot, we just need to draw points, where the indepedent variable is going to be 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, that is the position of each player.

So, all points we need to draw the plot are:

(1, 28)

(2, 13)

(3, 4)

(4, 8)

(5, 15)

(6, 10)

(7, 22)

(8, 7)

(9, 11)

(10, 15)

The image attached shows the plot that represents this data set.