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-Dominant- [34]

3 years ago

12

PART A: Write a system of linear equations to model this situation where x represents the pizza and y represents the bags of win

gs. Use the system to answer the following questions.

1 answer:

AleksAgata [21]3 years ago

6 0

Answer:

x -3

Step-by-step explanation:

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The measures of two angles of a triangle are given below. Find the measure of the third angle of the triangle in decimal form.

Sedbober [7]

Answer:

22.8°+65.3=88.1

180-88.1=91.9

5 0

2 years ago

How do I solve this problem

Harman [31]

X = # of posters, y = total cost
y = 2x + 250
y = 1000

1000 = 2x + 250
1000 - 250 = 2x
750 = 2x
750/2 = x
375 = x....so 375 posters can be printed
==========================
y = total cost, x = # of miles

y = 3x + 3
y = 117

117 = 3x + 3
117 - 3 = 3x
114 = 3x
114/3 = x
38 = x.....so the car was towed 38 miles

8 0

2 years ago

Help would be appreciated!

charle [14.2K]

C)
10hrs will get around 105 salaries
90hrs get 840+105=945
95 hours get 945+(105/2)=997.5
100 hours get 945+105=1050

6 0

3 years ago

Lisa wants to frame her little brother's drawing as a gift to her mother. the rectangular drawing is 43.5 centimeters by 934 mil

Alexxx [7]

934 mm = 93.4 cm
You have two 93.4 cm sides and two 43.5 cm sides
Add them together to get 273.8 cm of wood framing

8 0

3 years ago

Read 2 more answers

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