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SVEN [57.7K]
3 years ago
5

Robert is purchasing some equipment for his baseball team. He wants to buy some baseballs, each priced at m dollars. He has alre

ady spent b dollars on other equipment.
The following expression represents the total amount Robert will spend after buying x baseballs.



If the amount already spent by Robert on other equipment, b dollars, is doubled, then the value of m will______________________ .

If the cost of each baseball, m dollars, is tripled, then the value of b will_______________ and the value of mx will ________________________.
Mathematics
1 answer:
KiRa [710]3 years ago
4 0

Answer:

Part A: (1/2)m

Part B: (1/3rd)b and 3mx

Step-by-step explanation:

Price of each baseball is=$m

Price of other equipment are=$b collectively

Total amount He spent on purchasing equipment is =(m+b)$

Part A:

in part A the amount on purchasing other equipment than baseball is getting doubled i.e. $2b <u> then value of m will be (1/2)half of the total amount spent on purchasing baseball i.e. (1/2)m</u>

Part B:

in part B the amount spent on purchasing baseball is getting tripled i.e. <u>3m</u>, then value of b(amount of purchasing other equipment) is<u> (1/3rd) of total</u> <u>amount spent on purchasing other equipment</u> and value of mx will be<u> 3mx</u>

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Josh $600 into an account that pays simple interest at a rate of 4% per year. How much interest will he be paid in the first 4 y
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3 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]

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