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Oxana [17]
2 years ago
13

The volleyball team at West View High School is comparing T-shirt companies where they can purchase their practice shirts. The t

wo companies, Shirt Box and Just Tees, are represented by this system of equations where x is the number of T-shirts and y is the total cost of the T-shirts. y = 10.5x y = 7.5x + 30 How many T-shirts would the volleyball team need to purchase from each company for the total cost to be equal? For the total cost to be the same for both companies, the volleyball team would need to purchase T-shirts from each company for a total of $ for each company.
Mathematics
2 answers:
mina [271]2 years ago
5 0
10.5x = 7.5x + 30
10.5x - 7.5x = 30
3x = 30
x = 30/3
x = 10

10.5(10) = 105
7.5(10) + 30 = 75 + 30 = 105

they need to purchase 10 t-shirts from each company for a total of $105 for each company
Veronika [31]2 years ago
5 0

Answer:

On edu the correct answer is y=7.5+30 and y=10.5x

hope this helps you!

Step-by-step explanation:

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Which is equivalent to-2(4-3x)+ (5x-2)
almond37 [142]

Answer:

Step-by-step explanation:

-2(4-3x)+(5x-2)

-8+6x+5x-2

11x-10

3 0
2 years ago
Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i
jenyasd209 [6]

Answer:

If the limit that you want to find is \lim_{x\to \infty}\dfrac{P(x)}{Q(x)} then you can use the following proof.

Step-by-step explanation:

Let P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0} and Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0} be the given polinomials. Then

\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}

Observe that

\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}

and

\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}

Then

\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}

3 0
2 years ago
Sheila has $47 to buy graphing pads for her math class. Each pad costs $8,
bija089 [108]

Answer:

5 tablets can be bought

Step-by-step explanation:

47/8= 5

8 0
2 years ago
Arnold borrowed $7890 at 11.5 percent for five years. How much did Arnold pay in interest?
Rudiy27
937.5 is what I got.
5 0
2 years ago
Read 2 more answers
A length is measured as 21cm correct to 2 significant figures. what is the upper bound for the length?​
Leno4ka [110]

the upper bound for the length is  21.5  cm .

<u>Step-by-step explanation:</u>

Lower and Upper Bounds

  • The lower bound is the smallest value that will round up to the approximate value.
  • The upper bound is the smallest value that will round up to the next approximate value.

Ex:- a mass of 70 kg, rounded to the nearest 10 kg, The upper bound is 75 kg, because 75 kg is the smallest mass that would round up to 80kg.

Here , A length is measured as 21cm correct to 2 significant figures. We need to find what is the upper bound for the length . let's find out:

As discussed above , upper bound for any number will be the smallest value in decimals which will round up to next integer value . So , for 21 :

⇒ 21.5  cm

21.5 cm on rounding off will give 22 cm . So , the upper bound for the length is  21.5  cm .

7 0
2 years ago
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