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Nady [450]
3 years ago
11

Maya is taking her friends to a concert. Each ticket costs $25, and souvenir T-shirts are $12 each. There is a $4 service fee fo

r the entire purchase. She has $130. If she buys 3 tickets, how many T-shirts can she buy? If x = number of tickets and y = number of T-shirts, which of the following inequalities could you use to answer the problem?
a. 25x+12y 130
b. 25x+12y+4 130
c. 25x+12y+4 130
d. x+y < 130
Mathematics
1 answer:
evablogger [386]3 years ago
8 0

Answer:

25x+12y+4 \leq 130

Step-by-step explanation:

Here, x represents the number of tickets and y represents the number of T-shirts.

As per the statement:

Each ticket costs $25,  and souvenir T-shirts are $12 each

⇒Total cost of ticket = 25x and Total cost for T-shirt = 12y

It is also given that there is a $4 service fee for the entire purchase.

25x+12y+4

Since, she has $130

⇒25x+12y+4 \leq 130

Therefore, an inequality could you use to answer the problem is, 25x+12y+4 \leq 130

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  (3.346 x 10²⁶) (1.26 x 10²¹)

= (3.346 x 1.26) x 10²⁶⁺²¹

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Enter expression, e.g. (x^2-y^2)/(x-y)
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Which Equation could be used to find m, the number of miles he drove on his last trip?
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I think C is the current answer

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The cost of 2 kg of onions is ₹24. What will the cost of 12 kg of onions be?​
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Two particles travel along the space curves r1(t) = t, t2, t3 r2(t) = 1 + 2t, 1 + 6t, 1 + 14t . Find the points at which their p
GuDViN [60]

Answer:

A) points at which paths intersect : (1,1,1) ; (2,4,8)

B) DNE

Step-by-step explanation:

A) To find the points in which the particle paths intersect, it is necessary to find the values of t for which the three components of both vectors are equal:

t_1=1+2t_2\\\\t_1^2=1+6t_2\\\\t_1^3=1+14t_2

you replace t1 from the first equation in the second equation:

(1+2t_2)^2=1+6t_2\\\\1+4t_2+4t_2^2=1+6t_2\\\\4t_2^2-2t_2=0\\\\t_2(2t_2-1)=0\\\\t_2=0\\\\t_2=\frac{1}{2}

Then, for t2 = 0 and t2=1/2 you obtain for t1:

t_1=1+2(0)=1\\\\t_1=1+2(\frac{1}{2})=2

Hence, for t1=1 and t2=0 the paths intersect. Furthermore, for t1=2 and t2=1/2 the paths also intersect.

The points at which the paths  intersect are:

r_1(1)=(1,1,1)=r_2(0)=(1,1,1)\\\\r_1(2)=(2,4,8)=r_2(\frac{1}{2})=(2,4,8)

B) You have the following two trajectories of two independent particles:

r_1(t)=(t,t^2,t^3)\\\\r_2(t)=(1+2t,1+6t,1+14t)

To find the time in which the particles collide, it is necessary that both particles are in the same position on the same time. That is, each component of the vectors must coincide:

t=1+2t\\\\t^2=1+6t\\\\t^3=1+14t

From the first equation you have:

t=1+2t\\\\t=-1

This values does not have a physical meaning, then, the particle do not collide

answer: DNE

5 0
3 years ago
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