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I am Lyosha [343]
3 years ago
10

Halla la suma de 3, ocho veces un número, y tres veces el mismo número

Mathematics
1 answer:
prohojiy [21]3 years ago
6 0

Answer:

huh?

Step-by-step explanation:

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Samuel has to sell concert tickets worth at least $90. The price of a child ticket is $8, and the price of an adult ticket is $1
alina1380 [7]

Answer:

The second graph is correct, that the line intersects x at 6 and y at 11.5

Step-by-step explanation:

Samuel has to sell at least $90 for his concert. According to the graph, if Samuel sells only child ticket, he will have to sell 11.5 tickets.

If Samuel sells only adult tickets, he will have to sell at least 6.

The correct graph is the second one, that the line intersects x at 6 and y at 11.5

4 0
3 years ago
Tickets to a show cost $5.50 for adults and $4.25 for students. A family is purchasing 2 adult tickets and 3 student tickets. Wh
MArishka [77]

Answer:

$23.75

........................

8 0
3 years ago
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Please help solve these equations! Show steps(: thank you
Bond [772]
Y= x^2-58
X^2-58=0
X^2=58
X= sqrt of 58
X= 7.6

Y= x^2+13
X^2+13=0
X^2=-13
Get sq rt of 13 the negative becomes i
X= 3.6i

6 0
3 years ago
Explain the steps needed to determine the value of the ex below. Be sure to provide the correct value of the expression in your
yuradex [85]
The correct answer is -1 and 1/2.

First, you need to divide 1/2 by -4/5. To do this, you multiply 1/2 by -5/4. You will get -10/8.

Then, you add -10/8 with -2/8 to get -12/8, or -1 and 1/2.
6 0
3 years ago
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The fuel economy of a car, measured in miles per gallon, is modeled by the function ƒ(s) = –0.009s2 + 0.699s + 12 where s repres
Kisachek [45]

Answer:

The maximum fuel economy of the car is 25.57225

Step-by-step explanation:

we are given

f(s)=-0.009s^2+0.699s+12

where

where s represents the average speed of the car, measured in miles per hour

So, we have to maximize f(s)

So, firstly we will find derivative

f'(s)=-0.009*2s+0.699*1+0

f'(s)=-0.018s+0.699

now, we can set it to 0

and then we can solve for s

f'(s)=-0.018s+0.699=0

s=\frac{233}{6}

To find maximum fuel economy , we can plug s into f(s)

f(\frac{233}{6})=-0.009(\frac{233}{6})^2+0.699(\frac{233}{6})+12

f(\frac{233}{6})=25.57225

So,

The maximum fuel economy of the car is 25.57225

4 0
3 years ago
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