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BabaBlast [244]
3 years ago
12

The carnival charges $5 admission and $0.50 for each ride. Alex's dad gave him $15 to spend. How many rides can he go on at most

? Write and solve an inequality to answer the question.
Mathematics
2 answers:
Oxana [17]3 years ago
8 0

Let us suppose number of rides is x.One ride costs $0.50.

Cost of x rides is 0.50x. The carnival charges is $5. Alex has $15 to spend .

Number of rides at most Alex can have can be given by the inequality:

0.50x+5≤ 15

To solve for x we subtract 5 both sides

0.50x≤ 15-5

0.50x≤ 10

Dividing both sides by 0.50 we have :

x≤ 20

Number of rides at most Alex can have is 20.

8_murik_8 [283]3 years ago
5 0
He can ride on 20 rides. 15-5=10 then 10/.50=20. Hope that helps
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Answer:

5 apples

Step-by-step explanation:

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tanji creates a "square, circle" repeating pattern. Kenji creates a "Square, circle, trianglr, cricle" repeating pattern. if bot
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Both tanji & kenji have same number of circles i.e. 50 .

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

Here we have , tanji creates a "square, circle" repeating pattern. Kenji creates a "Square, circle, triangle, circle" repeating pattern. if both tanji and kenji have 100 shapes in their patterns,  . We need to find that which pattern contains more circles .Let's find out :

<u>tanji creates a "square, circle" </u>

it's given that tanji has 100 shapes and he is repeating this pattern . In this pattern circle comes 1 every time out of 2 shapes :

Total number of times circle = \frac{100}{2}(1)

Total number of times circle = 50

<u>Kenji creates a "Square, circle, triangle, circle" </u>

it's given that tanji has 100 shapes and he is repeating this pattern . In this pattern circle comes 2 every time out of 4 shapes :

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3 0
3 years ago
A cylindrical can without a top is made to contain 25 3 cm of liquid. What are the dimensions of the can that will minimize the
Basile [38]

Answer:

Therefore the radius of the can is 1.71 cm and height of the can is 2.72 cm.

Step-by-step explanation:

Given that, the volume of cylindrical can with out top is 25 cm³.

Consider the height of the can be h and radius be r.

The volume of the can is V= \pi r^2h

According to the problem,

\pi r^2 h=25

\Rightarrow h=\frac{25}{\pi r^2}

The surface area of the base of the can is = \pi r^2

The metal for the bottom will cost $2.00 per cm²

The metal cost for the base is =$(2.00× \pi r^2)

The lateral surface area of the can is = 2\pi rh

The metal for the side will cost $1.25 per cm²

The metal cost for the base is =$(1.25× 2\pi rh)

                                                 =\$2.5 \pi r h

Total cost of metal is C= 2.00 \pi r^2+2.5 \pi r h

Putting h=\frac{25}{\pi r^2}

\therefore C=2\pi r^2+2.5 \pi r \times \frac{25}{\pi r^2}

\Rightarrow C=2\pi r^2+ \frac{62.5}{ r}

Differentiating with respect to r

C'=4\pi r- \frac{62.5}{ r^2}

Again differentiating with respect to r

C''=4\pi + \frac{125}{ r^3}

To find the minimize cost, we set C'=0

4\pi r- \frac{62.5}{ r^2}=0

\Rightarrow 4\pi r=\frac{62.5}{ r^2}

\Rightarrow  r^3=\frac{62.5}{ 4\pi}

⇒r=1.71

Now,

\left C''\right|_{x=1.71}=4\pi +\frac{125}{1.71^3}>0

When r=1.71 cm, the metal cost will be minimum.

Therefore,

h=\frac{25}{\pi\times 1.71^2}

⇒h=2.72 cm

Therefore the radius of the can is 1.71 cm and height of the can is 2.72 cm.

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