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sashaice [31]
3 years ago
5

The distance between the school and the airport on a map is 7.25 cm. The scale of the map is 1 cm: 12 m. What is the actual dist

ance between the school and the airport?
Mathematics
1 answer:
pav-90 [236]3 years ago
8 0

Answer:  87m

Step-by-step explanation:

This problem can be solved by using a simple ratio and proportion.

The way you figure this out is:

Model = 12 m/ 1 cm = x/7.25 cm

Solve for x and you should receive this answer:

x = 87 m

Hope I helped :P

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You can tell if lines are perpendicular or not by seeing if meet in a right angle. A right angle is like a square, and the lines would have to be 90 degrees.

One example is when the two lines (horizontal and vertical) meet each other at the ends.

(You MUST make sure that they are directly horizontal and vertical. If not, they will either be acute or obtuse).

I hope this helped!

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Name the rule and find the next two numbers in the pattern 10,70,490
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Each of the numbers 10, 70 and 490 is separated from the next by a factor of 7.  10 times 7 is 70; 70 times 7 is 490; 7 times 490 is what?  7 times 7 times 490 is what?
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Cal is measuring temperature changes in four substances over different time periods as part of a school project. Which data show
djyliett [7]

Answer:

C) The third table. K= 5

C)\frac{10}{2}=\frac{15}{3}=\frac{25}{5}=\frac{40}{8}=k= 5

Step-by-step explanation:

1) Below there are the missing data:

A proportional relationship through a constant k. It is obtained when we divide:

k=\frac{y}{x}

2)In this case, when we divide the temperature (T) by the time (h).

k=\frac{T}{h}

3)So, examining the table below we are searching for a ratio k common to all measures (temperatures over hour).

A) \frac{12}{3}\neq\frac{25}{5}\neq\frac{36}{6}\neq\frac{81}{9}\\B) \frac{22.5}{3}\neq\frac{30}{4}\neq\frac{52.5}{7}\neq\frac{67.5}{8}\\C)\frac{10}{2}=\frac{15}{3}=\frac{25}{5}=\frac{40}{8}=k= 5\\D)\frac{8.5}{2}\neq\frac{22.5}{5}\neq\frac{28.5}{7}\neq\frac{36}{8}

3 0
3 years ago
The graph h = −16t^2 + 25t + 5 models the height and time of a ball that was thrown off of a building where h is the height in f
Thepotemich [5.8K]

Answer:

part 1) 0.78 seconds

part 2) 1.74 seconds

Step-by-step explanation:

step 1

At about what time did the ball reach the maximum?

Let

h ----> the height of a ball in feet

t ---> the time in seconds

we have

h(t)=-16t^{2}+25t+5

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

so

The x-coordinate of the vertex represent the time when the ball reach the maximum

Find the vertex

Convert the equation in vertex form

Factor -16

h(t)=-16(t^{2}-\frac{25}{16}t)+5

Complete the square

h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+5+\frac{625}{64}

h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}\\h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}

Rewrite as perfect squares

h(t)=-16(t-\frac{25}{32})^{2}+\frac{945}{64}

The vertex is the point (\frac{25}{32},\frac{945}{64})

therefore

The time when the ball reach the maximum is 25/32 sec or 0.78 sec

step 2

At about what time did the ball reach the minimum?

we know that

The ball reach the minimum when the the ball reach the ground (h=0)

For h=0

0=-16(t-\frac{25}{32})^{2}+\frac{945}{64}

16(t-\frac{25}{32})^{2}=\frac{945}{64}

(t-\frac{25}{32})^{2}=\frac{945}{1,024}

square root both sides

(t-\frac{25}{32})=\pm\frac{\sqrt{945}}{32}

t=\pm\frac{\sqrt{945}}{32}+\frac{25}{32}

the positive value is

t=\frac{\sqrt{945}}{32}+\frac{25}{32}=1.74\ sec

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3 years ago
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DaniilM [7]

that is not possible cause the first 2 indicate that the pattern is counting by 2s


5 0
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