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Stels [109]
3 years ago
13

How can you use bar graphs to compare quantities in real life situations

Mathematics
1 answer:
devlian [24]3 years ago
8 0
You can use bar graphs to compare quantities in real life situations by comparing different amounts, ex. you might use them for comparing different prices at a coffee shops around the world or different types of dogs you see at a park on a sunny day. You can really use bar graphs for comparing anything.
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What will the sale price be of a t-shirt that has an original price of $24.99 and is on sale for 30% off
Leviafan [203]

Original price $24.99

Now it's on sale with 30% off

So the price now will be 24.99 - 30% of 24.99

Or we can say that it will be 70% of the initial price

So:

24,99 . 70% = 24,99 . 70/100 = 17,493

The price will be $17,493

6 0
3 years ago
The function g(x) = 2x2 – 28x + 3 written in vertex form is g(x) = 2(x – 7)2 – 95. Which is one of the transformations applied t
mote1985 [20]
<span>3)shifted right 7 units</span>
3 0
3 years ago
Read 2 more answers
What are the intercepts?
klio [65]

Answer:

<em><u>In analytic geometry, using the common convention that the horizontal axis represents a variable x and the vertical axis represents a variable y, a y-intercept or vertical intercept is a point where the graph of a function or relation intersects the y-axis of the coordinate system.[1] As such, these points satisfy x = 0.</u></em>

7 0
3 years ago
A snail move 6 feet every 10 hours. how far it does in 15 hours
Anna007 [38]

6 feet:10 hours

multiply by 1.5 on both sides

9 feet:15 hours

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4 0
3 years ago
1. The mechanics at Lincoln Automotive are reborning a 6 in deep cylinder to fit a new piston. The machine they are using increa
Firdavs [7]

Answer:

0.0239\frac{in^{3}}{min}

Step-by-step explanation:

In order to solve this problem, we must start by drawing a diagram of the cylinder. (See attached picture)

This diagram will help us visualize the problem better.

So we start by determining what data we already know:

Height=6in

Diameter=3.8in

Radius = 1.9 in (because the radius is half the length of the diameter)

The problem also states that the radius will increase on thousandth of an inch every 3 minutes. We can find the velocity at which the radius is increasing with this data:

r'=\frac{1/1000in}{3min}

which yields:

r'=\frac{1}{3000}\frac{in}{min}

with this information we can start solving the problem.

First, the problem wants us to know how fast the volume is increasing, so in order to find that we need to start with the volume formula for a cylinder, which is:

V=\pi r^{2}h

where V is the volumen, r is the radius, h is the height and π is a mathematical constant equal approximately to 3.1416.

Now, the height of the cylinder will not change at any time during the reborning, so we can directly substitute the provided height, so we get:

V=\pi r^{2}(6)

or

V=6 \pi r^{2}

We can now take the derivative to this formula so we get:

\frac{dV}{dt}=2(6)\pi r \frac{dr}{dt}

Which simplifies to:

\frac{dV}{dt}=12\pi r \frac{dr}{dt}

We can now substitute the data provided by the problem to get:

\frac{dV}{dt}=12\pi (1.9) (\frac{1}{3000})

which yields:

\frac{dV}{dt}=0.0239\frac{in^{3}}{min}

3 0
3 years ago
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