All categories

3 years ago

What is the ratio of 24

Mathematics

1 answer:

dalvyx [7]3 years ago

5 0

It is 2,3,4,5 there are all of your answere

You might be interested in

Simplify the expression below: (-5x² + 2) - (4x² – 5x - 9)

lozanna [386]

Answer:

-9x² + 5x + 11

Step-by-step explanation:

-5x² + (-4x²) + 5x + 2 + 9

-9x² + 5x + 11

3 0

2 years ago

The expression x + 0.25x can be used to determine a 25% increase in the number of blog views from last month to this month. What

Drupady [299]

Answer:

1.25x

Step-by-step explanation:

Given that :

25% Increase in number of blog views can be expressed as (x + 0.25x)

Another way to make this representation is :

Total number of blog views + 25% of total

Total should add up to 100%

Let number of views = x

Hence,

(100% + 25%) of x

125% of x

125%x

= 1.25x

8 0

2 years ago

Romashka [77]

9.42 Units

Step-by-step explanation:

Since the angle shown is 90 degrees we can deduce that the rest of the arc is 270 degrees therefore we can make the fraction:

270/360; 360 being the total angle measurement of a circle

270/360 simplifies to 3/4

The fraction represents how much of the circle that arc covers

Now we have to find the circumference which would be

2(3.14)r

r = 2 therefore: 2(3.14)2 = 12.56

Now that we have the circumference of the WHOLE circle we multiply it by 3/4 to find out the arc length

Which gives us 9.42

8 0

3 years ago

A motor vehicle has a maximum efficiency of 39 mpg at a cruising speed of 40 mph. The efficiency drops at a rate of 0.1 mpg/mph

svp [43]

Answer:

38 mpg

Step-by-step explanation:

Initial efficiency = 39 mpg

Initial speed = 40 mph

Final speed = 50 mph

The efficiency drop between 40 mph and 50 mph is given by:

$E_d = 0.1 \frac{mpg}{mph}*\Delta V$

The total efficiency drop from 40 to 50 mph is:

$E_d = 0.1 \frac{mpg}{mph}*(50-40)mph\\E_d = 1\ mpg$

Therefore, the efficiency at 50 mph is:

$E_{50} = E_{40} -E_d\\E_{50} = 39 -1\\E_{50} = 38\ mpg$

5 0

3 years ago

How many gallons of a 80% antifreeze solution must be mixed with 80 gallons of 15% antifreeze to get a mixture that is 70% antif

konstantin123 [22]

440 gallons of 80 % antifreeze solution must be mixed with 80 gallons of 15% antifreeze to get a mixture that is 70% antifreeze

Solution:

Let "x" be the gallons of 80 % antifreeze added

Therefore, "x" gallons of 80 % antifreeze solution must be mixed with 80 gallons of 15% antifreeze

Final mixture is x + 80

Therefore, we can frame a equation as:

"x" gallons of 80 % antifreeze solution must be mixed with 80 gallons of 15% antifreeze to get (x + 80) gallons of 70 % antifreeze

Thus, we get,

x gallons of 80 % + 80 gallons of 15 % = (x + 80) gallons of 70 %

$x \times \frac{80}{100} + 80 \times \frac{15}{100} = (x+80) \times \frac{70}{100}\\\\0.8x + 80 \times 0.15 = (x+80) \times 0.7\\\\0.8x+12 = 0.7x+56\\\\0.8x-0.7x=56-12\\\\0.1x = 44\\\\x = \frac{44}{0.1}\\\\x = 440$

Thus 440 gallons of 80 % antifreeze solution must be mixed

4 0

3 years ago