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2 years ago

When are fractions useful and when are ratios useful?

2 answers:

7 0

There useful because all sorts of things are measured in fractions and you need ratios for all sorts of things like for income and stuff

kirill [66]2 years ago

3 0

Fractions are useful for splitting and ratios for sharing

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Hello I Need Some Help I Promise All Give Brainlist If You Are Corecct thank you

polet [3.4K]

The circle: x = 4 y = 0.5
The square: x = 3.5 y = 1.5
The triangle: x = 0 y = 2.5

5 0

1 year ago

Use the equation y=3x-8 to find y when x=8

Stels [109]

Answer: X = 8/3

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

The length width and height of a rectangular solid are 8,4,1 respectively what is the length of the longest line segment whose e

LUCKY_DIMON [66]

Answer:

9 units

Step-by-step explanation:

Given

$Length = 8$

$Width = 4$

$Height = 1$

Required

Determine the length of the longest segment

To do this, we use:

$Longest = \sqrt{Length^2 + Width^2 + Height^2}$

So, we have:

$Longest = \sqrt{8^2 + 4^2 + 1^2}$

$Longest = \sqrt{64 + 16 + 1}$

$Longest = \sqrt{81}$

Take positive square root

$Longest = 9$

<em>Hence, the length of the longest segment is 9 units</em>

3 0

2 years ago

The roof of the Terminal Tower is 709 feet above ground. A window washer dropped a bucket from the edge of the roof. How many se

Masja [62]

Answer:

6.64 seconds

Step-by-step explanation:

Given that:

Distance between roof of the terminal and ground = 709 feet

A bucket was dropped from the roof.

To find:

The time taken by the bucket to hit the ground?

Solution:

Distance traveled, $s$ = 709 feet

Initial velocity of the bucket, $u$ = 0 feet/sec (because initially it was at rest)

Acceleration of the bucket will be equal to acceleration due to gravity i.e.

$g$ or 32.1741 $ft/sec^{2}$

Let the time taken is $t$ seconds.

Formula for distance is given as:

$s = ut+\dfrac{1}{2}at^2$

Putting all the values:

$709 = 0 (t) + \dfrac{1}{2}(32.1741) t^2\\\Rightarrow 1418 = 32.1741 t^2\\\Rightarrow t^2 = 44.07\\\Rightarrow t \approx \bold{6.64\ sec}$

6 0

2 years ago

Compare the average score of a student at the beginning of a month and at the end of that month, if at first, it increased by 40

IceJOKER [234]

Answer:

Increased by 5%

Step-by-step explanation:

Let at the beginning the student's score be $s$ points.

At first, it increased by 40%. So, the new score is

$s+0.4s=1.4s$

Then it decreased by 25%, so the new score is

$1/4s-0.25\cdot 1.4s=1.4s-0.35s=1.05s$

At the beginning of a month the score was $s$ and at the end of that month the score was $1.05s$ . Since $1.05>1,$ the score increased by $1.05-1=0.05=5\%$

3 0

3 years ago