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3 years ago

What is four fifths more than a number is one third

Mathematics

2 answers:

LiRa [457]3 years ago

7 0

Answer:

1. 4/5+x=1/3

2. x=-7/15

Step-by-step explanation:

maks197457 [2]3 years ago

4 0

Answer:

a+4/5=1/3 because + means more than

Step-by-step explanation:

You might be interested in

Are 95/- 5, - 95/5, - 95/5 the same?

RSB [31]

Answer:

Step-by-step explanation:

Yes, 95/-5, -95/5, -95/5 are the same.

They are all equal to -19

3 0

3 years ago

PLLSS HELPPP itss due nowww

allsm [11]

Answer:

libk is given check it out

Step-by-step explanation:

https://www.slader.com/discussion/question/beginarray-l-triangle-t-a-m-has-vertices-t-05-a-41-and-m-36-find-the-gide-reflection-image-of-delta/

7 0

3 years ago

Mark ran28.2miles less than cody.mark ran12.4 miles how many miles cody run

Mama L [17]

Mark ran 28.2 miles less than Cody, but Mark ran 12.4 miles, so Cody ran 40.6 miles.
miles Cody ran - miles Mark ran = 28.2
C - 12.4 = 28.2
C = 40.6 miles

8 0

3 years ago

For a given input value r, the function h outputs a value q to satisfy the following equation.

balu736 [363]

h(r+1)

q could be 6 and it can be substituted with h

7 0

3 years ago

Ill give brainliest!!!

ehidna [41]

Given:

The coin wrapper volume = 27,480 mm³.

The coin diameter = 21.18 mm.

There are 50 such coins.

To find:

The thickness (height) of 1 coin.

Solution:

The 50 coins form a cylinder with a volume of 27,480 mm³.

If the diameter is 21.18 mm, the radius of a coin will be $\frac{21.18}{2} = 10.59$ mm.

The volume of a cylinder $= \pi r^{2} h.$

The volume of all 50 coins is given as 27,480 mm³. We need to determine the height of a single coin.

$50( \pi r^{2} h)=27,480.$

$50(3.1415)(10.59^{2} )(h) = 27,480.$

$(17,615.662)(h)=27,480.$

$h= \frac{27,480}{17,615.662} = 1.5599$ mm.

So the thickness of 1 coin is 1.5599 mm.

7 0

3 years ago