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jocelynmarquillo1
23.01.2020 •
Mathematics
Arain barrel can hold 12 gallons of water . before a storm 2 1/5 gallons of water were in the barrel . the storm added another 6 3/5 gallons of water to the barrel . how many more gallons of water can that barrel hold
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Ответ:
The rain barrel an hold
of water more.
Step-by-step explanation:
Given:
Amount of water barrel can hold = 12 gallons
Amount of water in the barrel before storm =![2\frac{1}{5}\ gallons](/tpl/images/0467/8806/7702f.png)
Amount of water in the barrel before storm =![\frac{11}{5}\ gallons](/tpl/images/0467/8806/2711b.png)
Amount of water storm added =![6\frac{3}{5}\ gallons.](/tpl/images/0467/8806/1cab4.png)
Amount of water storm added =![\frac{33}{5}\ gallons.](/tpl/images/0467/8806/42125.png)
we need to find the amount of water barrel can hold more.
Solution:
Now we can say that;
the amount of water barrel can hold more can be calculated by Subtracting the sum of Amount of water in the barrel before storm and Amount of water storm added from Amount of water barrel can hold.
framing in equation form we get;
the amount of water barrel can hold more =![12-(\frac{11}{5}+\frac{33}{5})= 12-\frac{11+33}{5}= 12- \frac{44}{5}](/tpl/images/0467/8806/54e1a.png)
Now we can see that 1 number is whole number and other is fraction.
So we will make the whole number into fraction by multiplying the numerator and denominator with the number in the denominator of the fraction.
so we can say that;
the amount of water barrel can hold more =![\frac{12\times5}{5}-\frac{44}{5} = \frac{60}{5}-\frac{44}{5}](/tpl/images/0467/8806/8dd04.png)
Now we can see that denominator is common so we can subtract the numerator.
the amount of water barrel can hold more =![\frac{60-44}{5}=\frac{16}{5}\ gallons \ OR \ \ 3\frac{1}{5}\ gallons](/tpl/images/0467/8806/d87a6.png)
Hence The rain barrel an hold
of water more.
Ответ:
cscΘ = -![\frac{5}{3}](/tpl/images/0356/6787/dcb5a.png)
Step-by-step explanation:
Using the trigonometric identities
tan x =
and csc x = ![\frac{1}{sinx}](/tpl/images/0356/6787/b5806.png)
sinΘ = tanΘ × cosΘ
=
× -
= - ![\frac{3}{5}](/tpl/images/0356/6787/62d38.png)
Hence
cscΘ =
= - ![\frac{5}{3}](/tpl/images/0356/6787/dcb5a.png)