What is 720 centiliters to 1 liter

Mathematics

2 answers:

mr_godi [17]2 years ago

8 0

For this question the answer is about 7.2 liters :)

KonstantinChe [14]2 years ago

8 0

The Answer Is 7.2 That's What I Got.

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A manager bought 12 pounds of peanuts for $30. He wants to mix $5 per pound cashews with the peanuts to get a batch of mixed nut

seraphim [82]

Answer:

18 pounds of cashews are needed.

Step-by-step explanation:

Given;

A manager bought 12 pounds of peanuts for $30.

Price of peanut per pound P = $30/12 = $2.5

Price of cashew per pound C = $5

Price of mixed nut per pound M = $4

Let x represent the proportion of peanut in the mixed nut.

The proportion of cashew will then be y = (1-x), so;

xP + (1-x)C = M

Substituting the values;

x(2.5) + (1-x)5 = 4

2.5x + 5 -5x = 4

2.5x - 5x = 4 -5

-2.5x = -1

x = 1/2.5 = 0.4

Proportion of cashew is;

y = 1-x = 1-0.4 = 0.6

For 12 pounds of peanut the corresponding pounds of cashew needed is;

A = 12/x × y

A = 12/0.4 × 0.6 = 18 pounds

18 pounds of cashews are needed.

4 0

2 years ago

Please I need this ASAP so can someone helpppppp

iris [78.8K]

AC = 8.9
BC = 6.5
(90% angle)
8.9 x 6.5 = <span>57.85
57.85/2 = 28.925
Area of ^ABC = 28.925 cm squared

DE = 55.1
55.1/ 11.02 = 5
DEF is 5 times as big as ABC
DF = 44.5
FE = 32.5
44.5 x 32.5 = 1446.25
1446.25/2 = 723.125
I am not positive but I believe DEF = 723.125 cm squared
Good Luck on the Test!! :D

</span>

7 0

3 years ago

How to solve this?<br>\int \frac { 4 - 3 x ^ { 2 } } { ( 3 x ^ { 2 } + 4 ) ^ { 2 } } d x

ivanzaharov [21]

$\Large \mathbb{SOLUTION:}$

$\begin{array}{l} \displaystyle \int \dfrac{4 - 3x^2}{(3x^2 + 4)^2} dx \\ \\ = \displaystyle \int \dfrac{4 - 3x^2}{x^2\left(3x + \dfrac{4}{x}\right)^2} dx \\ \\ = \displaystyle \int \dfrac{\dfrac{4}{x^2} - 3}{\left(3x + \dfrac{4}{x}\right)^2} dx \\ \\ \text{Let }u = 3x + \dfrac{4}{x} \implies du = \left(3 - \dfrac{4}{x^2}\right)\ dx \\ \\ \text{So the integral becomes} \\ \\ = \displaystyle -\int \dfrac{du}{u^2} \\ \\ = -\dfrac{u^{-2 + 1}}{-2 + 1} + C \\ \\ = \dfrac{1}{u} + C \\ \\ = \dfrac{1}{3x + \dfrac{4}{x}} + C \\ \\ = \boxed{\dfrac{x}{3x^2 + 4} + C}\end{array}$