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Soloha48 [4]
3 years ago
7

EXPERTS/ACE/GENIUSES PLZ HELP QUICK!!:) THANKS.

Mathematics
1 answer:
Rina8888 [55]3 years ago
7 0
Hey there Hailey!

So, to start this all of, what formula are we even going to use.

The formula that we will be using in this question would be . . .

\left[\begin{array}{ccc}\boxed{V= \pi r^2 \frac{h}{2}} \end{array}\right]

So, now that we know what formula that we will use, we will now plug in those number's into the formula.

Justification:

We do, 9 = (height)*3=(radius) \ but \ using \ the \ formula \ also.


\left[\begin{array}{ccc}27*3.14= \boxed{84.78} \\ Your \ correct \ answer \ would \ be \ \boxed{\boxed{27}}\end{array}\right]

Hope this helps you!
~Jurgen
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its 1 or -5

Step-by-step explanation:

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3 0
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Practice
viva [34]

Answer:

B.

Step-by-step explanation:

x = small box

(small box + 5 ounces) = large box.

B shows:

x + (x + 5) = 27

Small box + (small box + 5 ounces) = 27 ounces.

Small box + large box = 27 ounces.

————————————————————

Why A and C are incorrect.

A describes:

x + 5x = 27

Small box + 5 multiplied by small box = 27

(5 multiplied by small box is equivalent to 5 small boxes.)

Small box + 5 small boxes = 27

C describes:

x + 5 = 27

small box + 5 ounces = 27

Large box = 27

(We know that a small box + 5 ounces is a large box)

7 0
3 years ago
Atul has 2/3 lb of candy. Jose has 3/5 lb and Maria has 1/2 lb less then José. How many more pounds of candy does Atul have than
liraira [26]

Given:

Atul has \dfrac{2}{3} lb of candy.

Jose has \dfrac{3}{5} lb of candy.

Maria has \dfrac{1}{2} lb less than Jose.

To find:

How many more pounds of candy does Atul have than Maria?

Solution:

Since, Maria has \dfrac{1}{2} lb less than Jose, therefore

Maria has = \dfrac{3}{5}-\dfrac{1}{2} lb

Maria has = \dfrac{6-5}{10} lb

Maria has = \dfrac{1}{10} lb

Difference between the candies Atul and Maria have.

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Difference = \dfrac{17}{30} lb

Therefore,  Atul have \dfrac{17}{30} lb of candy more than Maria.

6 0
3 years ago
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