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Murljashka [212]
3 years ago
9

How do you find the greatest common factor of 175 and 25?

Mathematics
1 answer:
gulaghasi [49]3 years ago
7 0
The greatest common factor of 175 and 25?

Find out if 175 can be divided by 25. If it can, then it becomes the greatest common factor
_____
25)175. = 7
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At the sewing store ashly bought a bag of mixed buttons she got 25 buttons in all 5 of the buttons were large what percentage of
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The graph of f '(x) is continuous and decreasing with an x-intercept at x = 2. Which of the following statements must be true? (
bixtya [17]

Answer:

The graph of f is always concave down ⇒ the first answer

Step-by-step explanation:

* Lets explain how to solve the problem

- Remember that : If  f(x)  is a function then the solutions to the

 equation f′(x) = 0 gives the maximum and minimum values to f(x)

- The value of  x  gives maximum if f′′(x) is negative and minimum if

  f′′(x) is positive.

- Inflection points of the function  f(x) are found the solutions of the

 equation  f′′(x) = 0

* Lets solve the problem

- The graph of f'(x) is continuous means that the graph is unbroken line

- The graph of f'(x) decreasing with an x-intercept at x = 2 means

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- The differentiation of a function equal to zero at the critical point

  (minimum or maximum) of the function

∵ f'(x) = 0 at x = 2

∴ The x-coordinate of the critical point of f(x) is 2

- If the differentiation of the function is decreasing, then the critical

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∵ f'(x) is decreasing

∴ The critical point of the f(x) is maximum point

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5 0
3 years ago
A rectangular box has length 20cm, width 6cm and height 4cm. find how many cubes of size 2cm that will fit into the box.​
Papessa [141]

<u>Answer:</u>

\boxed{\pink{\sf The \ number \ of \ cubes \ that \ can \ be \ fitted \ is 60 .}}

<u>Step-by-step explanation:</u>

Given dimensions of the box = 20cm × 6cm × 4cm .

Dimension of the cube = 2cm × 2cm × 2cm .

Therefore the number of cubes that can be fitted into the box will be equal to the Volume of box divided by the Volume of the cube. So ,

\boxed{\red{\bf \implies No. \ of \ cubes \ = \dfrac{Volume \ of \ box}{Volume \ of \ cube }}}

\bf \implies n_{cubes} = \dfrac{20cm \times  6cm \times 4cm .}{2cm  \times 2cm  \times 2cm } \\\\\bf\implies n_{cubes}  = 10 \ times 3cm \times 2cm \\\\\implies \boxed{\bf n_{cubes}= 60 }

<h3><u>Hence</u><u> the</u><u> </u><u>number</u><u> </u><u>of</u><u> </u><u>cubes</u><u> </u><u>that</u><u> </u><u>can</u><u> </u><u>be</u><u> </u><u>fitted</u><u> </u><u>in</u><u> the</u><u> </u><u>box </u><u>is</u><u> </u><u>6</u><u>0</u><u> </u><u>.</u></h3>

6 0
3 years ago
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