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WINSTONCH [101]
3 years ago
7

The hcf of 55 and 99 is expressible in the form of 55m - 99 then the value of m is​

Mathematics
1 answer:
GuDViN [60]3 years ago
5 0

Answer:

2

Step-by-step explanation:

HCF(55,99)= 11

  • 55m-99=11
  • 55m=110
  • m=110/2
  • m=2
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Answer:

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Step-by-step explanation:

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3 years ago
Match each exponential function to the description of its percent rate of change. 22% growth 12% decay 2% decay 20% growth 20% d
makvit [3.9K]

The exponential function is defined as y = a(1+r)^x, where "a" represents the original account and "r" the rate of growth or decay.

Then we have the following:

1) 22% grow

y = a( 1 + 22%/100 )^x = a(1.22)^x

So the solution is: 124(1.22)^x

2) 12% decay

y = a( 1 - 12%/100 )^x = a(0.88)^x

So the solution is: y = f(x) = 44(0.88)^x

3) 20% decay

y = a( 1 - 20%/100 )^x = a(0.8)^x

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4) 12% Groth

y = a( 1 + 12%/100 )^x = a(1.12)^x

So the solution is: f(x) = 42(1.12)^x

4 0
3 years ago
Pls help with this question 1-5 thanks
belka [17]

maybe 1/5 (409) ::(880) 1+2

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3 years ago
Hello people ~
Luden [163]

Cone details:

  • height: h cm
  • radius: r cm

Sphere details:

  • radius: 10 cm

================

From the endpoints (EO, UO) of the circle to the center of the circle (O), the radius is will be always the same.

<u>Using Pythagoras Theorem</u>

(a)

TO² + TU² = OU²

(h-10)² + r² = 10²                                   [insert values]

r² = 10² - (h-10)²                                     [change sides]

r² = 100 - (h² -20h + 100)                       [expand]

r² = 100 - h² + 20h -100                        [simplify]

r² = 20h - h²                                          [shown]

r = √20h - h²                                       ["r" in terms of "h"]

(b)

volume of cone = 1/3 * π * r² * h

===========================

\longrightarrow \sf V = \dfrac{1}{3}  * \pi  * (\sqrt{20h - h^2})^2  \  ( h)

\longrightarrow \sf V = \dfrac{1}{3}  * \pi  * (20h - h^2)  (h)

\longrightarrow \sf V = \dfrac{1}{3}  * \pi  * (20 - h) (h) ( h)

\longrightarrow \sf V = \dfrac{1}{3} \pi h^2(20-h)

To find maximum/minimum, we have to find first derivative.

(c)

<u>First derivative</u>

\Longrightarrow \sf V' =\dfrac{d}{dx} ( \dfrac{1}{3} \pi h^2(20-h) )

<u>apply chain rule</u>

\sf \Longrightarrow V'=\dfrac{\pi \left(40h-3h^2\right)}{3}

<u>Equate the first derivative to zero, that is V'(x) = 0</u>

\Longrightarrow \sf \dfrac{\pi \left(40h-3h^2\right)}{3}=0

\Longrightarrow \sf 40h-3h^2=0

\Longrightarrow \sf h(40-3h)=0

\Longrightarrow \sf h=0, \ 40-3h=0

\Longrightarrow \sf  h=0,\:h=\dfrac{40}{3}<u />

<u>maximum volume:</u>                <u>when h = 40/3</u>

\sf \Longrightarrow max=  \dfrac{1}{3} \pi (\dfrac{40}{3} )^2(20-\dfrac{40}{3} )

\sf \Longrightarrow maximum= 1241.123 \ cm^3

<u>minimum volume:</u>                 <u>when h = 0</u>

\sf \Longrightarrow min=  \dfrac{1}{3} \pi (0)^2(20-0)

\sf \Longrightarrow minimum=0 \ cm^3

6 0
2 years ago
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Answer:

v = 9

Step-by-step explanation:

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