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3 years ago

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:

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

So the solution is: f(x) = 22(0.8)

4) 12% Groth

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

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

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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.

Using Pythagoras Theorem

(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)

First derivative

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

apply chain rule

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

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

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

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

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

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

$\Longrightarrow \sf h=0,\:h=\dfrac{40}{3}$

maximum volume: when h = 40/3

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

$\sf \Longrightarrow maximum= 1241.123 \ cm^3$

minimum volume: when h = 0

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

$\sf \Longrightarrow minimum=0 \ cm^3$

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2 years ago

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Pleaseee help geometry

Ad libitum [116K]

Answer:

v = 9

Step-by-step explanation:

50 = 6v - 4

50 + 4 = 6v

54 = 6v

6v/6 = 54/6

v = 9

7 0

2 years ago

Read 2 more answers

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

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