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Gemiola [76]
2 years ago
6

Find the product of theL.C.M and H.C.F of 8 and 12.​

Mathematics
2 answers:
NikAS [45]2 years ago
6 0

Answer:

24*4=96

Step-by-step explanation:

lcm of 8 and 12 is 24

hcf of 8 and 12 is 4

24 times 4 = 96

V125BC [204]2 years ago
3 0

Answer:

<h3>96 is the answer. </h3>

Step-by-step explanation:

<h3>The L.C.M of 8 and 12 is 24</h3><h3>The H.C.F of 8 and 12 is 4</h3>

<h3>We multiply </h3><h3>24 *4 we shall get 96 as the answer. </h3><h3 />
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For f(x)=4sin(x2) between x=0 and x=3, find the coordinates of all intercepts, critical points, and inflection points to two dec
telo118 [61]

Answer:

Intercepts:

x = 0, y = 0

x = 1.77, y = 0

x = 2.51, y = 0

Critical points:

x = 1.25, y = 4

x = 2.17 , y = -4

x = 2.8, y = 4

Inflection points:

x = 0.81, y = 2.44

x = 1.81, y = -0.54

x = 2.52, y = 0.27

Step-by-step explanation:

We can find the intercept by setting f(x) = 0

4sin(x^2) = 0

sin(x^2) = 0

x^2 = n\pi where n = 0, 1, 2,3, 4, 5,...

x = \sqrt(n\pi)

Since we are restricting x between 0 and 3 we can stop at n = 2

So the function f(x) intercepts at y = 0 and x:

x = 0

x = 1.77

x = 2.51

The critical points occur at the first derivative = 0

f^{'}(x) = 4cos(x^2)2x = 8xcos(x^2) = 0

xcos(x^2) = 0

x = 0 or

cos(x^2) = 0

x^2 = \frac{\pi}{2} + n\pi where n = 0, 1, 2, 3

x = \sqrt{\pi(n+1/2)}

Since we are restricting x between 0 and 3 we can stop at n =  2

So our critical points are at

x = 1.25, y = f(1.25) = 4sin(1.25^2) = 4

x = 2.17 , y = f(2.17) = 4sin(2.17^2) = -4

x = 2.8, y = f(2.8) = 4sin(2.8^2) = 4

For the inflection point, we can take the 2nd derivative and set it to 0

f^[''}(x) = 8(cos(x^2) - xsin(x^2)2x) = 8cos(x^2) - 16x^2sin(x^2) = 0

cos(x^2) = 2x^2sin(x^2)

tan(x^2) = \frac{1}{2x^2}

We can solve this numerically to get the inflection points are at

x = 0.81, y = f(0.81) = 4sin(0.81^2) = 2.44

x = 1.81, y = f(1.81) = 4sin(1.81^2) = -0.54

x = 2.52, y = f(2.52) = 4sin(2.52^2) = 0.27

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Write the equation for a line in standard form (Ax+By+C) that is perpendicular to y = 3x -
salantis [7]

Answer:

x+3y-6=0

Step-by-step explanation:

given eqn is y=3x-2 which is 3x-y-2=0

the eqn of line perpendicular to given eqn is -x+3y+k=0

it passes through (6,4)

-6+3*4+k=0

or,. -6+12+k=0

or, k= -6

therefore, the eqn of line perpendicular to given eqn is x+3y-6=0

8 0
3 years ago
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