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Nadusha1986 [10]
3 years ago
14

Find the common difference of the sequence shown 1/6, 1/4, 1/3

Mathematics
2 answers:
Anvisha [2.4K]3 years ago
8 0

Answer:

Common difference of the sequence is = \frac{1}{12}

Step-by-step explanation:

Common difference of any sequence is defined by the difference between a term and its successive term.

Sequence is \frac{1}{6}, \frac{1}{4},\frac{1}{3}.....

Common difference = T_{2}-T_{1}=\frac{1}{4}-\frac{1}{6}

                                 = \frac{2}{24}

                                 = \frac{1}{12}

Now we will do the same for T_{3} and T_{2}

T_{3}-T_{2} = \frac{1}{3}-\frac{1}{4}

= \frac{4-3}{12}

= \frac{1}{12}

Therefore, answer is \frac{1}{12}

Damm [24]3 years ago
7 0

1/4 - 1/6 = (6-4)/24 = 2/24 = 1/12
1/3 - 1/4 = (4-3)/12 = 1/12



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omeli [17]

Answer:

y = (-x)^3 - 4

Step-by-step explanation:

Ok, for the function:

y = x^3

When x = 0, we have:

y = 0^3  = 0

So the original graph passes through the point (0, 0)

If we look at the given graph, we can see that the y-intercept (the value of y when x = 0) is:

y = -4

So, this is the graph of y = x^3 moved down 4 units.

You can also see that the graph goes downward as x increases (and up as x decreases) while for the function:

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as x increases, we should see that y also increases.

Then we have a reflection across the x-axis.

Ok, now let's describe a vertical shift.

For a general function f(x), a vertical shift of N units is written as:

g(x) = f(x) + N

if N is positive, the shift is upwards

if N is negative, the shift is downwards.

And for a function f(x), a reflection across the x-axis is written as:

g(x) = - f(x)

Here we first apply the reflection across the x-axis, so we get:

g(x) = -f(x)

now we apply the shift 4 units downwards

g(x) = - f(x) - 4

replacing f(x) by our function, x^3

we get:

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Complex numbers zj and zz are given by zj = 3- j2, zz = - 4 + j3. (a) express zj and zz in polar form. (b) find lz1l by first ap
inna [77]

Solution:

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Part a:

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Part c:

With the help of part a:

z1z2 = 3.6e^{-j33.7^{\circ}}.5e^{j143.1^{\circ}}

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