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dmitriy555 [2]
3 years ago
11

Consider the pattern shown in the table below

Mathematics
1 answer:
irina [24]3 years ago
4 0

Answer:

there is not table

Step-by-step explanation:

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Help needed! <br>-Thanks​
solong [7]

Answer:

the answer is C.

Step-by-step explanation:

the sum of the problem is 8.164.... so 8/C is the answer

8 0
3 years ago
Read 2 more answers
Which statement describes the graph of this polynomial function?
Artist 52 [7]

Answer:

Option (3)

Step-by-step explanation:

Given function is,

f(x) = x⁴ + x³ - 2x²

     = x²(x² + x - 2)

     = x²(x² + 2x - x - 2)

     = x²[x(x + 2) - 1(x + 2)]

     = x²(x + 2)(x - 1)

So the factored form of the polynomial function is,

f(x) = x²(x + 2)(x - 1)

For x - intercepts,

F(x) = x²(x + 2)(x - 1) = 0

x = -2, 1

This function has even multiplicity = 2 at x = 0.

Therefore, graph of the function will touch the x-axis at x = 0

And at other roots x = -2, 1 has odd multiplicity = 1, so the graph will cross the x-axis.

Option (3) will be the correct option.

3 0
3 years ago
Which fraction has a value that's equal to 7⁄8?
Katen [24]
B is the answer to the question
3 0
3 years ago
Find the period, range, and amplitude of the cosine function y= 3/2cos t/2
Rudik [331]

Answer:

Amplitude = /\frac{3}{2}/

Range = [\frac{-3}{2} , \frac{3}{2}]

Period = \pi

Step-by-step explanation:

Given: y = \frac{3}{2}cos\frac{t}{2}

Comparing the equation with the standard form of cosine function :

y = A cos(Bx- C)

where:

A = Amplitude

Formula for calculating Amplitude is given as:

Amplitude = /A/

The formula for calculating Period is given as :

Period =\frac{2\pi }{B}

B = \frac{1}{2}

Therefore , with the comparison

A = /\frac{3}{2}/

which means that:

Range = \frac{-3}{2} , \frac{3}{2}

Period = \frac{2\pi }{2}

Period = \pi

7 0
3 years ago
The length of a rectangular plot is (m- 1) and width is (m+1). What is the area​
pentagon [3]

Step-by-step explanation:

Area of rectangle = l x b

= (m-1) (m+1)

This is the difference of 2 squares

So

=m^2-1

8 0
3 years ago
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