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Natalija [7]
3 years ago
7

The nth term of a quadratic sequence is:

Mathematics
1 answer:
anzhelika [568]3 years ago
7 0
We can work out the first three terms of this quadratic equation with substitution.
The first term of this sequence is where n=1, we can substitute 1 in for n
y=1^2+2(1)-4
y=1+2-4
y=-1
We can do this for the next two terms as well
y=2^2+2(2)-4
y=4+4-4
y=4
For the last term:
y=3^2+2(3)-4
y=9+6-4
y=11
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Answer:

Step-by-step explanation: you have to divide 112 by 14

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galina1969 [7]
Answer: y=5x+3
Step by step explanation:
Perpendicular meaning that the product of the slope/gradient is -1 hence ...
-1/5 * x = -1
x = -1/-1/5
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The equation perpendicular to y=-1/5x-3 is y=5x-c
Now it contains the points (1,2)
x variable is 1 and y variable is 2 plug them into the line equation
2=5(1)-c
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5 0
2 years ago
Simplify 27^2.<br><br> please (thxx)
adoni [48]

Answer:

729

Step-by-step explanation:

Raise 27 to the power of 2.

729

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3 years ago
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What is the midpoint of the vertical line segment graphed below? (2, 4), (2, -9) A. (4, -5). B. (2, -5/2). C. (2, -5). D. (4, 5/
omeli [17]

Answer:

B. (2, -\frac{5}{2})

Step-by-step explanation:

Given:

(2, 4) and (2, -9)

Required:

Midpoint of the vertical line with the above endpoints

Solution:

Apply the midpoint formula, which is:

M(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})

Where,

(2, 4) = (x_1, y_1)

(2, -9) = (x_2, y_2)

Plug in the values into the equation:

M(\frac{2 + 2}{2}, \frac{4 + (-9)}{2})

M(\frac{4}{2}, \frac{-5}{2})

M(2, -\frac{5}{2})

8 0
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