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olchik [2.2K]
3 years ago
14

SOMEONE PLEASE HELP ME

Advanced Placement (AP)
2 answers:
IrinaVladis [17]3 years ago
8 0

Question 1:

For this case we must indicate the graph corresponding to:

y = x ^ 2 + 6x + 9

We factor the expression by looking for two numbers that multiplied give as a result 9 and added as a result 6. These numbers are 3 and 3:

3 + 3 = 6\\3 * 3 = 9

So:

y = (x + 3) (x + 3) = (x + 3) ^ 2

We use the form of vertex of a parabola:

y = a (x-h) ^ 2 + k\\y = (x + 3) ^ 2

So:

a = 1

h = -3, is moved 3 units to the left

k = 0

The vertex of the parabola is given by: (h, k) = (- 3,0)

Since a = 1> 0 then the parabola is concave upwards.

With these data we can conclude that the correct option is the option H.

Answer:

Option H

Question 2:

For this case we have by definition, that the area of a rectangle is given by:

A = a * b

Where a and b are the sides of the rectangle.

We have as data that:

a = 4x + 5\\b = 2x + 3

Then, the area is given by:

A = (4x + 5) (2x + 3)

We must apply distributive property, which by definition establishes that:

(a + b) (c + d) = ac + ad + bc + bd

Then the area of the rectangle is given by:

A = 8x ^ 2 + 12x + 10x + 15\\A = 8x ^ 2 + 22x + 15

Answer:

Option D

nekit [7.7K]3 years ago
5 0

You can use a graphing calculator (Desmos for an example) to solve 36. And you would multiply 2x+3 • 4x+5 to find the area by the foil/distribution method (2x+3)(4x*5)

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Answer:

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Required:

Explain how to solve the 2020th term

Solve the 2020th term

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Assume n = 5

T_{5} = \frac{T_1 + T_2 + T_3 +T_4}{4}

T_{5} = \frac{(T_1 + T_2) + T_3 +(T_4)}{4}

Substitute 2T_{3} = T_1 + T_2 and T_{4} = T_3

T_{5} = \frac{2T_3 + T_3 +T_3}{4}

T_{5} = \frac{4T_3}{4}

T_{5} = \frac{(5-1)T_3}{5-1}

<em>Replace 5 with n</em>

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<em>(n-1) will definitely cancel out (n-1); So, we're left with</em>

T_{n} = T_3

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T_{2020} = T_3

Calculating T_3

T_{3} = \frac{10 + 20}{2}

T_{3} = \frac{30}{2}

T_{3} = 15

Recall that T_{2020} = T_3

T_{2020} = 15

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