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Xelga [282]
3 years ago
13

–20, - 12,5, 22, . . . An= A50=

Mathematics
1 answer:
marshall27 [118]3 years ago
8 0

Answer:

an = 17n - 46

a50 = 804

Step-by-step explanation:

-29, -12, 5, 22, ...

Subtract each number from the next one.

-12 - (-29) = 17

5 - (-12) = 17

22 - 5 = 17

The common difference is 17. This is an arithmetic sequence which starts with -29, and in which each subsequent value is 17 more than the previous value.

a1 = -29

a2 = -29 + 17

a3 = -29 + 2(17)

a4 = -29 + 3(17)

Notice that for each term, you have -29 and something added to it. What you add to -29 is 17 multiplied by 1 less than the number of the term. For term 1, 1 less than 1 is 0. You add 0 * 17 to -29 and get -29. term 1 is -29. For term 2, 1 less than 2 is 1. You add 1 * 17 to -29 and get -12, etc.

For term n, 1 less than n is n - 1. Add (n - 1) * 17 to -29 to get term n.

an = -29 + 17(n - 1)

This formula can be simplified.

an = -29 + 17n - 17

an = -46 + 17n

an = 17n - 46

a50 = 17(50) - 46

a50 = 804

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The area of a parallelogram is 8x^2+2x-6 and it's height is 2x+2. What is the base of the parallelogram?​
ZanzabumX [31]

Answer:

4x - 6

Step-by-step explanation:

Remember the formula for finding the area of a parallelogram;(base)*(height)=(area)

Manipulate the equation such that it is solved for the base;

(base)=\frac{(area)}{(height)}

So the equation that needs to be solved is;

\frac{(8x^{2} + 2x - 6 )}{(2x + 2)}

Set up the division problem and solve;

              _____<u>4x - 6</u>___

2x + 2     | 8x^{2} + 2x - 6

             -(8x^{2}+8x)<u />

                ________

                       -6x - 6

                      + (6x + 6 )

                      ________

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4 0
3 years ago
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The two triangles exist congruent if they contain two congruent corresponding sides and their contained angles exist congruent.

Let $&\overline{A B} \cong \overline{D E} \\ and $&\overline{A C} \cong \overline{D F}

Angle between $\overline{A B}$ and $\overline{A C}$ exists $\angle A$.

Angle between $\overline{D E}$ and $\overline{D F}$ exists $\angle D$.

Therefore, $\triangle A B C \cong \triangle D E F$ by SAS, if $\angle A \cong \angle D$$.

<h3>What is SAS congruence property?</h3>

Given:

$&\overline{A B} \cong \overline{D E} \\ and

$&\overline{A C} \cong \overline{D F}

According to the SAS congruence property, two triangles exist congruent if they contain two congruent corresponding sides and their contained angles exist congruent.

Let $&\overline{A B} \cong \overline{D E} \\ and $&\overline{A C} \cong \overline{D F}

Angle between $\overline{A B}$ and $\overline{A C}$ exists $\angle A$.

Angle between $\overline{D E}$ and $\overline{D F}$ exists $\angle D$.

Therefore, $\triangle A B C \cong \triangle D E F$ by SAS, if $\angle A \cong \angle D$$.

To learn more about SAS congruence property refer to:

brainly.com/question/19807547

#SPJ9

4 0
2 years ago
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