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vekshin1
4 years ago
9

Nth term of 7 16 25 34 43

Mathematics
2 answers:
Bogdan [553]4 years ago
8 0

Here are the terms:

7, 16, 25, 34, 43, 52, 61, 70, 79

As you can see the pattern here is add 9.

The 9th term will be 79.

DiKsa [7]4 years ago
3 0

Okay, the first step is to find what number is added up to 7 for instance, to make 16 and so on. To find that you ask yourself what number plus 7 equals 16. The answer to that would be 9.

Lets add up the numbers,

7 + 9 = 16

16 + 9 = 25

25 + 9 = 34

34 + 9 = 43

43 + 9 = 52

52 + 9 = 61

61 + 9 = 70

70 + 9 = 79

79 + 9 = 88

When you're constantly counting how much numbers you've went through, then you'll find that you have found the 9th number of your question.

Your answer is 79

Hope this helped!~

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Answer:

The Proof for

Part C , Qs 9 and Qs 10  is below.

Step-by-step explanation:

PART C .

Given:

AD || BC ,

AE ≅ EC

To Prove:

ΔAED ≅ ΔCEB

Proof:

Statement                             Reason

1. AD || BC                           1. Given

2. ∠A ≅ ∠C                        2. Alternate Angles Theorem as AD || BC

3. ∠AED ≅ ∠CEB               3. Vertical Opposite Angle Theorem.

4. AE ≅ EC                        4. Given

5. ΔAED ≅ ΔCEB              5. By A-S-A congruence test....Proved

Qs 9)

Given:

AB ≅ BC ,

∠ABD ≅ ∠CBD

To Prove:

∠A ≅ ∠C

Proof:

Statement                             Reason

1. AB ≅ BC                        1. Given

2. ∠ABD ≅ ∠CBD            2. Given      

3. BD ≅ BD                       3. Reflexive Property

4. ΔABD ≅ ΔCBD             4. By S-A-S congruence test

5. ∠A ≅ ∠C                       5. Corresponding parts of congruent Triangles Proved.

Qs 10)

Given:

∠MCI ≅ ∠AIC

MC ≅ AI

To Prove:

ΔMCI ≅ ΔAIC

Proof:

Statement                             Reason

1. ∠MCI ≅ ∠AIC       1. Given

2. MC ≅ AI              2. Given

3. CI ≅ CI                3. Reflexive Property

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1. I'm assumig that the paths are perfect parabolas

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it's easier to find vertex form first then expand to get general form

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2.

same as last time

vertex is (10,72) so (h,k)=(10,72) so h=10 and k=72

equation is h_2=a(t-10)^2+72

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from the graph, we can see that the lowest possible height is 0yd and the highest height is 50yd

so range is 0 to 50 or 0≤h≤50


domain is the numbers that t is allowed to be

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it will stop flying when it hits the ground or at t=20

it starts flying at t=0

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8 0
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