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djyliett [7]
3 years ago
10

How would I do this

Mathematics
1 answer:
andriy [413]3 years ago
6 0
Subtract top exponent from bottom. answer is a^2 b^7 c
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Hello i would like help and the answer to this please; what is the nth term rule of
Paraphin [41]

Answer:

The nth term of the given sequence

a_{n} = a_{n-1} + (2n-1)

Step-by-step explanation:

<u><em>Explanation:-</em></u>

Given sequence  -5,-4,-1,4,11,20,31

a₀ = -5

a₁ = a₀ +1 = -5 +1   = -4

a₂ = a₁ + 3 = -4+3  = -1

a₃ = a₂ + 5 = -1 +5 = 4

a₄ = a₃ + 7 = 4 + 7  =11

a₅ = a₄ + 9 = 11+9 = 20

a₆ = a₅ + 11 =  20+11 = 31

a₇ = a₆ + 13 = 31 +13 =44

and so on

The nth term of the given sequence

a_{n} = a_{n-1} + (2n-1)

7 0
3 years ago
What is the scientific notation for number of kilometers light will travel in 60 seconds?
Natalka [10]
Travel distance by light in a second = 3*10^5 kms
Travel distance by light in 60 seconds = 
3*10^5 * 60
180 * 10^5
1.8 * 10^7 kms 
3 0
3 years ago
Which point is on the circle centered at the origin with a radius of 5 units?
Aleonysh [2.5K]

Answer:

(2,√21)

Step-by-step explanation:

The circle centered at the origin has equation:

{x}^{2}  +  {y}^{2}  = 25

Any point that satisfies this equation lie on this circle.

When x=2, we substitute and solve for y.

{2}^{2}  +  {y}^{2}  = 25 \\ 4 +  {y}^{2}  = 25 \\  {y}^{2}  = 25 - 4 \\ {y}^{2}  = 21

Take square root to get:

y =  \pm \sqrt{21}

Therefore (2,-√21) and (2,√21) are on this circle.

From the options, (2,√21) is the correct answer

3 0
3 years ago
Read 2 more answers
Use the zeros of the following quadratic to find the x-value of the vertex.
Ivan
B I got this by graphing it into my graphing calculator. Hope this helps
5 0
4 years ago
In triangle △ABC, ∠ABC=90°, BH is an altitude. Find the missing lengths. AB=4 and BC=3, Find AH, CH and BH.
IRISSAK [1]

Answer:

  • AH = 3.2
  • CH = 1.8
  • BH = 2.4

Step-by-step explanation:  

It can be convenient to compute the length of the hypotenuse of this triangle (AC). The Pythagorean theorem tells you ...

   AC^2 = AB^2 + CB^2

   AC^2 = 4^2 + 3^2 = 16 + 9 = 25

   AC = √25 = 5  

The altitude divides ∆ABC into similar triangles ∆AHB and ∆BHC. The scale factor for ∆AHB is ...

  scale factor ∆ABC to ∆AHB = AB/AC = 4/5 = 0.8  

And the scale factor to ∆BHC is ...  

  scale factor ∆ABC to ∆BHC = BC/AC = 3/5 = 0.6  

Then the side AH is 0.8·AB = 0.8·4 = 3.2

And the side CH is 0.6·BC = 0.6·3 = 1.8  

These two side lengths should add to the length AC = 5, and they do.  

The remaining side BH can be found from either scale factor:    

  BH = AB·0.6 = BC·0.8 = 4·0.6 = 3·0.8 = 2.4

_____  

The sides of interest are ...  

  AH = 3.2

  CH = 1.8

  BH = 2.4

3 0
3 years ago
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