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Rasek [7]
2 years ago
15

What constant term should be used to complete the square?

Mathematics
1 answer:
natita [175]2 years ago
8 0
X2-9x+ ___= 6

x2-9x+81/4=6

(This is because you need to do 2*x*a=-9x) Hence 2a=-9

So a=-9/2 hence a2= 81/4

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J. Reexamine the sequence 20, 14, 8, 2, ... from the problem
DENIUS [597]

Answer:

The n th of the given sequence is t_{n} = 26-6 n

Step-by-step explanation:

<u>Step 1</u> :-

Given sequence is 20,14,8,2,.......this sequence in arithmetic progression but this sequence is decreasing sequence.

given first term is 20 and difference isd = second term- first term = 14-20=-6

now the nth term of given sequence is

by using formula t_{n}=a+(n-1)d

t_{n}= 20+(n-1)(-6)

t_{n}= 20-6 n+6

final answer:-

t_{n} = 26-6 n

<u>verification</u>:-

t_{n} = 26-6 n

put n=1 we get first term is 20

put n=2 we get second term is 14

put n=3 we get third term is 8

put n=4 we get fourth term is 2

so the n th term of sequence is

t_{n} = 26-6 n

3 0
3 years ago
HELP ME!!! 4- 5/8 =?
BlackZzzverrR [31]

Answer:

27/8

Step-by-step explanation:

1.) 4-5/8

Convert element to fraction

2.) 4 x 8/8-5/8

3.) 4 x 8-5/8

4.) 27/8

7 0
2 years ago
To solve this system using the addition method, you would need to multiply the first equation by what number in order for the y'
yanalaym [24]
You multiply 3x - y = 3.
So then it's:
6x - 2y = 3
-2x + 2y =6

which equals 4x = 9
8 0
3 years ago
You have two exponential functions. One function has the formula g(x) = 5 x . The other function has the formula h(x) = 5-x . Wh
AVprozaik [17]
Given the exponential functions g(x)= 5^{x} and h(x)= 5^{-x}

k(x)=(g-h)(x)
k(x)=g(x)-h(x)&#10;
k(x)= 5^{x} - 5^{-x}
k(x)= 5^{x} - \frac{1}{ 5^{x} }
k(x)= \frac{ 5^{x} 5^{x}  }{ 5^{x} } - \frac{1}{ 5^{x} }
k(x)= \frac{ 5^{2x} }{ 5^{x} } - \frac{1}{ 5^{x} }
k(x)= \frac{ 5^{2x}-1 }{ 5^{x} }
4 0
2 years ago
Read 2 more answers
PLZ HELP IF YOU DO THIS FOR POINTS I REPORT
lawyer [7]

Answer: you r not showing the info I need to complete this

Step-by-step explanation: please put more info so I can help :)

3 0
3 years ago
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