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jasenka [17]
3 years ago
5

What value(s) of x make the equation x2 - 18x + 81 = 0 true?

Mathematics
2 answers:
Sergeu [11.5K]3 years ago
8 0

Answer:

9

Step-by-step explanation:

mel-nik [20]3 years ago
5 0
We can factorise this to get (x-9)(x-9). To find the value, each bracket needs to equal 0, so the value would be x=9
You might be interested in
-14, -6, 2, 0,...<br> Is it geometric
marysya [2.9K]

Answer:

No it is not.

Step-by-step explanation:

Reason.

For geometric: common ratio r should be equal

r = T2/T1 = T3/T2 but is not true here as -6/-14 is not equal to 2/-6

5 0
3 years ago
Read 2 more answers
A jewelry sells gold and platinum rings. Each ring is available in eight styles and is fitted with one of ten gemstone
katrin [286]

Answer: There are 160 different rings.

Step-by-step explanation:

The question is missing, i guess you want to know the number of different rings that the jewelry has:

For the material we have 2 options.

For the stile we have 8 options

For the gem we have 10 options.

Now, the total number of possible combinations is equal to the product of the number of options for each selection, this is:

Combinations = 2*8*10 = 160

4 0
3 years ago
I don't know what I am doing help please (C:
NeX [460]

Answer:

Step-by-step explanation:

The far right side of the triangle is 17 cm, which is the height.

The base would be 34 minus 17 because whole length of the rectangle is 34 and the triangle covers all but 17 of it.

So then you would take 1/2 of 17.

Then multiply it by 17 to get 160.5

7 0
3 years ago
Divide. Yes rectangular model to record the partial quotient <br> 246/3=
ra1l [238]
82... because 24/3 is 8 and 6/3 is 2
Therefore you get 82
5 0
3 years ago
What are the coordinates of the inflection point on the graph of y=(x+1)arctanx
Dennis_Churaev [7]

Inflection point is the point where the second derivative of a graph is zero.

y = (x+1)arctan xy' = (x+1)(arctan x)' + (1)arctan xy' = (x+1)/(x^2+1) + arctan xy'' = (x+1)(1/(1+x^2))' + 1/(1+x^2) + 1/(1+x^2)y'' = (x+1)(-1/(1+x^2)^2)(2x)+2/(1+x^2)y'' = ((x+1)(-2x)+1+x^2)/(1+x^2)^2y'' = (-2x^2-2x+2+2x^2)/(1+x^2)^2y'' = (-2x+2)/(1+x^2)^2

Solving for point of inflection: y'' = 00 = (-2x+2)/(1+x^2)^20 = -2x+2x = 1y(1) = (1+1)arctan(1) = 2 * pi/4 = pi/2

Therefore, E(1, pi/2).


I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

5 0
3 years ago
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