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harina [27]
3 years ago
11

Solve x2 - 8x = 3 by completing the square. Which is the solution set

Mathematics
1 answer:
Simora [160]3 years ago
3 0

Answer:

4\pm\sqrt{19}

Step-by-step explanation:

Move everything to left side

x^{2} -8x -3 =0

Now to solve this square equation we need to find Descriminant

D=b^{2} -4ac=64+12=76

As Descriminant D is greater than 0 then we have 2 solutions which we can calculate by this formula

x_{1,2}=\frac{-b\pm\sqrt{D}}{2a} =\frac{8\pm2\sqrt{19}}{2}=4\pm\sqrt{19}

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Simplify the expression
vekshin1

Answer:

the answer is c

Step-by-step explanation:

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4 years ago
If NTM is congruent to OUM, SP is congruent to SQ, S is the center of the circle, and OM = 18, find NT. Round the answer to the
Aleks [24]
OM=18, so OQ=QM=18/2=9.
 Given QU=8
from figure OQU is a right angled triangle , so OU^2=OQ^2 + QU^2
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3 years ago
An 18-slice pizza was made with only pepperoni and mushroom toppings, and every slice has at least one topping. Exactly ten slic
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3 years ago
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Read 2 more answers
How many permutations can be formed from all the letters in the word engineering?
laila [671]

Answer:

277,200

Step-by-step explanation:

To find the number of permutation we can form from the letters of the word "engineering", we first need to find the frequencies of the different letters present.

E = 3

G = 2

N= 3

I = 2

R = 1

Now that we have the frequencies, we count the number of letters in the word "engineering".

E N G I N E E R I N G

11 letters

Now we take the factorial of total number of letters and divide it by the number of repeats and their factorial

So we get:

\dfrac{11!}{3!2!3!2!1!}

We remove the 1! because it will just yield 1.

\dfrac{11!}{3!2!3!2!}

So the total number of permutations from the letters of the word "engineering" will be:

Total number of permutations = \dfrac{39,916,800}{144}

Total number of permutations = 277,200

3 0
3 years ago
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