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shutvik [7]
3 years ago
15

How do you solve quadratic equations?

Mathematics
1 answer:
masha68 [24]3 years ago
8 0
In any equation there is a variable (usually x). In a quadratic equation there is a lot involved.

-B (+or-) sqrt of b^2 - 4 x A x C
________________________  divide it by
               2 x A
                                                   There is usually an original equation that looks like:  Ax^2 + Bx + C = 0    Use the variables from the equation to the left to fill the upper equation.  (You can also look up a better formula if confused).

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3 years ago
What is the measure of < BCD ? Justify your answer.​
Likurg_2 [28]

Answer:

  • please mark brainlist please

Step-by-step explanation:

4π/3 radians. Triangle A B C has angles labeled as follows: A, (101x + 2) degrees; B, 34x degrees; C, unlabeled. The measure of ∠BCD is 120°. )

8 0
3 years ago
(secx)dy/dx=e^(y+sinx), please help me solve the differential equation. Thanks :)
nalin [4]
First, you must know these formula  d(e^f(x) = f'(x)e^x dx, e^a+b=e^a.e^b, and d(sinx) = cosxdx, secx = 1/ cosx

(secx)dy/dx=e^(y+sinx), implies  <span>dy/dx=cosx .e^(y+sinx), and then 
</span>dy=cosx .e^(y+sinx).dx, integdy=integ(cosx .e^(y+sinx).dx, equivalent of 
integdy=integ(cosx .e^y.e^sinx)dx, integdy=e^y.integ.(cosx e^sinx)dx, but we know that   d(e^sinx) =cosx e^sinx dx,
so integ.d(e^sinx) =integ.cosx e^sinx dx,
and e^sinx + C=integ.cosx e^sinxdx
 finally, integdy=e^y.integ.(cosx e^sinx)dx=e^2. (e^sinx) +C
the answer is 
y = e^2. (e^sinx) +C, you can check this answer to calculate dy/dx
7 0
3 years ago
Read 2 more answers
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