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Zina [86]
3 years ago
9

What is the answer to the problem. 4c-6=2

Mathematics
2 answers:
stich3 [128]3 years ago
8 0
The answer Will be 1 c=1
Olin [163]3 years ago
7 0
C=1 is your answer is you simplify
You might be interested in
Can anyone help me integrate :
worty [1.4K]
Rewrite the second factor in the numerator as

2x^2+6x+1=2(x+2)^2-2(x+2)-3

Then in the entire integrand, set x+2=\sqrt3\sec t, so that \mathrm dx=\sqrt3\sec t\tan t\,\mathrm dt. The integral is then equivalent to

\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt
=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\sec^2t-1}}\,\mathrm dt
=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt
=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{|\tan t|}\,\mathrm dt

Note that by letting x+2=\sqrt3\sec t, we are enforcing an invertible substitution which would make it so that t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3} requires 0\le t or \dfrac\pi2. However, \tan t is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which \tan t>0 so that |\tan t|=\tan t. This allows us to write

=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt
=\displaystyle\int(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\csc t\,\mathrm dt

We can show pretty easily that

\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C
\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C
\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C
\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C

which means the integral above becomes

=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C
=3\sqrt3\sec^2t-18\sec t+6\sqrt3\ln|\tan t|+12\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|+C

Back-substituting to get this in terms of x is a bit of a nightmare, but you'll find that, since t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}, we get

\sec t=\dfrac{x+2}{\sqrt3}
\sec^2t=\dfrac{(x+2)^2}3
\tan t=\sqrt{\dfrac{x^2+4x+1}3}
\cot t=\sqrt{\dfrac3{x^2+4x+1}}
\csc t=\dfrac{x+2}{\sqrt{x^2+4x+1}}
\csc2t=\dfrac{(x+2)^2}{2\sqrt3\sqrt{x^2+4x+1}}

etc.
3 0
3 years ago
What is the value of X?
sleet_krkn [62]

Answer:

  11/3 = 3 2/3 ≈ 3.66667

Step-by-step explanation:

The external angle is half the difference of the intercepted arcs:

  3x = (1/2)((9x +25) -36)

  6x = 9x -11

  11 = 3x . . . . . . . add 11-6x

  11/3 = x

_____

The attached figure is drawn to scale.

5 0
3 years ago
HELP PLEASE ANSWER 7
ololo11 [35]

Answer:

f(x) = 7

Step-by-step explanation:

we use the bottom equation (4x+3) because

1 > 0

so

4(1) +3 = 7

7 0
3 years ago
There are two squares with integer side lengths a and b. The perimeter of the entire figure is 86 cm, and the sum of the areas o
7nadin3 [17]

Answer:

24

Step-by-step explanation:

4a+2b= 86 cm

a²+b²= 386 cm²

a+b=?

----------

  • 4a+2b= 86  
  • 2a+b= 43
  • (2a+b)²=43²
  • 4a²+4ab+b²=1849

b²= 386- a², b= 43- 2a

  • 4a²+4a(43-2a)+386-a²-1849=0
  • 4a²+172a-8a²-a²-1463=0
  • -5a²+172a-1463=0
  • 5a²-172a+1463=0

solving quadratic equation we get

a=19 - integer solution

b= 43-2*19= 5

a+b=19+5=24

3 0
3 years ago
What is -2.8 + 0.9 - 12 - 4 simplified. if you know the answer it would help me greatly. thxs
Mumz [18]

Answer:

-17.9

Step-by-step explanation:

7 0
2 years ago
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