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Marat540 [252]
2 years ago
6

Show how to solve 2x+5=13, justify the steps

Mathematics
2 answers:
liberstina [14]2 years ago
8 0

Answer:

The answer is 4

Step-by-step explanation:

2x + 5 = 13

Now, Subtract 5 from both side we get,

2x + 5 - 5 = 13 - 5

2x = 8

Then, Divide both side by 2 we get,

2x/2 = 8/2

x = 4

Thus, The answer is 4

<u>-TheUnknownScientist</u><u> 72</u>

vichka [17]2 years ago
8 0

Answer:

x = 4

Step-by-step explanation:

2x + 5 = 13

We must solve to find the value of 'x'. Take  5 to the right hand side to find x.

2x  = 13 - 5

2x = 8

Take 2 in the '2x' to the right hand side to finally solve and find the value of x.

x = 8 ÷ 2

x = 4

<em>[ Note: when you take a term to the other side of equation, the sign will be the opposite of what it was before. Fore example of 2x was positive on the left hand side of the equation, it will be negative on the right hand. Same goes with multiplications and divisions. If 2x was multiplied with another term on the left hand, it will divided when shifted to the right hand. ]</em>

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Step-by-step explanation:

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3 years ago
A cheetah ran 230 feet in 1.932 seconds. What is the cheetah's speed in miles per hour?
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Answer:

The answer to your question is speed = 81.5 mi/h

Step-by-step explanation:

Data

distance = 230 ft

time = 1.932 s

speed = ? mi/h

Process

1.- Convert the distance to mi

                1 mile ----------------- 5280 ft

                 x        ----------------- 230 ft

                 x = (230 x 1) / 5280

                 x = 0.044 mi

2.- Convert time to hours

                 1 h    ------------------ 3600 s

                  x      ------------------ 1.932 s

                        x = (1.932 x 1) / 3600

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3.- Calculate the speed

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3 years ago
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
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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
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\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
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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}}

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