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3 years ago

What does x+7=2 what does x mean

Mathematics

1 answer:

Ksenya-84 [330]3 years ago

5 0

Answer:-5

Step-by-step explanation:

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Help !!! with 14 please :))

BartSMP [9]

Since p is an obtuse angle
& that it's between two acute angles

3 0

3 years ago

What's the exact value for this trigonometric function sin 60 degrees

Elodia [21]

$\sin60^\circ=\dfrac{\sqrt{3}}{2}$

3 0

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

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

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

A 3 kg toy car rests st the top of a 0.45 meter ramp. What is its gravitational potential energy?

salantis [7]

<span>gravitational potential energy : P
Gravity : g
Mass : m
height : h

P = mgh = 3 x 9.8 x 0.45 = 13.23 Joule

Potential energy is work , from the known formula

W = Fd ( work = force x distance )

W = P ( in case of potential energy height change)

F is the force acting on the body in case of ideal ramp , the only force acting is the weight of the body

F = mg ( not just <m> as the force is mg (Newton) gravity effect)

d is the displacement in direction of force, as we have considered the force to be the weight not it's component in direction of the ramp , the change in displacement is the change in height so

d = h

W = Fd = (F = mg) x (d = h) = mgh

W = mgh = P

</span>

7 0

3 years ago

An airplane ascends such that its gain h in altitude is proportional to the square root of the change x in horizontal distance t

Mazyrski [523]

Answer:velocity=185 m/s

Step-by-step explanation:

So make k the subject of the formula by divide both by sqrt(x)

h= k × sqrt(x)

k=h/sqrt(x)

k=256 / sqrt(256)

k=256 / 16

k=16

So now substitute the value of k :

h= 16 × sqrt (x)

Then differentiate:

=(1/2 × sqrt(x))× 16

=16/(2×sqrt(x)

=8/sqrt(x)

Then

= 8/sqrt(x) × 370

=8/ sqrt(256) × 370

=8/16 × 370

velocity=185 m/s

6 0

3 years ago