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3 years ago

I need help with this one

Mathematics

2 answers:

borishaifa [10]3 years ago

8 0

Answer:

x = 82.5

Step-by-step explanation:

The sum of the 3 angles in a triangle = 180°

Sum the 3 angles and equate to 180, that is

x + x - 45 + 60 = 180, that is

2x + 15 = 180 ( subtract 15 from both sides )

2x = 165 ( divide both sides by 2 )

x = 82.5

andrew11 [14]3 years ago

3 0

Answer:

--------------------------------------------------

Step-by-step explanation:

60+x-45+x=180

60-45+2x=180

15+2x=180

2x=180-15

2x=165

x=165/2

x=82.5

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In the diagram shown, /Il m, m

Dmitry_Shevchenko [17]

Answer:

29°

Step-by-step explanation:

Angles 1 and 8 are alternate exterior angles where a transversal crosses parallel lines, so are congruent.

m∠1 = m∠8

3x+25 = 4x-17

42 = m . . . . . . . . add 17-3x

Then the measure of ∠1 is 3·42 +25 = 151 (degrees), and the measure of its supplement, ∠2 is ...

m∠2 = 180° -151° = 29°

5 0

3 years ago

Write an essay explaining the importance of getting to know people before forming an opinion about them.

Andrej [43]

Nobody's going to write an essay for you dude

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3 years ago

Use cylindrical coordinates. find the volume of the solid that is enclosed by the cone z = x2 + y2 and the sphere x2 + y2 + z2 =

sashaice [31]

Let $R$ be the solid. Then the volume is

$\displaystyle\iiint_R\mathrm dV=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=\sqrt8}\int_{\zeta=r^2}^{\zeta=\sqrt{72-r^2}}r\,\mathrm d\zeta\,\mathrm dr\,\mathrm d\theta$

which follows from the facts that

$\begin{cases}x=r\cos\theta\\y=r\sin\theta\\z=\zeta\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=r\,\mathrm dr\,\mathrm d\theta\,\mathrm d\zeta$
(by computing the Jacobian)

and

$z=x^2+y^2=r^2\implies z+z^2=72\implies z=-9\text{ or }z=8$
(we take the positive solution, since it's clear that $R$ lies above the $x$ - $y$ plane)
$r^2+z^2=72\implies z=\pm\sqrt{72-r^2}$
(again, taking the positive root for the same reason)
$z=r^2\implies 8=r^2\implies r=\sqrt8$

$\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=\sqrt8}\int_{\zeta=r^2}^{\zeta=\sqrt{72-r^2}}r\,\mathrm d\zeta\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle2\pi\int_{r=0}^{r=\sqrt8}\int_{\zeta=r^2}^{\zeta=\sqrt{72-r^2}}r\,\mathrm d\zeta\,\mathrm dr$
$=\displaystyle2\pi\int_{r=0}^{r=\sqrt8}r(\sqrt{72-r^2}-r^2)\,\mathrm dr$
$=\displaystyle2\pi\int_{r=0}^{r=\sqrt8}(r\sqrt{72-r^2}-r^3)\,\mathrm dr$
$=\dfrac{32(27\sqrt2-35)\pi}3$