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Mathematics 50 Online
OpenStudy (anonymous):

Find the exact value of tan(M).

OpenStudy (anonymous):

do we know anything at all about \(M\)?

OpenStudy (anonymous):

OpenStudy (anonymous):

\[\tan(M)=\frac{\text{opposite}}{\text{adjacent}}\]

OpenStudy (anonymous):

|dw:1443665613699:dw|

OpenStudy (anonymous):

oh 5/12

OpenStudy (anonymous):

yup

OpenStudy (anonymous):

can I ask more?

OpenStudy (anonymous):

:)

OpenStudy (anonymous):

sure why not?

OpenStudy (anonymous):

btw you have my second favorite name on OS

OpenStudy (anonymous):

The diagonals of kite KITE intersect at point P. Which statement is true?

OpenStudy (sweetburger):

to find m wouldnt you need to take tan-1(5/12)

OpenStudy (sweetburger):

tan^-1(5/12) sorry typo

OpenStudy (anonymous):

Thank you @satellite73 what's your first favorite :)

OpenStudy (anonymous):

@sweetburger no

OpenStudy (anonymous):

i just said second because i didn't want to totally commit

OpenStudy (sweetburger):

alright

OpenStudy (anonymous):

btw i am going to take a stab at this geometry, but no guarantees what are the options for this "which of the following"

OpenStudy (anonymous):

A) <KIE=<TPE B) <TIP=<TPE C) <KIT=<KET D) <KPE=TPE

OpenStudy (anonymous):

going to take a minute to go through all of these

OpenStudy (anonymous):

the last one is true for sure

OpenStudy (anonymous):

i think that is the only one

OpenStudy (anonymous):

i was hoping for more trig

OpenStudy (anonymous):

If TUVW is an isosceles trapezoid with TU/ parallel to WV/, name a pair of similar triangles. Explain.

OpenStudy (anonymous):

WTV and WUV are similar the diagonals have the same length, TW equals UV and angle W equals angle V

OpenStudy (anonymous):

A. triangle UXV= triangle WXT by the SAS similarity theorem. B. triangle TXU= triangle WXV by the SAS similarity theorem. C. triangle WUV= triangle TVW by the AA similarity postulate. D. triangle TXU= triangle VXW by the AA similarity postulate.

OpenStudy (anonymous):

ughh... that took a long time.

OpenStudy (anonymous):

yeah now let me check

OpenStudy (anonymous):

ok first one looks good, but let me keep checking

OpenStudy (anonymous):

not B

OpenStudy (anonymous):

Ok :)

OpenStudy (anonymous):

Verify the identity. cos 4x + cos 2x = 2 - 2 sin2 2x - 2 sin2 x

OpenStudy (anonymous):

hold the phone, not done with the previous one (i hate geometry)

OpenStudy (anonymous):

same... I hate it too.

OpenStudy (anonymous):

cause C looks good too damn lets move on

OpenStudy (anonymous):

\[\cos (4x )+ \cos (2x) = 2 - 2 \sin^2( 2x) - 2 \sin^2(x)\] i am a bit confused with the 2's can you post a screen shot or is that it?

OpenStudy (anonymous):

hey could we hold off on this problem for a sec, we'll come back to it. Is that fine?

OpenStudy (anonymous):

sure

OpenStudy (anonymous):

let me guess, more shape...

OpenStudy (anonymous):

In the figure,line AB is parallel to line CD. If the area of triangle CED/ is 425, the area of BEA / is 68, and BE = 10, find CE.

OpenStudy (anonymous):

yup lol.

OpenStudy (anonymous):

A. 25 B. 36 C. 48 D. 62.5

OpenStudy (anonymous):

the big triangle has area 425 and the small one area 68 is that right?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

ok we can do this

OpenStudy (anonymous):

the ratio of the big area to the small area is \(425:68\) which reduces to \(25:4\) or \(\frac{25}{4}\) the ratio of the sides is therefore the square root of that, namely \( \frac{5}{2}\)

OpenStudy (anonymous):

since \(BE=10\) multply it by \(\frac{5}{2}\) and get \(25\)

OpenStudy (anonymous):

Yea I did that.

OpenStudy (anonymous):

ok good

OpenStudy (anonymous):

what do I do next?

OpenStudy (anonymous):

have a beer?

OpenStudy (anonymous):

pick A we are done, the answer is 25

OpenStudy (anonymous):

Find the equation of the circle that has a diameter with endpoints located at (-5,-3) and (-11,-3)

OpenStudy (anonymous):

this is much easier

OpenStudy (anonymous):

first find the midpoint you got that?

OpenStudy (anonymous):

do I have to type in the answer choices...

OpenStudy (anonymous):

no we can knock this out in a heartbeat

OpenStudy (anonymous):

what is the average of \(-5\) and \(-11\)?

OpenStudy (anonymous):

you know what i mean? add them up, divide by 2

OpenStudy (anonymous):

lol no

OpenStudy (anonymous):

\[-5-11=-16\] \[-16\div 2=-8\]

OpenStudy (anonymous):

ok in any case that makes the midpoint \((-8,-3)\) so the equation is \[(x+8)^2+(y+3)^2=r^2\]now we need \(r\)

OpenStudy (anonymous):

the distance between \((-8,-3)\) and \((-5,-3)\) is evidently 3 (from -8 to -5 is 3 units) so \(r=3\)

OpenStudy (anonymous):

final answer \[(x+8)^2+(y+3)^2=9\]

OpenStudy (anonymous):

OpenStudy (anonymous):

OpenStudy (anonymous):

i hate to break it to you but i have no $%#@ing idea

OpenStudy (anonymous):

Lol that alright, hold on.

OpenStudy (anonymous):

repost i am sure someone knows this and i am sure it is not me

OpenStudy (anonymous):

Triangle GHI is similar to triangle JKL. If IH=36 and KP=18, LK=24 then MH=_____. A. 12 B. 20 C. 27 D. 48

OpenStudy (anonymous):

it take goddamn forever just to read it hold on

OpenStudy (anonymous):

@satellite73 That's fine for now we'll move on from that problem :) ^

OpenStudy (anonymous):

no it is doable give me a sec

OpenStudy (anonymous):

Lol alrighty.

OpenStudy (anonymous):

\[\frac{18}{24}=\frac{3}{4}\] and so \[\frac{MH}{36}=\frac{3}{4}\] making \[MH=\frac{3\times 36}{4}\]

OpenStudy (anonymous):

i hope 27 is a choice

OpenStudy (anonymous):

it is ty!

OpenStudy (anonymous):

whew did i mention that i hate geometry?

OpenStudy (anonymous):

Lol yes you did.

OpenStudy (anonymous):

The diagonals of rectangle NOPQ intersect at point R. If OR=3x-4 / and NP=5x+20/, solve for x. A. 2 B. 3 C. 12 D. 28

OpenStudy (anonymous):

\[2\times OR=NP\] pr \[2(3x-4)=5x+20\] solve for \(x\)

OpenStudy (anonymous):

*or

OpenStudy (anonymous):

6x-8=5x+20 -5x -5x x-8=20 +8 +8 x=28 yay!

OpenStudy (anonymous):

yay indeed

OpenStudy (anonymous):

Find the volume of a sphere with a diameter 40 cm in length. Approximate pi as 3.14 and round your answer to the nearest tenth. A. 5,026.5 cm^3 B. 33,493.3 cm^3 C. 42,090.0 cm^3 D. 268,082.6 cm^3

OpenStudy (anonymous):

google

OpenStudy (anonymous):

Lol fine with me.

OpenStudy (anonymous):

it comes right up

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