All Answers 358087 Answers

WHO WON'T KNOW THAT?IT IS 19!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Four tasks

\(\large A=\frac{1}{2}\ ab\ sin\gamma\)

(a)

What is the area of this triangle. Only round your final answer to the nearest hundredth.

\(\large A=\frac{1}{2}\cdot 13\cdot7.5\ inch^2\ sin \ 94°=48.63\ inch^2\)

(b)

What is the area of this panel? Round your final answer to the nearest tenth.

\(\large A=\frac{1}{2}\cdot 12\cdot14\ ft^2\ sin \ 20°=28.7\ ft^2\)

(c)

A triangular pennet has two sides that are 90 cm long each with an included angle of 25dg. What is the area of this pennant? Round only your final answer to the nearest tenth.

\(\large A=\frac{1}{2}\cdot 90\cdot90\ cm^2\ sin \ 25°=1711.6cm^2\)

(d)

What is the area of Triangle RST? Round only your final answer to the nearest tenth. 

\(\large A=\frac{1}{2}\cdot 8\cdot9\ yd^2\ sin \ 47°=26.3\ yd^2\)

  !

7a^2r^5        

 ----------     =  

 8d^4r^-6

A negative exponent on a variable brings that variable "across the bar" with a positive exponent...!!!

(7/8) (a^2) (r^5)(r^6) / d^4  =

7a^2r^11

_______

8d^4

All of these are exactly the same....

Area =  (1/2)(product of two given sides) (sine of the included angle)

Area of 1st one  =   (1/2)(7.5)(13)sin(94)   ≈ 48.63 in^2

You should be able to do the rest......pay attention to the units given......!!!!!

Thanks so much!

You can go to this link and see the solution to your problem:

https://www.triangle-calculator.com/?what=ssa&a=5&b=7&b1=80&submit=Solve

Just copy and paste in the address bar of your Browser.

I am sorry what I said is actually not good, I should have said slant height...

But here is my explanation. You can see how that it is because if you take the triangles and lay them out like this:

Since the base of each triangle is the same length,

area of one blue triange = (1/2)(1/4 * perimeter of base)(slant height)

There are 4 triangles total that area all the same size.. so,

area of all blue triangles = 4 * (1/2)(1/4 * perimeter of base)(slant height)

area of all blue triangles = (1/2)(perimeter of base)(slant height)

numerator / denominator

fe, ½ = 1 / 2

weren't you taught this in school?

No.

That's called cheating.

Solve for v:
8.09×10^-17 = 4.55×10^-31 v^2

8.09×10^-17 = 4.55×10^-31 v^2 is equivalent to 4.55×10^-31 v^2 = 8.09×10^-17:
4.55×10^-31 v^2 = 8.09×10^-17

Multiply both sides by 2197802197802197697429956984832:
v^2 = 177802197802197

Take the square root of both sides:
Answer: | v = sqrt(177802197802197) or v = -sqrt(177802197802197) = ± 1.33342×10^7 m/s.

You need the bases and heights of the triangles to find the areas.

the calculator mode of what?

(whole + numerator / denominator)

fe, ...1 ⅓... = ...(1 + 1 / 3)...

Hi tertre, thank you for your Wow! asinus :- )  

a fractional exponent is the same as a root. n^1/2 = sqrt (n), n^1/3 = sqrt3 (n) and so on

sqrt3 (343) = 7 because 7^3 = 343

If cos(t) = 1/3 and sin(t) is negative,

this is a picture of the triangle ∠t makes:

Now we can use the Pythagorean theorem to find sin(t).

sin²(t) + (1/3)² = 1²

sin²(t) = 1 - 1/9

sin(t) = \(\pm\sqrt{\frac89}=\pm\frac{\sqrt8}{3}\)

The prolem says that sin < 0 , so we know that it is the negative option.

sin(t) = \(-\frac{2\sqrt2}{3}\)

tan = sin / cos

tan(t) = sin(t) / cos(t)

tan(t) = \(-\frac{2\sqrt2}{3}/\frac13\)

tan(t) = \(-\frac{2\sqrt2}{3}\cdot\frac31\)

tan(t) = \(-2\sqrt2\)

Stop! You now what I mean!

Oh, I understand.

4...............

19............

lol mine shall now be :o)

one.

just dont get any blood on the carpet. it's hard to get it out.

Note that R and O   intercept the same minor arc, NQ

But, since O is a central angle intercepting the same arc as R,  it will have twice the measure of inscribed angle R

So   2R   = 2 (32)  =   64   =  O

Note, NSS, that drawing a radius to either end of the chord will produce a right triangle   with legs of 5 and   (1/2)21  = 10.5   [the fact that the segment marked a "5"  meets the chord at a right angle means that the chord is bisected at that point of intersection....and the bisected part forms the other leg]

So the hypotenuse of this triangle  will equal the radii, x

x =  sqrt ( 10.5^2 + 5^2 )   =  sqrt (135.25)  ≈ 11.6

[ Looks like you got the correct answer....!!  ]