The answer should not depend on which machine or which pencil you use to
find it. If you work a problem two different ways and get two different answers,
then at least one of them is wrong, and there's a pretty good chance that both
of them are.
(9.99 of anything) + (1.11 of the same thing) = 11.1 of them
9.99 (x 10^-2) + 1.11 (x 10^-2) = <em>11.1 (x 10^-2)</em> .
Can we do any more with that ?
10^-2 = 1 / 10^2 = 1 / 100 .
11.1 x 10^-2 = 11.1 / 100 = <em>0.111</em>
From 6,1 to 7,4
We can determine the slope of the line
rise/run
3/1=3
Now we know the slope of the line, a line parallel to it must obtain the same slope, therefore, I tried to see if the increase in y is three times the increase in x
I found the true pair to be A
(-4,1) and (-2,7)
Answer:
The shorter piece x = 8.4 in
The longer one y = 15.6 in
Step-by-step explanation:
If we cut a string length L in to unequal pieces we get two pieces
first one with length x and the other one with length y
Let call x the smaller piece, then y = L - x will be the longer one
If x = 35 % then
x = 0,35* L and
y = L - x = L - 0,35*L ⇒ y = 0,65 L ⇒ y = 0.65*24
Now if L = 24 in
x = 0,35*24 ⇒ x = 8.4 in
and
L - x = 0.65*L ⇒ L - x = 0.65*24 y = L - x = 15.6 in
as way of verification you can add the length of the two pieces and find:
8.4 + 15.6 = 24 in
Answer:
123 cm²
Step-by-step explanation:
Refer to attachment.
<em>Hope</em><em> </em><em>it</em><em> </em><em>helps</em><em> </em><em>:</em><em>)</em>
bearing in mind that, on the III Quadrant, sine as well as cosine are both negative, and that hypotenuse is never negative, so, if the sine is -4/5, the negative number must be the numerator, so sin(x) = (-4)/5.
![\bf sin(x)=\cfrac{\stackrel{opposite}{-4}}{\stackrel{hypotenuse}{5}}\impliedby \textit{let's find the \underline{adjacent}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{5^2-(-4)^2}=a\implies \pm\sqrt{9}=a\implies \pm 3=a \\\\\\ \stackrel{III~Quadrant}{-3=a}~\hfill cos(x)=\cfrac{\stackrel{adjacent}{-3}}{\stackrel{hypotenuse}{5}} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20sin%28x%29%3D%5Ccfrac%7B%5Cstackrel%7Bopposite%7D%7B-4%7D%7D%7B%5Cstackrel%7Bhypotenuse%7D%7B5%7D%7D%5Cimpliedby%20%5Ctextit%7Blet%27s%20find%20the%20%5Cunderline%7Badjacent%7D%7D%20%5C%5C%5C%5C%5C%5C%20%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%20%5C%5C%5C%5C%20c%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20%5Csqrt%7Bc%5E2-b%5E2%7D%3Da%20%5Cqquad%20%5Cbegin%7Bcases%7D%20c%3Dhypotenuse%5C%5C%20a%3Dadjacent%5C%5C%20b%3Dopposite%5C%5C%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20%5Cpm%5Csqrt%7B5%5E2-%28-4%29%5E2%7D%3Da%5Cimplies%20%5Cpm%5Csqrt%7B9%7D%3Da%5Cimplies%20%5Cpm%203%3Da%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7BIII~Quadrant%7D%7B-3%3Da%7D~%5Chfill%20cos%28x%29%3D%5Ccfrac%7B%5Cstackrel%7Badjacent%7D%7B-3%7D%7D%7B%5Cstackrel%7Bhypotenuse%7D%7B5%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf tan\left(\cfrac{\theta}{2}\right)= \begin{cases} \pm \sqrt{\cfrac{1-cos(\theta)}{1+cos(\theta)}} \\\\ \cfrac{sin(\theta)}{1+cos(\theta)}\qquad \leftarrow \textit{let's use this one} \\\\ \cfrac{1-cos(\theta)}{sin(\theta)} \end{cases} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20tan%5Cleft%28%5Ccfrac%7B%5Ctheta%7D%7B2%7D%5Cright%29%3D%20%5Cbegin%7Bcases%7D%20%5Cpm%20%5Csqrt%7B%5Ccfrac%7B1-cos%28%5Ctheta%29%7D%7B1%2Bcos%28%5Ctheta%29%7D%7D%20%5C%5C%5C%5C%20%5Ccfrac%7Bsin%28%5Ctheta%29%7D%7B1%2Bcos%28%5Ctheta%29%7D%5Cqquad%20%5Cleftarrow%20%5Ctextit%7Blet%27s%20use%20this%20one%7D%20%5C%5C%5C%5C%20%5Ccfrac%7B1-cos%28%5Ctheta%29%7D%7Bsin%28%5Ctheta%29%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
