let's bear in mind that sin(θ) in this case is positive, that happens only in the I and II Quadrants, where the cosine/adjacent are positive and negative respectively.
![\bf sin(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{hypotenuse}{6}}\qquad \impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{6^2-5^2}=a\implies \pm\sqrt{36-25}\implies \pm \sqrt{11}=a \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20sin%28%5Ctheta%20%29%3D%5Ccfrac%7B%5Cstackrel%7Bopposite%7D%7B5%7D%7D%7B%5Cstackrel%7Bhypotenuse%7D%7B6%7D%7D%5Cqquad%20%5Cimpliedby%20%5Ctextit%7Blet%27s%20find%20the%20%5Cunderline%7Badjacent%20side%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%5Cpm%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%7B6%5E2-5%5E2%7D%3Da%5Cimplies%20%5Cpm%5Csqrt%7B36-25%7D%5Cimplies%20%5Cpm%20%5Csqrt%7B11%7D%3Da%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
Answer:
Ya can t at least I don't think you can
Step-by-step explanation:
Answer:
Step-by-step explanation:
A system of linear equations is one which may be written in the form
a11x1 + a12x2 + · · · + a1nxn = b1 (1)
a21x1 + a22x2 + · · · + a2nxn = b2 (2)
.
am1x1 + am2x2 + · · · + amnxn = bm (m)
Here, all of the coefficients aij and all of the right hand sides bi are assumed to be known constants. All of the
xi
’s are assumed to be unknowns, that we are to solve for. Note that every left hand side is a sum of terms of
the form constant × x
Solving Linear Systems of Equations
We now introduce, by way of several examples, the systematic procedure for solving systems of linear
equations.
Here is a system of three equations in three unknowns.
x1+ x2 + x3 = 4 (1)
x1+ 2x2 + 3x3 = 9 (2)
2x1+ 3x2 + x3 = 7 (3)
We can reduce the system down to two equations in two unknowns by using the first equation to solve for x1
in terms of x2 and x3
x1 = 4 − x2 − x3 (1’)
1
and substituting this solution into the remaining two equations
(2) (4 − x2 − x3) + 2x2+3x3 = 9 =⇒ x2+2x3 = 5
(3) 2(4 − x2 − x3) + 3x2+ x3 = 7 =⇒ x2− x3 = −1
Answer:
Part A: A
Part B: A
Step-by-step explanation:
1)
the inequality should be less than or equal to, since we need to find how many are left
now, the answer B is wrong, since there can't be negative 56, since Xander collected some cans, though he still must collect a few
thus, the answer is A
2)
A again is correct, since it represents:c ≤ 56
B represents c ≥ 56
C represents c ≤ –56
D represents c ≥ -56
There are no visible parenthasees or exponents
so pemdas
mutipcation or division whichever comes first (because division is multiplying by a fraction)
61/2
so fist is division
D is answer
