To solve this question, we simply need to divide the total amount of students by 8% to find out how many students were absent and how many were present.
However, we can't simply multiply it by 8%, so we need to turn that into a decimal.
8% - 0.08
<em>Multiply:</em>
<em>355 x 0.08</em>
<em>= 28.4</em>
<em>Round:</em>
<em>28 kids were absent</em>
<em>Subtract:</em>
<em>355 - 28</em>
<em>= 327</em>
<em />
<em>This means that </em><em>28 kids were absent</em><em>, and </em><em>327 kids were present</em><em>.</em><em> </em>
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Answer:
Isosceles
Step-by-step explanation:
Hope this helps you
Ok, so you are given the value of P=q+2
The substitution method tells us that we must insert the value we know, into the second equation, 4P+3q= -27
Doing so will give us 4(q+2)+3q= -27
For right now, lets just focus on the first part, 4(q+2)
We can simplify this by distributing(multiplying) the 4 to whats inside the variables.
This will give us 4q+8
now lets add this back to the rest of the equation >>> 4q+8+3q = -27
We can further simplify by adding like terms >>> 7q+8 = -27
subtract the 8 from both sides >>> 7q = -35
now divide both sides by 7 >>> q = -35/7
Therefor q = -5
EDIT*
now that we know q = -5 we can put q into the equation for P !
we know that p=q+2
so lets put q in now >>> p=(-5)+2
and simplify>>> p = -3
I hope this helps:)
Answer:
5*3=15
Step-by-step explanation:
Now, the cosecant of θ is -6, or namely -6/1.
however, the cosecant is really the hypotenuse/opposite, but the hypotenuse is never negative, since is just a distance unit from the center of the circle, so in the fraction -6/1, the negative must be the 1, or 6/-1 then.
we know the cosine is positive, and we know the opposite side is -1, or negative, the only happens in the IV quadrant, so θ is in the IV quadrant, now
![\bf csc(\theta)=-6\implies csc(\theta)=\cfrac{\stackrel{hypotenuse}{6}}{\stackrel{opposite}{-1}}\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-(-1)^2}=a\implies \pm\sqrt{35}=a\implies \stackrel{IV~quadrant}{+\sqrt{35}=a}](https://tex.z-dn.net/?f=%5Cbf%20csc%28%5Ctheta%29%3D-6%5Cimplies%20csc%28%5Ctheta%29%3D%5Ccfrac%7B%5Cstackrel%7Bhypotenuse%7D%7B6%7D%7D%7B%5Cstackrel%7Bopposite%7D%7B-1%7D%7D%5Cimpliedby%20%5Ctextit%7Blet%27s%20find%20the%20%5Cunderline%7Badjacent%20side%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%5C%5C%5C%5C%0Ac%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20%5Cpm%5Csqrt%7Bc%5E2-b%5E2%7D%3Da%0A%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0Ac%3Dhypotenuse%5C%5C%0Aa%3Dadjacent%5C%5C%0Ab%3Dopposite%5C%5C%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Cpm%5Csqrt%7B6%5E2-%28-1%29%5E2%7D%3Da%5Cimplies%20%5Cpm%5Csqrt%7B35%7D%3Da%5Cimplies%20%5Cstackrel%7BIV~quadrant%7D%7B%2B%5Csqrt%7B35%7D%3Da%7D)
recall that
![\bf sin(\theta)=\cfrac{opposite}{hypotenuse} \qquad\qquad cos(\theta)=\cfrac{adjacent}{hypotenuse} \\\\\\ % tangent tan(\theta)=\cfrac{opposite}{adjacent} \qquad \qquad % cotangent cot(\theta)=\cfrac{adjacent}{opposite} \\\\\\ % cosecant csc(\theta)=\cfrac{hypotenuse}{opposite} \qquad \qquad % secant sec(\theta)=\cfrac{hypotenuse}{adjacent}](https://tex.z-dn.net/?f=%5Cbf%20sin%28%5Ctheta%29%3D%5Ccfrac%7Bopposite%7D%7Bhypotenuse%7D%0A%5Cqquad%5Cqquad%20%0Acos%28%5Ctheta%29%3D%5Ccfrac%7Badjacent%7D%7Bhypotenuse%7D%0A%5C%5C%5C%5C%5C%5C%0A%25%20tangent%0Atan%28%5Ctheta%29%3D%5Ccfrac%7Bopposite%7D%7Badjacent%7D%0A%5Cqquad%20%5Cqquad%20%0A%25%20cotangent%0Acot%28%5Ctheta%29%3D%5Ccfrac%7Badjacent%7D%7Bopposite%7D%0A%5C%5C%5C%5C%5C%5C%0A%25%20cosecant%0Acsc%28%5Ctheta%29%3D%5Ccfrac%7Bhypotenuse%7D%7Bopposite%7D%0A%5Cqquad%20%5Cqquad%20%0A%25%20secant%0Asec%28%5Ctheta%29%3D%5Ccfrac%7Bhypotenuse%7D%7Badjacent%7D)
therefore, let's just plug that on the remaining ones,
![\bf sin(\theta)=\cfrac{-1}{6} \qquad\qquad cos(\theta)=\cfrac{\sqrt{35}}{6} \\\\\\ % tangent tan(\theta)=\cfrac{-1}{\sqrt{35}} \qquad \qquad % cotangent cot(\theta)=\cfrac{\sqrt{35}}{1} \\\\\\ sec(\theta)=\cfrac{6}{\sqrt{35}}](https://tex.z-dn.net/?f=%5Cbf%20sin%28%5Ctheta%29%3D%5Ccfrac%7B-1%7D%7B6%7D%0A%5Cqquad%5Cqquad%20%0Acos%28%5Ctheta%29%3D%5Ccfrac%7B%5Csqrt%7B35%7D%7D%7B6%7D%0A%5C%5C%5C%5C%5C%5C%0A%25%20tangent%0Atan%28%5Ctheta%29%3D%5Ccfrac%7B-1%7D%7B%5Csqrt%7B35%7D%7D%0A%5Cqquad%20%5Cqquad%20%0A%25%20cotangent%0Acot%28%5Ctheta%29%3D%5Ccfrac%7B%5Csqrt%7B35%7D%7D%7B1%7D%0A%5C%5C%5C%5C%5C%5C%0Asec%28%5Ctheta%29%3D%5Ccfrac%7B6%7D%7B%5Csqrt%7B35%7D%7D)
now, let's rationalize the denominator on tangent and secant,