Simplify the expression 12k-2+6k A. 16k2 B. 10k+6 C. 18k-2 D. 16k

Mathematics

2 answers:

WITCHER [35]3 years ago

5 0

12k-2+6k

find like terms and put them next to each other

12k+6k-2

combine like terms

18k-2

C

lubasha [3.4K]3 years ago

3 0

Answer:

16k so it is D

Step-by-step explanation:

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After t seconds, a ball tossed in the air from the ground level reaches a height of h feet given by the equation h = 144t-16t^2

lesantik [10]

144t-16t^2= 16t(9-t)
after 3 seconds (just substitute t=3):
16(3)(9-3)=
16(3)(6)=
16*18=
288
maximum height: find the average of the roots:
the roots of 16t(9-t) are t=0 or t=9
since it's a parabola, the maximum is at t=4.5, at 324 ft
if its height is 224 feet, the equation is 224=144t-16t^2
16t^2-144t+224=0
divide by 16: t^2-9t+14=0
this can be factored as (t-2)(t-7)=0
the roots are t=2 and t=7, so the ball has been in the air for either 2 seconds or 7 seconds
the roots to 144t-16t^2 are 0 and 9, so the ball will hit the ground 9 seconds after being thrown

4 0

3 years ago

Read 2 more answers

Suppose theta is an angle in the standard position whose terminal side is in quadrant 4 and cot theta = -6/7. find the exact val

zimovet [89]

First off, let's notice that the angle is in the IV Quadrant, where sine is negative and the cosine is positive, likewise the opposite and adjacent angles respectively.

Also let's bear in mind that the hypotenuse is never negative, since it's simply just a radius unit.

$\bf cot(\theta )=\cfrac{\stackrel{adjacent}{6}}{\stackrel{opposite}{-7}}\qquad \impliedby \textit{let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{6^2+(-7)^2}\implies c=\sqrt{36+49}\implies c=\sqrt{85} \\\\[-0.35em] ~\dotfill$

$\bf tan(\theta)=\cfrac{\stackrel{opposite}{-7}}{\stackrel{adjacent}{6}} ~\hfill csc(\theta)=\cfrac{\stackrel{hypotenuse}{\sqrt{85}}}{\stackrel{opposite}{-7}} ~\hfill sec(\theta)=\cfrac{\stackrel{hypotenuse}{\sqrt{85}}}{\stackrel{adjacent}{6}} \\\\\\ sin(\theta)=\cfrac{\stackrel{opposite}{-7}}{\stackrel{hypotenuse}{\sqrt{85}}}\implies \stackrel{\textit{and rationalizing the denominator}}{sin(\theta)=\cfrac{-7}{\sqrt{85}}\cdot \cfrac{\sqrt{85}}{\sqrt{85}}\implies sin(\theta)=-\cfrac{7\sqrt{85}}{85}}$

$\bf cos(\theta)=\cfrac{\stackrel{adjacent}{6}}{\stackrel{hypotenuse}{\sqrt{85}}}\implies \stackrel{\textit{and rationalizing the denominator}}{cos(\theta)=\cfrac{6}{\sqrt{85}}\cdot \cfrac{\sqrt{85}}{\sqrt{85}}\implies cos(\theta)=\cfrac{6\sqrt{85}}{85}}$