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insens350 [35]
2 years ago
7

What is the remainder when the polynomial 8x2 4x−3 is divided by 2x−1?.

Mathematics
1 answer:
stiks02 [169]2 years ago
3 0
Whaaaaaaaaaaaaaaaaaaat
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77+ 2(3x – 5) = 8 – 3(2x + 1)<br> 78 2 + 3(2 -7) – 0 - 13 I 1).
nika2105 [10]

Answer:

-31/6

Step-by-step explanation:

77+2(3x-5)=8-3(2x+1)

77+6x-10=8-6x-3

77-10+6x=8-3-6x

67+6x=5-6x

67+6x-(-6x)=5

67+6x+6x=5

67+12x=5

12x=5-67

12x=-62

x=-62/12

simplify

x=-31/6

3 0
3 years ago
Find the remaining trigonometric ratios of θ if csc(θ) = -6 and cos(θ) is positive
VikaD [51]
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}}&#10;\\\\\\&#10;\textit{using the pythagorean theorem}\\\\&#10;c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a&#10;\qquad &#10;\begin{cases}&#10;c=hypotenuse\\&#10;a=adjacent\\&#10;b=opposite\\&#10;\end{cases}&#10;\\\\\\&#10;\pm\sqrt{6^2-(-1)^2}=a\implies \pm\sqrt{35}=a\implies \stackrel{IV~quadrant}{+\sqrt{35}=a}

recall that 

\bf sin(\theta)=\cfrac{opposite}{hypotenuse}&#10;\qquad\qquad &#10;cos(\theta)=\cfrac{adjacent}{hypotenuse}&#10;\\\\\\&#10;% tangent&#10;tan(\theta)=\cfrac{opposite}{adjacent}&#10;\qquad \qquad &#10;% cotangent&#10;cot(\theta)=\cfrac{adjacent}{opposite}&#10;\\\\\\&#10;% cosecant&#10;csc(\theta)=\cfrac{hypotenuse}{opposite}&#10;\qquad \qquad &#10;% secant&#10;sec(\theta)=\cfrac{hypotenuse}{adjacent}

therefore, let's just plug that on the remaining ones,

\bf sin(\theta)=\cfrac{-1}{6}&#10;\qquad\qquad &#10;cos(\theta)=\cfrac{\sqrt{35}}{6}&#10;\\\\\\&#10;% tangent&#10;tan(\theta)=\cfrac{-1}{\sqrt{35}}&#10;\qquad \qquad &#10;% cotangent&#10;cot(\theta)=\cfrac{\sqrt{35}}{1}&#10;\\\\\\&#10;sec(\theta)=\cfrac{6}{\sqrt{35}}

now, let's rationalize the denominator on tangent and secant,

\bf tan(\theta)=\cfrac{-1}{\sqrt{35}}\implies \cfrac{-1}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{-\sqrt{35}}{(\sqrt{35})^2}\implies -\cfrac{\sqrt{35}}{35}&#10;\\\\\\&#10;sec(\theta)=\cfrac{6}{\sqrt{35}}\implies \cfrac{6}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{6\sqrt{35}}{(\sqrt{35})^2}\implies \cfrac{6\sqrt{35}}{35}
3 0
3 years ago
Is (2,1) a solution of 3y&lt;2x+5
Nikitich [7]

Answer:

No

Step-by-step explanation:

Plug in the x and y values and you should get 3<9

<h2><em><u>**Please rate 5 give thanks and vote brainliest**</u></em></h2>
3 0
3 years ago
What is the radius of a cylinder if it has a volume of 769.3 cubic meters and a height of 5 meters? (Use 3.14 for π.)
katen-ka-za [31]
769.3= (3.14) r^2 *5
153.86=(3.14) r^2
49= r^2
The square root of 49 is 7

The answer is 7.
6 0
3 years ago
Read 2 more answers
What is the sum?? need help
SOVA2 [1]
43+35=78 the answer is 78
3 0
3 years ago
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