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cluponka [151]
3 years ago
10

Plzzzzz helpppppp meeeee

Mathematics
2 answers:
Zigmanuir [339]3 years ago
7 0
The answer is for this question is 1/4
aleksklad [387]3 years ago
3 0
The answere to the 3rd qu is 1/6 because there is only i option for TT out of the possible 6
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If pie-0 =3.14159265359 what is pie time 160=
Zepler [3.9K]

Answer:

Hey!

To start this problem, we'll most likely need a calculator. Handling \pi can get a little messy, and tricky.

(also note that "pie" is a food, the mathematical term \pi, is pronounced "pie", but spelled "pi.")

Anyway, multiplying 3.14159265359..... (\pi) \cdot 160 = 502.654824574...... going on and on.

In general, these types of problems are better for a calculator if you want an exact answer. Otherwise, just resort to estimation. \pi is pretty close to 3.14.

160 \cdot 3.14 = 502.4, which is decently close to our other answer.

Hope this helps!

- sahkfam

7 0
3 years ago
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Miguel's teacher asks him to color 4/8 of his grid. He must use 3 colors: red, blue, and green. There must be more green section
Julli [10]
Miguel must color miguel must color 2/8 green, 1/8 red, and 1/8 blue.

i hoped i helped



6 0
3 years ago
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PLEASE HELP I’LL MAKE YOU BRAINLIEST! (PLEASE WRITE THE CALCULATIONS ASWELL!)
Elena-2011 [213]

Answer:

Multiply each as in like 120 multiply by 1.2. If 60 multiply by 0.6.

Step-by-step explanation:

6 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Cboxed%7B%5Csf%202x%2B31%3D5%7D" id="TexFormula1" title="\boxed{\sf 2x+31=5}" alt="\boxed{\s
AleksAgata [21]

2x + 31 = 5

2x = 5 - 31

2x = -26

x = -26/2

x = -13.

_______

꧁✿ ᴿᴬᴵᴺᴮᴼᵂˢᴬᴸᵀ2222 ✬꧂

3 0
3 years ago
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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
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