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mario62 [17]
2 years ago
15

Which of the following terms is best described as the point halfway between

Mathematics
1 answer:
zysi [14]2 years ago
6 0
The answer should be A midpoint
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YZ has endpoints Y(0,5) and Z (12,3). Find the length of YZ to the nearest tenth.
Alex Ar [27]
Using the formula;
 d(Y, Z) =\sqrt{(xZ - xY)^2 + (yZ - yY)^2

You should get 12.1655 or 12.2 units.

Hope this helps you with Coordinate Solving!

4 0
3 years ago
Read 2 more answers
PLEASE HELP ME!!!!!
Irina18 [472]

Answer:

f(6) = 1

Step-by-step explanation:

Function:

f(x) = -2/3x + 5

f(6) = -2/3(6) + 5

f(6) = -4 + 5

f(6) = 1

4 0
3 years ago
Read 2 more answers
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}}
\\\\\\
\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}

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}

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}}

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}
\\\\\\
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
Which one is right ?
sergey [27]

Answer:

6/10

Step-by-step explanation:

all can be reduced to 2/5, except 6/10:

2/5

8/20=2/5   (divide out 4)

12/30=2/5 (divide out 6)

but not

6/10 = 3/5!


7 0
3 years ago
Find the area of a circle with a radius of 6 centimeters. Round to the nearest tenth
coldgirl [10]

Answer: 113.1 cm^2

Step-by-step explanation: The formula to find the area of a circle is 3.14r^2. R stands for radius which is 6 in this case so just plug that in for R. 3.14(6)^2=113.1

8 0
3 years ago
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