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KiRa [710]
3 years ago
12

I need help asap plz I will mark branlist

Mathematics
2 answers:
Gnoma [55]3 years ago
8 0

Find the mean. There will be 6 numbers, so divide by 6

14 = (9 + 12 + 17 + 15 + 13 + x)/6

Simplify

14 = (x + 66)/6

Multiply 6 to both sides

14(6) = (x + 66)/6(6)

84 = x + 66

Subtract 66 from both sides

84 (-66) = x + 66 (-66)

x = 84 - 66

x = 18

18 is your answer for x

hope this helps

Phoenix [80]3 years ago
7 0

The answer to your question would be 15

You find out what the median is (13) and then you look for the bigger number than what your looking for (15)  add it to the list then find the median with the set of numbers and 15 then you got your answer


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Matthew bought snacks for soccer team. He bought a bag of apple slices for $5.65, and he bought a 12 pack of Gatorade bottles. T
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Answer:

6.48/12=B

Step-by-step explanation:

B=6.48

6.48/12=0.54

Each Gatorade bottle costs 54 cents.

5 0
3 years ago
Read 2 more answers
Determine which lines are parallel and which are not parallel. Explain your reasoning.
NeX [460]

Answer:

If M and N are parallel, then the correspondent angles generated by the intersection with line R (or line S) should be equal.

So, if M and N are parallel, we should get that angles 2 and angle 4 are equal, as both of these are on the fourth quadrant of the intersection with the same line.

Similarly, if R and S are parallel, then correspondent angles generated by the intersection with line M (or line N) should be equal, this means that:

angle 3 and angle 7 should be equal.

Also remember that if we have two lines intersecting, generating 4 angles, any pair of two adjacent angles will always add to 180°.

Here we have two cases:

1)  m∠3=69°, m∠7=71°

Here we can see that:

m∠3 ≠ m∠7

Thus, lines  R and S are not parallel.

And here we do not have any information on the angles 1, 2, 5, and 6, so we can't compare the angles generated by line N with the ones generated by line M, so we can not know if lines N and M are parallel or not.

2) Now we have that:

m∠3=76°  , m∠8=114°

Notice that angle 8 and angle 7 are adjacent, then we have that:

∠7 + ∠8 = 180°

And we know that:

∠8 = 114°

Then:

∠7 + 114° = 180°

∠7 = 180° - 114° = 76°

Then we have:

∠7 = ∠3

From this we can conclude that lines R and S are parallel.

(again, we can not do anything with lines N and M)

6 0
2 years ago
What is the equation in a point-slope form of the line that passes through the point (-8, 5) and has a slope of 12?
Anton [14]

Answer:

y-5=12(x+8)

Step-by-step explanation:


4 0
3 years ago
Find the area of the shaded region. Round your answer to the nearest tenth.
Alex
Check the picture below on the left-side.

we know the central angle of the "empty" area is 120°, however the legs coming from the center of the circle, namely the radius, are always 6, therefore the legs stemming from the 120° angle, are both 6, making that triangle an isosceles.

now, using the "inscribed angle" theorem, check the picture on the right-side, we know that the inscribed angle there, in red, is 30°, that means the intercepted arc is twice as much, thus 60°, and since arcs get their angle measurement from the central angle they're in, the central angle making up that arc is also 60°, as in the picture.

so, the shaded area is really just the area of that circle's "sector" with 60°, PLUS the area of the circle's "segment" with 120°.

\bf \textit{area of a sector of a circle}\\\\
A_x=\cfrac{\theta \pi r^2}{360}\quad 
\begin{cases}
r=radius\\
\theta =angle~in\\
\qquad degrees\\
------\\
r=6\\
\theta =60
\end{cases}\implies A_x=\cfrac{60\cdot \pi \cdot 6^2}{360}\implies \boxed{A_x=6\pi} \\\\
-------------------------------\\\\

\bf \textit{area of a segment of a circle}\\\\
A_y=\cfrac{r^2}{2}\left[\cfrac{\pi \theta }{180}~-~sin(\theta )  \right]
\begin{cases}
r=radius\\
\theta =angle~in\\
\qquad degrees\\
------\\
r=6\\
\theta =120
\end{cases}

\bf A_y=\cfrac{6^2}{2}\left[\cfrac{\pi\cdot 120 }{180}~-~sin(120^o )  \right]
\\\\\\
A_y=18\left[\cfrac{2\pi }{3}~-~\cfrac{\sqrt{3}}{2} \right]\implies \boxed{A_y=12\pi -9\sqrt{3}}\\\\
-------------------------------\\\\
\textit{shaded area}\qquad \stackrel{A_x}{6\pi }~~+~~\stackrel{A_y}{12\pi -9\sqrt{3}}\implies 18\pi -9\sqrt{3}

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A personal balance sheet
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