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2 years ago

(16 - 6) + f - 9 = 12

Mathematics

2 answers:

CaHeK987 [17]2 years ago

6 0

Answer:

f = 11

Step-by-step explanation:

STALIN [3.7K]2 years ago

5 0

Answer:

F=11

Step-by-step explanation:

parenthesis first

10+f-9=12

combine like terms

1+f=12

-1 from both sides to solve for f

f=12-1

f=11

<em>Hope this helps have a great day</em>

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All of the expressions that have the same value as LaTeX: 892\div8892 ÷ 8. Group of answer choices LaTeX: 8920\div808920 ÷ 80 La

algol13

Answer:

$(a)$ $8920\div80$

$(e)$ $0.892\div0.008$

Step-by-step explanation:

Given

$892\div8$

Required

Select equivalent expression

$8920\div80$

$894\div10$

$89.2\div0.08$

$8.92\div0.8$

$0.892\div0.008$

Start by solving the given expression (using a calculator)

$892\div8$ $= 111.5$

Then we solve the given options (using a calculator)

$(a)$ $8920\div80$

$8920\div80$ $= 111.5$

$(b)$ $894\div10$

$894\div10$ $= 89.4$

$(c)$ $89.2\div0.08$

$89.2\div0.08$ $= 1115$

$(d)$ $8.92\div0.8$

$8.92\div0.8$ $= 11.15$

$(e)$ $0.892\div0.008$

$0.892\div0.008$ $= 111.5$

The equivalent expressions are (a) and (e)

5 0

2 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°.

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

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

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