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3 years ago

Help plz ASAP for 20pts

Mathematics

1 answer:

Paul [167]3 years ago

4 0

Answer:

Not sure if this is correct but this is what I found.

area of sector=74.84

radius = 8

angle of sector=?

74.84=1/2×8²×angle of sector

74.84=32 × angle of sector

74.84/32= angle of sector

2.33875=angle of sector

Step-by-step explanation:

Sorry if it is incorrect.

You might be interested in

200-[(50-4)×3+5] <img src="https://tex.z-dn.net/?f=200%20-%20%28%2850%20-%204%29%20%5Ctimes%203%20%2B%205%29" id="TexFormula1

lbvjy [14]

200 - [ ( 50 - 4 ) × 3 + 5 ]
= 200 - [ 46 × 3 + 5 ]
<Put brackets around the 2 number that enclose the times sign as a reminder to do that first>
= 200 - [ ( 46 × 3 ) + 5 ]
= 200 - [ 138 + 5 ]
= 200 - (143)
= 57

Hope this helps!

3 0

3 years ago

I'm not understanding this can you help me

hoa [83]

Answer:

gawin mo pag isipan mo wag puro brainly

3 0

3 years ago

Read 2 more answers

Adding the fractions 3/14+2/21+1/6

NARA [144]

Answer:

$\frac{10}{21}$

Step-by-step explanation:

The LCM of 14, 21 and 6 is 42

We require to change the fractions to fractions with a denominator of 42

$\frac{3(3)}{14(3)}$ + $\frac{2(2)}{21(2)}$ + $\frac{1(7)}{6(7)}$

= $\frac{9}{42}$ + $\frac{4}{42}$ + $\frac{7}{42}$ ← add the numerators, leaving the denominator

= $\frac{9+4+7}{42}$

= $\frac{20}{42}$ ← divide both values by 2

= $\frac{10}{21}$ ← in simplest form

6 0

3 years ago

Write an equation that represents the line. Use exact numbers. (-2,-1) (4,6)

vazorg [7]

Check the picture below.

to get the equation of any straight line, we simply need two points off of it, so let's use those in the picture

$(\stackrel{x_1}{-2}~,~\stackrel{y_1}{1})\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{6})~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{6}-\stackrel{y1}{1}}}{\underset{run} {\underset{x_2}{4}-\underset{x_1}{(-2)}}} \implies \cfrac{6 -1}{4 +2} \implies \cfrac{ 5 }{ 6 }$

$\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{1}=\stackrel{m}{\cfrac{5}{6}}(x-\stackrel{x_1}{(-2)})\implies y-1=\cfrac{5}{6}(x+2) \\\\\\ y-1=\cfrac{5}{6}x+\cfrac{5}{3}\implies y=\cfrac{5}{6}x+\cfrac{5}{3}+1\implies y=\cfrac{5}{6}x+\cfrac{8}{3}$