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abruzzese [7]
3 years ago
12

The circumference of a circle is 407 cm. Find the diameter, the radius, the length of an arc of 100°, and the area of a sector w

ith central angle 100°.​
Mathematics
1 answer:
iogann1982 [59]3 years ago
4 0

Answer: 407cm =

100° =

100° =

I hope this helps :)

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Write an equation of the circle with center (-7,4) and radius 11.
olga_2 [115]

Answer: (x+7)^2 + (y-4)^2 = 121  or  (x+7)^2 + (y-4)^2 = 11^2

Step-by-step explanation:

Equation of the circle is:

(x-h)^2+(y-k)^2=r^2 , which is the center (h,k) and the radius r^2

-Place the center (-7,4) and radius 11 onto that equation:

(x+7)^2+(y-4)^2=121 (if simplified radius needed)

or

(x+7)^2(y-4)^2=11^2

7 0
3 years ago
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Answer now pls I’ll give you brainliest just pls answer
hjlf

Answer:

7,460 (or 7460 as some non-native English speakers would write it) since you always round a number down when it ends in 1 through 4, and always round it up when it ends in 5 through 9.

Step-by-step explanation:

pls give me brainliest plssssssssssssssssssssss.

5 0
3 years ago
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Judah deposited $450 into an account that earns simple interest, and after 4 years, the interest in the account was 90$. What wa
guapka [62]

Answer:

$22.50 I think I'm not 100% sure, Im sorry

Step-by-step explanation:

90/4 is 22.5 a year

4 0
2 years ago
The points A(1, 4), B(5,1) lie on a circle. The line segment AB is a chord. Find the equation of a diameter of the circle.
tangare [24]

Check the picture below.

well, we want only the equation of the diametrical line, now, the diameter can touch the chord at any several angles, as well at a right-angle.

bearing in mind that <u>perpendicular lines have negative reciprocal</u> slopes, hmm let's find firstly the slope of AB, and the negative reciprocal of that will be the slope of the diameter, that is passing through the midpoint of AB.

\bf A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{1}}}\implies \cfrac{-3}{4} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{slope of AB}}{-\cfrac{3}{4}}\qquad \qquad \qquad \stackrel{\textit{\underline{negative reciprocal} and slope of the diameter}}{\cfrac{4}{3}}

so, it passes through the midpoint of AB,

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{5+1}{2}~~,~~\cfrac{1+4}{2} \right)\implies \left(3~~,~~\cfrac{5}{2} \right)

so, we're really looking for the equation of a line whose slope is 4/3 and runs through (3 , 5/2)

\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{\frac{5}{2}}) \stackrel{slope}{m}\implies \cfrac{4}{3} \\\\\\ \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}{\cfrac{5}{2}}=\stackrel{m}{\cfrac{4}{3}}(x-\stackrel{x_1}{3})\implies y-\cfrac{5}{2}=\cfrac{4}{3}x-4 \\\\\\ y=\cfrac{4}{3}x-4+\cfrac{5}{2}\implies y=\cfrac{4}{3}x-\cfrac{3}{2}

4 0
3 years ago
30 POINTS<br> The question is the picture above<br> Help PLEASE <br> Giving 30 points
Papessa [141]

.73 x .05= .0365

.733333333 x .0000000555555555= .036555555555

i think this is correct hope this helps


5 0
2 years ago
Read 2 more answers
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