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liraira [26]
3 years ago
10

Estimate the area of the circle. Round to the nearest whole number. A) 3 square units B) 6 square units C) 9 square units D) 12

square units
Mathematics
2 answers:
ra1l [238]3 years ago
8 0

Answer:

Please attach a pic :)

Brainliest would be appreciated

erica [24]3 years ago
6 0

Answer:

3 square units branliest pls

Step-by-step explanation:

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ohaa [14]

Step-by-step explanation:

hope it's help u jajajn s

8 0
2 years ago
What is the solution to the following equation? (4 points) 5(2x − 6) + 20 = 10 Group of answer choices
vichka [17]

Answer:

x=2

Step-by-step explanation:

6 0
3 years ago
Number 1d please help me analytical geometry
lesantik [10]
For a) is just the distance formula

\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
A&({{ x}}\quad ,&{{ 1}})\quad 
%  (c,d)
B&({{ -4}}\quad ,&{{ 1}})
\end{array}\qquad 
%  distance value
\begin{array}{llll}

d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
\sqrt{8} = \sqrt{({{ -4}}-{{ x}})^2 + (1-1)^2}
\end{array}
-----------------------------------------------------------------------------------------
for b)  is also the distance formula, just different coordinates and distance

\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
A&({{ -7}}\quad ,&{{ y}})\quad 
%  (c,d)
B&({{ -3}}\quad ,&{{ 4}})
\end{array}\ \ 
\begin{array}{llll}

d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
4\sqrt{2} = \sqrt{(-3-(-7))^2+(4-y)^2}
\end{array}
--------------------------------------------------------------------------
for c)  well... we know AB = BC.... we do have the coordinates for A and B
so... find the distance for AB, that is \bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
A&({{ -3}}\quad ,&{{ 0}})\quad 
%  (c,d)
B&({{ 5}}\quad ,&{{ -2}})
\end{array}\qquad 
%  distance value
\begin{array}{llll}

d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}\\\\
d=\boxed{?}

\end{array}

now.. whatever that is, is  = BC, so  the distance for BC is

\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
B&({{ 5}}\quad ,&{{ -2}})\quad 
%  (c,d)
C&({{ -13}}\quad ,&{{ y}})
\end{array}\qquad 
%  distance value
\begin{array}{llll}

d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}\\\\
d=BC\\\\
BC=\boxed{?}

\end{array}

so... whatever distance you get for AB, set it equals to BC, BC will be in "y-terms" since the C point has a variable in its ordered points

so.. .solve AB = BC for "y"
------------------------------------------------------------------------------------

now d)   we know M and N are equidistant to P, that simply means that P is the midpoint of the segment MN

so use the midpoint formula

\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
M&({{-2}}\quad ,&{{ 1}})\quad 
%  (c,d)
N&({{ x}}\quad ,&{{ 1}})
\end{array}\qquad
%   coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)=P
\\\\\\


\bf \left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)=(1,4)\implies 
\begin{cases}
\cfrac{{{ x_2}} + {{ x_1}}}{2}=1\leftarrow \textit{solve for "x"}\\\\
\cfrac{{{ y_2}} + {{ y_1}}}{2}=4
\end{cases}

now, for d), you can also just use the distance formula, find the distance for MP, then since MP = PN, find the distance for PN in x-terms and then set it to equal to MP and solve for "x"


7 0
3 years ago
PLEASE I WILL GIVE BRAINLIEST AND I WIL REPORT ANSWERS THAT HAVE NOTHING RI DO WITH MATH THANKS ​
Rudiy27

Answer:

26.1533936612

Step-by-step explanation:

bc

h= 6 square root 3

24^2+6 square root 3^2=684

square root 684=26.1533936612

7 0
2 years ago
What is the missing information in the paragraph proof?
loris [4]
Astt: Angle Sum Tri Theorem
7 0
3 years ago
Read 2 more answers
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