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2 years ago

Solve ( 2 4/16 + 3 2/3 )

Mathematics

2 answers:

elena-s [515]2 years ago

7 0

Answer:

5 11/12

Step-by-step explanation:

2 4/16 = 2 1/4

2 1/4 + 3 2/3 = 9/4 + 11/3

27/12 + 44/12 = 71/12 = 5 11/12

jeyben [28]2 years ago

7 0

Answer:

Step-by-step explanation:

2 2/8 + 3 2/3

18/8 + 11/3

54/24 + 88/24

142/24

5.91

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What is the probability that out of 125 babies born, at least 70 will be girls?

mote1985 [20]

Answer:

Well just looking at the choices can lead you to the answer right away.

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3 years ago

The base of a triangle is 6 inches more than 3 times the height. If the area of the triangle is 120 square inches.

wariber [46]

Answer:

2160

Step-by-step explanation:

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2 years ago

Find the equation of the circle that has a diameter with endpoints located at (7, 3) and (7, –5).

posledela

So hmm check the picture below, that's about the circle and the endpoints, but notice, the endpoints make up a segment, namely the diameter of the circle, well.... let's see how long that is, because, the radius is half the diameter

$\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ 7}}\quad ,&{{ 3}})\quad 
% (c,d)
&({{ 7}}\quad ,&{{ -5}})
\end{array}\qquad 
% distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
d=\sqrt{(7-7)^2+(-5-3)^2}\implies d=\sqrt{0+(-8)^2}\implies d=8
\\\\\\
\textit{the radius is half that, so is }\boxed{r=4}$

now.. hmmm notice, the midpoint of the diameter, is the center of the circle, let's check that one out

$\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ 7}}\quad ,&{{ 3}})\quad 
% (c,d)
&({{ 7}}\quad ,&{{ -5}})
\end{array}\qquad 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)
\\\\\\
\left(\cfrac{{{ 7+7}}}{2}\quad ,\quad \cfrac{{{ -5}} + {{ 3}}}{2} \right)\implies \left( \cfrac{14}{2}\ ,\ \cfrac{-2}{2} \right)\implies \boxed{(7,-1)}$

now.. that we know what the center is, and what the radius is, well

$\bf \textit{equation of a circle}\\\\ 
(x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2
\qquad 
\begin{array}{lllll}
center\ (&{{ h}},&{{ k\quad }})\qquad 
radius=&{{ r}}\\
&7&-1&4
\end{array}$

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3 years ago