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Vera_Pavlovna [14]
3 years ago
5

6. Find the distance between the two points: (3,-1) and (7.-5). Round to the nearest tenth. *

Mathematics
1 answer:
NikAS [45]3 years ago
7 0
Idk it might be 8.6 and 2.0
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(4+2i)+(-3-5) need help plz
ivolga24 [154]

4+2=6

-3-5 = -8

6 + -8  = -2.

8 0
2 years ago
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Please help me fast I still nee to finish 2 more ixl's by monday!!!!
sleet_krkn [62]

Answer:No

Step-by-step explanation:

Given system of equation is:

2x + 3 = y

2x + y = 15

To check whether (2,7) is solution to this system or not, we will put x=2 and y=7 in both equations.

Putting x=2 and y=7 in Eqn 1

2(2) + 3 = 7

4 + 3 = 7

7 = 7

Thus the ordered pair satisfies the equation

Putting x=2 and y=7 in Eqn 2

2(2) + 7 = 15

4 + 7 = 15

11 ≠ 15

The ordered pair do not satisfy the second equation.

Hence,

(2,7) is not a solution to the given system of equations.

3 0
2 years ago
You have a new pool and want to know it’s volume. The pool is 5 feet deep and has a radius of 7 feet. About how much water can t
kow [346]

Answer:

it has 48 gallons of water in the pool.

Step-by-step explanation:

5 0
3 years ago
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What is the slope of a line perpendicular to the line whose equation is x - y = 6. Fully simplify your answer.​
Marat540 [252]

Answer:

-1

Step-by-step explanation:

first put the equation in slop int form:

y= x-6

to find the slope of a line that is perpendicular take the negative reciprical of the slope

so the answer is

-x

3 0
2 years ago
What is the equation for a circle with a center at (-2,-4) that passes through the point (3,8)?
Margarita [4]
Check the picture below.

so, the center of the circle is the midpoint of that diametrical segment, and half that length is the radius.

\bf ~~~~~~~~~~~~\textit{middle point of 2 points }
\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&&(~ -2 &,& -4~) 
%  (c,d)
&&(~ 3 &,& 8~)
\end{array}\qquad
%   coordinates of midpoint 
\left(\cfrac{ x_2 +  x_1}{2}\quad ,\quad \cfrac{ y_2 +  y_1}{2} \right)
\\\\\\
\left( \cfrac{3-2}{2}~~,~~\cfrac{8-4}{2} \right)\implies \left( \frac{1}{2}~,~2 \right)\impliedby center

\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&&(~ -2 &,& -4~) 
%  (c,d)
&&(~ 3 &,& 8~)
\end{array}~~~ 
%  distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
d=\sqrt{[3-(-2)]^2+[8-(-4)]^2}\implies d=\sqrt{(3+2)^2+(8+4)^2}
\\\\\\
d=\sqrt{25+144}\implies d=\sqrt{169}\implies d=13\qquad\qquad \qquad  \stackrel{radius}{\frac{13}{2}}

\bf \textit{equation of a circle}\\\\ 
(x- h)^2+(y- k)^2= r^2
\qquad 
center~~(\stackrel{\frac{1}{2}}{ h},\stackrel{2}{ k})\qquad \qquad 
radius=\stackrel{\frac{13}{2}}{ r}
\\\\\\
\left( x-\frac{1}{2} \right)^2+(y-2)^2=\left( \frac{13}{2} \right)^2\implies \left( x-\frac{1}{2} \right)^2+(y-2)^2=\frac{169}{4}

3 0
3 years ago
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