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ioda
4 years ago
13

Help please I need it ASAP

Mathematics
1 answer:
andreyandreev [35.5K]4 years ago
3 0
Clearer picture please
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a line segment with one end at C(6,5) has midpoint m(4,2) determine the coordinates of the other endpoint d
meriva

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ C(\stackrel{x_1}{6}~,~\stackrel{y_1}{5})\qquad D(\stackrel{x_2}{x}~,~\stackrel{y_2}{y}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{x+6}{2}~~,~~\cfrac{y+5}{2} \right)~~=~~\stackrel{\stackrel{Midpoint}{M}}{(4~,~2)}\implies \begin{cases} \cfrac{x+6}{2}=4\\[1em] x+6=8\\ \boxed{x= 2}\\ \cline{1-1} \cfrac{y+5}{2}=2\\[1em] y+5=4\\ \boxed{y=-1} \end{cases}

5 0
4 years ago
20 POINTS FOR STEP BY STEP!!
kobusy [5.1K]
BAD=BCD(base angle ,isosceles triangle)
ABD=CBD(given)
BAD+BCD+ABC=180(angle sum of triangle)
36+36+ABC=180
ABC=108
ABC=ABD+CBD
CBD=108/2
=54
therefore
x=54
8 0
3 years ago
Read 2 more answers
Y=2x+3 using function equation
RoseWind [281]
Slope: 2
Y-intercept: -3
5 0
3 years ago
Given A-<br> 2 2<br> 1 3<br> what is A-¹?
devlian [24]

The inverse of matrix A = \left[\begin{array}{cc}2&2&1&3\\\end{array}\right] is A^{-1} = \frac{1}{4} \left|\begin{array}{cc}3&-2&-1&2\\\end{array}\right|      

The given matrix is A = \left[\begin{array}{cc}2&2&1&3\\\end{array}\right]

The inverse of a matrix is given by

A^{-1} = \frac{1}{|A|} adjA                   ...(1)

Now, determinant of matrix A is

|A| = \left|\begin{array}{cc}2&2&1&3\\\end{array}\right|

= (3)(2) - (2)(1)

= 6 - 2

= 4

Now, adjoint A will be

A_{11} = 3

A_{12} = -1

A_{21} = -2

A_{22} = 2

AdjA = \left|\begin{array}{cc}3&-2&-1&2\\\end{array}\right|

Now, substituting the values of |A| and AdjA in equation (1), we get

A^{-1} = \frac{1}{4} \left|\begin{array}{cc}3&-2&-1&2\\\end{array}\right|        

Therefore, A^{-1} = \frac{1}{4} \left|\begin{array}{cc}3&-2&-1&2\\\end{array}\right|      

Know more about Inverse of Matrix: -brainly.com/question/12442362

#SPJ9

     

7 0
2 years ago
What is the value of (tan 65° - tan 35°) / 1 + tan 65°tan 35°?<br> 0<br> 0 V3/3
goldenfox [79]

Answer:

\frac{ \sqrt{3} }{3}

Step-by-step explanation:

We know that tangent angle addition formula is

\tan(a   -  b)  =  \frac{ \tan(a)  -  \tan(b) }{1 +  \tan(a)  \tan(b) }

This is in form already. Based on the value, a is 65 and b is 35 so let just subtract that

\tan(65 - 35)  =  \tan(30)

W e know that tan 30 is

\tan(30)  =  \frac{ \sin(30) }{ \cos(30) }  =  \frac{ \frac{1}{2} }{ \frac{ \sqrt{3} }{2} }  =  \frac{1}{ \sqrt{3} }  =  \frac{ \sqrt{3} }{3}

6 0
2 years ago
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