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lina2011 [118]
4 years ago
13

M Part A: m Part B: M

Mathematics
1 answer:
tiny-mole [99]4 years ago
5 0
M part a:a part:b answer
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Which equation can be used to find the answer?
o-na [289]
Well if we look at these equations we can see that D is the only appropriate answer considering 23 + 112 = 135 bracelets.
Yes the equation works as well
b - 23 = 112
we add 23 to both sides
b - 23 + 23 = 112 + 23
b = 135
8 0
4 years ago
Will mark brainliest:) helppppp -7x + y = -19 -2x+ 3y = -19 solve the system of equations
Luba_88 [7]

Answer:

X=2, Y=-15

Step-by-step explanation:

-7x+y=-19

-2x+3y=-19

------------------

-3(-7x+y)=-19

   -2x+3y=-19

---------------------

21x-3y=57

-2x+3y=-19

---------------------

19x     = 38

19x/19=38/19

x=2

now plug 2 into x of the second equation to get y=-15

7 0
4 years ago
The octagon is a regular polygon. What is the measure of each interior angle in the polygolon
MrRissso [65]
(N-2)*180/n
Octagon has 8 sides
(8-2)*180/8
=135
5 0
3 years ago
Marcus has x songs downloaded on his phone. He downloads 20 more songs to his phone.
GREYUIT [131]
I believe the answer is x+20. 
6 0
3 years ago
A point K is on the perpendicular bisector of a segment with endpoints at H and J. What must be true about point K? It is equidi
kykrilka [37]

Answer:

K is equidistant from H and J.

Step-by-step explanation:

Given that the point K which is on the perpendicular bisector of the line segment having endpoints at H and J.

The given situation can be represented as the diagram as attached in the answer area.

Referring to the \triangle HOK, \triangle JOK:

\angle HOK = \angle JOK=90^\circ (As it is the perpendicular bisector)

OH = OJ (As it is the perpendicular bisector)

Also, the side OK is the common side.

Therefore by S-A-S congruence, \triangle HKO\cong \triangle JKO

As per the properties of congruent triangles:

Side HK = Side JK

HK and JK are nothing but the distance of the point K from the end points H and J which are proved to be equal to each other.

Therefore, we can conclude that:

K is equidistant from H and J.

5 0
3 years ago
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