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TiliK225 [7]
3 years ago
6

What is -3 2/3 times (-2 1/2)? A. -8 1/4 B. -6 1/6 C. 8 1/4 D. 8 3/4

Mathematics
2 answers:
uranmaximum [27]3 years ago
7 0
None of the answers you have are correct. The right answer is 9 1/6. Sorry, I hope I was able to help.
Ivenika [448]3 years ago
3 0
The answer would be c I think
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1/2 x (48 divided by 6) - 2 + 5
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Step-by-step explanation:

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2 years ago
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Write a coordinate proof to prove that the segment that joins the vertex angle of an isosceles triangle to the midpoint of its b
Natalija [7]

Answer:

Slope of the base segment line is zero, hence base segment is horizontal

Slope of the segment that joins the vertex angle to the midpoint of its base is undefined hence the line is vertical

Therefore, angle between base segment line and the segment line from the vertex angle to the midpoint is perpendicular

Therefore, the segment that joins the vertex angle to an isosceles triangle to the midpoint of its base is perpendicular to the base

Step-by-step explanation:

Here we prove the required relation as follows;

Let the isosceles be ABC

The coordinates of the points are

C = (0, 0) (Vertex)

A =  (-a, b)

C = (a, b)

P = (0, b) (Midpoint of base)

Therefore, the gradient or slope of the base AC is presented as follows;

Slope, m \, of \. base, \, AC= \frac{dy}{dx}  = \frac{b - b}{a - (-a)}  = \frac{0}{2\cdot a} = 0

Hence, segment AC is vertical

Slope, m \, of \, segment \ CP \,  joining \, vertex, \, to \, midpoint, P, on  \,AC = \frac{0 - b}{0 - 0}  = \frac{-b}{0} = Undifined.

Hence, segment CP is vertical

Therefore, the segment that joins the vertex angle to an isosceles triangle to the midpoint of its base is perpendicular to the base.

6 0
3 years ago
YS is the perpendicular bisector of XYZ and YS is a shared side of XYS and ZYS. which of the following must be congrent in order
german

Answer:

The triangles SYX and SYZ are congrient so by CPCTC XY\cong YZ, XS\cong ZS and \angle X\cong \angle Z.

Step-by-step explanation:

Given information: YS is the perpendicular bisector of XYZ.

In triangle YXS and triangle YSZ,

\angle SYX=\angle SYZ                 (YS is the perpendicular bisector of XYZ)

YS=YS                                            (Common side, Reflexive property)

\angle YSX=\angle YSZ=90^{\circ}   (YS is the perpendicular bisector of XYZ)

By ASA postulate,

\triangle SYX\cong \angle SYZ

The corresponding parts of congruent triangles are congruent. so,

XY\cong YZ

XS\cong ZS

\angle X\cong \angle Z

6 0
3 years ago
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