Consider the equation

alina1380 [7]

A perpendicular bisector
this means the length IJ=JK and the angles IJB, BJK are 90° (and their congruent angles too)

but this does not mean AB=IK, because their lengths can be arbitrary
for a similar reason AJ=IJ is wrong, essentially the same statement but with half lengths
BI=BK is true, because it is a bisector of IK this means the sides IJ=JK and JB is a shared side, therefore also having the same length. We also know IJB and BJK are congruent angles, so all in all these are congruent triangle and therefore also share the side length
BI=AK is false, because we don't know where the intersection point j is on AB, it could split the length into any arbitrary ratio, in case these are two equal sides then it would be true, but it doesn't hold for all other possibilities

so BI=BK is the solution

3 0

3 years ago

One of the tables shows a proportional relationship.

beks73 [17]

Answer:

See attached

Step-by-step explanation:

The graph of a proportional linear relationship is a line that passes through the origin (0, 0).

From inspection of the given tables, the linear equations for each table of points is:

Table 1: y = x + 1
Table 2: y = x/2
Table 3: y = x + 2
Table 4: y = 2x + 1

The only equation for which y = 0 when x = 0 is y = x/2 → Table 2.

Given points from Table 2:

(2, 1)
(4, 2)
(6, 3)
(8, 4)

To graph the line, plot the given points and draw a line through them (see attached).

5 0

2 years ago

Math nation section 10 test yourself

Aleksandr [31]

Huhnsnmdmfqkwkwnnzfnmfkw

8 0

3 years ago

What is the y-intercept of this quadratic function f(x) = x^2 + 2x + 3 A. (0,1) B. (0,-1) C. (0,-3) D. (0,3)

Firdavs [7]

Answer:

d=(0,3)

Step-by-step explanation:

8 0

3 years ago

student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te

alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = $\frac{1}{4}$ .

Thus, the probability that the student does not receive version A = $1-\frac{1}{4}$ = $\frac{3}{4}$ .

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

$\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}$