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2 years ago

Please help asap I need this for a test and I’m failing

Mathematics

2 answers:

Nady [450]2 years ago

5 0

Answer:

perpendicular bisector

RideAnS [48]2 years ago

3 0

Answer:

perpendicular bisector

Step-by-step explanation:

sorry if im wrong

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Help with this please

Alik [6]

Answer:

Yes, the given parallelogram is a rectangle.

Step-by-step explanation:

The vertices of parallelogram are J(-5,0), K(1,4), L(3,1) and M(-3,-3).

The slope formula is

$m=\frac{y_2-y_1}{x_2-x_1}$

$JK=\frac{4-0}{1-(-5)}=\frac{4}{6}=\frac{2}{3}$

$KL=\frac{1-4}{3-1}=\frac{-3}{2}$

$LM=\frac{-3-1}{-3-3}=\frac{-4}{-6}=\frac{2}{3}$

$JM=\frac{-3-0}{-3-(-5)}=\frac{-3}{2}$

The slopes of opposites sides are same it means they are parallel to each other.

The product of slopes of two consecutive sides is

$\frac{2}{3}\times \frac{-3}{2}=-1$

Since the product of slopes of two consecutive sides is -1, therefore the consecutive sides are perpendicular to each other.

Yes, the given parallelogram is a rectangle.

7 0

3 years ago

Points P and Q belong to segment AB . If AB = a, AP = 2PQ = 2QB, find the distance: midpoints between AP QB

Digiron [165]

Answer: The distace between midpoints of AP and QB is $\frac{a}{8}$ .

Step-by-step explanation: Points P and Q are between points A and B and the segment AB measures a, then:

AP + PQ + QB = a

According to the question, AP = 2 PQ = 2QB, so:

PQ = $\frac{AP}{2}$

QB = $\frac{AP}{2}$

Substituing:

AP + 2*( $\frac{AP}{2}$ ) = a

2AP = a

AP = $\frac{a}{2}$

Since the distance is between midpoints of AP and QB:

2QB = AP

QB = $\frac{AP}{2}$

QB = $\frac{a}{2}*\frac{1}{2}$

QB = $\frac{a}{4}$

MIdpoint is the point that divides the segment in half, so:

<u>Midpoint of AP</u>:

$\frac{AP}{2} = \frac{a}{2}*\frac{1}{2}$

$\frac{AP}{2} = \frac{a}{4}$

<u>Midpoint of QB</u>:

$\frac{QB}{2} = \frac{a}{4}*\frac{1}{2}$

$\frac{QB}{2} = \frac{a}{8}$

The distance is:

d = $\frac{a}{4} - \frac{a}{8}$

d = $\frac{a}{8}$

4 0

2 years ago

The probability all of the students in this class will be present and on time is 3/10. The probability all of the students are p

cestrela7 [59]

Answer:

3/4

Step-by-step explanation:

A: students present
B: students on time

P(all students present and on time) = P(A and B) = 3/10
P(all students present) = P(A) = 2/5

P(A and B) = P(A).P(B|A) where P(B|A) is the probability of everyone being on time given that everyone is present

So P(B|A) = P(A and B) /P(A) = 3/10 ÷ 2/5 = 3/10 * 5/2 = 15/20 which can be reduced to 3/4 by dividing numerator and denominator by 5

7 0

1 year ago

Solve the proportional equation below 3/4 =2/v-2

Vinil7 [7]

For this case we have the following equation:

$\frac {3} {4} = \frac {2} {v-2}$

Resolving we have:

$3 (v-2) = 2 * 4\\3 (v-2) = 8$

We apply distributive property to the terms of parentheses:

$3v-6 = 8$

Adding 6 to both sides of the equation:

$3v = 8 + 6\\3v = 14$

Dividing between 3 on both sides of the equation:

$v = \frac {14} {3}$

ANswer:

$v = \frac {14} {3}$

3 0

2 years ago

Read 2 more answers

The quadrilateral ABCD has an area of 58 in2 and diagonal AC = 14.5 in. Find the length of diagonal BD if AC ⊥ BD.

diamong [38]

So the quadrilateral is a rhombus. RSM Is hard and etc...
Anyways if the area = $58^{2}$ and we know that one diagonal is 14.5 inches and we know that the area of a rhombus is
$\frac{D^{1} * D^{2}}{2} =$
then we can do inverse operation and do
$\frac{54}{14.5} = 4$
4 * 2 because of the inverse operation has to be done to get the right answer and then you get the length of diagonal 2 is 8 in

4 0

2 years ago