Ray BD is the angle bisector of Given that D<br> Om m-ABD=42<br> m-ABD=21<br> Om ABD=96"

It costs Danny $5 to park his car at a parking meter for 2 hours. What is the

Andre45 [30]

Answer:

$2.50

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Will give correct answer brainliest

Elodia [21]

Answer:

2. Division property of equality

3. x+1 = -4

4. Subtraction property of equality

5. x = -5

Step-by-step explanation:

8 0

2 years ago

Three vertices of a parallelogram are shown in the figure below. Give the coordinates of the fourth vertex.

topjm [15]

Given:

The three vertices of a parallelogram are (-3,8), (4,5), (2,-5).

To find:

The fourth vertex of the parallelogram.

Solution:

Let the vertices of the parallelogram are A(-3,8), B(4,5), C(2,-5) and D(a,b).

We know that, diagonals of a parallelogram bisect each other. It means midpoints of both diagonals are same.

Midpoint formula:

$Midpoint=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$

Two diagonals of ABCD are AC and BD.

Midpoint of AC = Midpoint of BD

$\left(\dfrac{-3+2}{2},\dfrac{8-5}{2}\right)=\left(\dfrac{4+a}{2},\dfrac{5+b}{2}\right)$

$\left(\dfrac{-1}{2},\dfrac{3}{2}\right)=\left(\dfrac{4+a}{2},\dfrac{5+b}{2}\right)$

On comparing both sides, we get

$\dfrac{4+a}{2}=\dfrac{-1}{2}$

$4+a=-1$

$a=-1-4$

$a=-5$

And,

$\dfrac{5+b}{2}=\dfrac{3}{2}$

$5+b=3$

$b=3-5$

$b=-2$

Therefore, the coordinates of fourth vertex are (-5,-2).

3 0

3 years ago

The lateral surface area of cone A is equal to the lateral surface area of cylinder B.

Natasha2012 [34]

Answer:

TRUE

Step-by-step explanation:

Lateral area of cone is given by: πrl

where r is the radius and l is the slant height

Here r=r and l=2h

Hence, lateral area of cone A= π×r×2h

= 2πrh

Lateral area of cylinder is given by: 2πrh

where r is the radius and h is the height

Lateral area of cylinder B=2πrh

Clearly, both the lateral areas are equal

Hence, the statement that:The lateral surface area of cone A is equal to the lateral surface area of cylinder B. is:

True

4 0

3 years ago

A bag contains 6 red apples and 5 yellow apples. 3 apples are selected at random. Find the probability of selecting 1 red apple

tester [92]

Answer: The required probability of selecting 1 red apple and 2 yellow apples is 36.36%.

Step-by-step explanation: We are given that a bag contains 6 red apples and 5 yellow apples out of which 3 apples are selected at random.

We are to find the probability of selecting 1 red apple and 2 yellow apples.

Let S denote the sample space for selecting 3 apples from the bag and let A denote the event of selecting 1 red apple and 2 yellow apples.

Then, we have

$n(S)=^{6+5}C_3=^{11}C_3=\dfrac{11!}{3!(11-3)!}=\dfrac{11\times10\times9\times8!}{3\times2\times1\times8!}=165,\\\\\\n(A)\\\\\\=^6C_1\times^5C_2\\\\\\=\dfrac{6!}{1!(6-1)!}\times\dfrac{5!}{2!(5-2)!}\\\\\\=\dfrac{6\times5!}{1\times5!}\times\dfrac{5\times4\times3!}{2\times1\times3!}\\\\\\=6\times5\times2\\\\=60.$

Therefore, the probability of event A is given by

$P(A)=\dfrac{n(A)}{n(S)}=\dfrac{60}{165}=\dfrac{4}{11}\times100\%=36.36\%.$

Thus, the required probability of selecting 1 red apple and 2 yellow apples is 36.36%.

4 0

3 years ago

Ray BD is the angle bisector of Given that D Om m-ABD=42 m-ABD=21 Om ABD=96"