Simplify 9/2 - 3√7 +8 - 28 0.11√2-5√7 11√4-5√14 6 5 6 9

How many integers are there that consist of four digits (from 1000 to 9999) where the first digit is a 5 or an 8 and the third d

Ostrovityanka [42]

Answer:

1600 integers

Step-by-step explanation:

Since we have a four digit number, there are four digit placements.

For the first digit, since there can either be a 5 or an 8, we have the arrangement as ²P₁ = 2 ways.

For the second digit, we have ten numbers to choose from, so we have ¹⁰P₁ = 10.

For the third digit, since it neither be a 5 or an 8, we have two less digit from the total of ten digits which is 10 - 2 = 8. So, the number of ways of arranging that is ⁸P₁ = 8.

For the last digit, we have ten numbers to choose from, so we have ¹⁰P₁ = 10.

So, the number of integers that can be formed are 2 × 10 × 8 × 10 = 20 × 80 = 1600 integers

5 0

3 years ago

Find the area of the given quadrilateral. Pls :)

Tasya [4]

I think your teacher may have made a typo. Refer to the diagram below. If we ignore segment ZA for now, we can see a rectangle forms with angle W = angle Y = 90. Also, we can see that WX = 6 is the horizontal base and WZ = 8 is the vertical height.

However, the side opposite WZ (which is XY) should also be 8 cm tall, and not 6 cm. If XY was 6 cm, then point Y would move to where point A is located (and the quadrilateral would be WXAZ instead of WXYZ). This new quadrilateral would not have a 90 degree angle at point Y's new location.

In other words, if we want angle Y to be 90 degrees, then side XY must be 8 units long. I have a feeling this is where the typo is, but your teacher might have been intending to say something else.