All Answers 358087 Answers

Let's represent the number as �x. The problem can be expressed as an equation:

5��+7=0x+75x​=0

For this equation to equal zero, the numerator (5�5x) must be zero. Therefore, the original number is �=0x=0.

The total students in either club (or both) equals the sum of math and science club members minus those in both clubs. Total = Math + Science - Both.

Draw a line from the center to the point of tangency.  

This is one leg of a right triangle, its length is 8.  

Draw a line from the center to one end of the chord.  

This is that triangle's hypotenuse, its length is 17.  

By Pythagoras' Theorem the other leg of the triangle, 

aka half the chord, is the square root of (17² – 8²).  

Half the chord is sqrt(225) which is 15.  

Therefore the whole chord is 30.  

_.

You have 5 different mathematics textbooks and 4 different psychology textbooks that you want to place on a bookshelf. However, you want to keep all the psychology textbooks together, and all the mathematics textbooks together. This means that you have two groups of textbooks that you need to place on the shelf in a specific order. 

To solve this problem, first, you need to determine the number of ways you can arrange the mathematics textbooks and the number of ways you can arrange the psychology textbooks. Since you have 5 different mathematics textbooks, you can arrange them in 5! or 120 different ways. Similarly, you have 4 different psychology textbooks, and you can arrange them in 4! or 24 different ways.

Once you have arranged the mathematics and psychology textbooks separately, you need to place them together in a specific order. Since there are only two ways you can group the textbooks (all mathematics textbooks followed by all psychology textbooks or all psychology textbooks followed by all mathematics textbooks), you have only two possible arrangements. 

Therefore, the total number of ways you can place all 9 textbooks on the bookshelf, in a row, if all the psychology textbooks must be together, and all the mathematics textbooks must be together, is:

2 × 5! × 4! = 2 × 120 × 24 = 5,760

So, there are 5,760 ways to place all 9 textbooks on the bookshelf under the given conditions.

The total number of ways that the numbers can be chosen is 40c4. Exactly 10 of these possibilities result in the four slips having the same number.

Now we need to determine the number of ways that two slips can have a number  a and the other two slips have a number b , with \($b\ne a$\) . There are  10c2 ways to choose the distinct numbers a and b . For each value of a , there are 10c2 ways to choose the two slips with a  and  for each value of b , there are 4C2 ways to choose the two slips with b . Hence the number of ways that two slips have some number  and the other two slips have some distinct numberb  is

(10C2)(4c2)(4c2)=45*6*6=1620

So the probabilities  p and q  are 10/40c4 and 1620/40c4 , respectively, which implies that

q/p=1620/10=162

16 -5 = 11

Here is one way:

log1/2 (11)  means   1/2^x  = 11    LOG both sides

                                 x LOG (1/2) = LOG 11

                                  x = - 3.4595 

Depending on your calculator you may be able to calculate it directly     log_1/2 (11) =   -3.4595    

where are the 2 sides? I can't see them. If they are x and y, then it is all the integers from (the smallest integer larger than x-y), all the way to (the largest integer smaller than x+y)

b: (ax+b)^6 has a x^4 coefficient of 15*a^4*b^2, so in this case a = 2, b = -5, so 15*16*25 = 3*8*2*125 = 6000

so, basically, say x is the amount of days 1 person can do the task in. then x/5-x/6=8, meaning x=240, and 5 workers take 48 days, so if you want to take 20 days, you have to have 12 workers. 12-5 = 7, so you need 7 extra workers!

simplify to x^2+10x-10 = 0

plug it in the quadratic formula!

$8ab - 12a - 12b + 18$.

A particle's position in the plane t seconds after it starts moving is

x = 6 cos kt

y = 6 sin kt

for some constant k.  If the particle moves a distance of 120 pi for every second, then what is k?

The distance traveled by the particle in t seconds is given by the perimeter of the ellipse it traces out. The semi-major and semi-minor axes of the ellipse are 6 and 6, respectively, so the distance traveled is 2π⋅6⋅6=72π.

Since the distance traveled is 120 pi for every second, we have k⋅72π=120π. Solving for k, we find k=35​​.