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4 years ago

Someone please answer at least one of the questions i beg you

Mathematics

1 answer:

bazaltina [42]4 years ago

7 0

Answer:

12: 21

13: A: 16/x^12

B: 9/x^8

Step-by-step explanation:

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Please helpppppppppppppp

Helen [10]

Answer:

90/23

Step-by-step explanation:

First, you want to start in the parentheses. 5-6=-1, then multiply it by -2. Since a negative multiplied by a negative is a positive, -2*-1=2. -3^2=-9. 2+8=10. Then -9 times 10= -90. Then, we move on to the bottom. 5--2=7. 4^2=16, and -2(2)=-4. So -5(7) + 16-4=-23. Now, the equation should look like this, -90/-23. Put the negatives together and you get 90/23.

6 0

3 years ago

Which quadratic function has the largest maximum over the set of REAL numbers?

koban [17]

The answer is number 1, all you have to do is graph the first 2 and look at the highest numbers in the table

7 0

4 years ago

Read 2 more answers

Rebekka had two lengths of string, 3 1/10 inches and 4 8/9 inches. About how many inches did she

kondor19780726 [428]

Answer:

Rebekka has about 10 inches of string. (The exact number is 7.988888889 inches of string)

Step-by-step explanation:

7 0

3 years ago

Help me on 23 please

ElenaW [278]

In this situation you would use cos
which is adjacent over hypotenuse
so I take the cos50 and have it equal to 17 / x
cos(50)= 17/x
which is the same as ...
17 divided by cos(50) =x
x=26.44730506
which rounds to 26.45
So your answer would be A)

I hope everything makes sense, enjoy! <span />

4 0

4 years ago

A store sells 8 colors of balloons with at least 28 of each color. How many different combinations of 28 balloons can be chosen?

Len [333]

Answer:

(a) $Selection = 6724520$

(b) $At\ most\ 12 = 6553976$

(c) $At\ most\ 8 = 6066720$

(d) $At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 5896638$

Step-by-step explanation:

Given

$Colors = 8$

$Balloons = 28$ --- at least

Solving (a): 28 combinations

From the question, we understand that; a combination of 28 is to be selected. Because the order is not important, we make use of combination.

Also, because repetition is allowed; different balloons of the same kind can be selected over and over again.

So:

$n => 28 + 8-1$ $= 35$

$r = 28$

$Selection = ^{35}^C_{28$

$Selection = \frac{35!}{(35 - 28)!28!}$

$Selection = \frac{35!}{7!28!}$

$Selection = \frac{35*34*33*32*31*30*29*28!}{7!28!}$

$Selection = \frac{35*34*33*32*31*30*29}{7!}$

$Selection = \frac{35*34*33*32*31*30*29}{7*6*5*4*3*2*1}$

$Selection = \frac{33891580800}{5040}$

$Selection = 6724520$

Solving (b): At most 12 red balloons

First, we calculate the ways of selecting at least 13 balloons

Out of the 28 balloons, there are 15 balloons remaining (i.e. 28 - 13)

So:

$n => 15 + 8 -1 = 22$

$r = 15$

Selection of at least 13 =

$At\ least\ 13 = ^{22}C_{15}$

$At\ least\ 13 = \frac{22!}{(22-15)!15!}$

$At\ least\ 13 = \frac{22!}{7!15!}$

$At\ least\ 13 = 170544$

Ways of selecting at most 12 =

$At\ most\ 12 = Total - At\ least\ 13$ --- Complement rule

$At\ most\ 12 = 6724520- 170544$

$At\ most\ 12 = 6553976$

Solving (c): At most 8 blue balloons

First, we calculate the ways of selecting at least 9 balloons

Out of the 28 balloons, there are 19 balloons remaining (i.e. 28 - 9)

So:

$n => 19+ 8 -1 = 26$

$r = 19$

Selection of at least 9 =

$At\ least\ 9 = ^{26}C_{19}$

$At\ least\ 9 = \frac{26!}{(26-19)!19!}$

$At\ least\ 9 = \frac{26!}{7!19!}$

$At\ least\ 9 = 657800$

Ways of selecting at most 8 =

$At\ most\ 8 = Total - At\ least\ 9$ --- Complement rule

$At\ most\ 8 = 6724520- 657800$

$At\ most\ 8 = 6066720$

Solving (d): 12 red and 8 blue balloons

First, we calculate the ways for selecting 13 red balloons and 9 blue balloons

Out of the 28 balloons, there are 6 balloons remaining (i.e. 28 - 13 - 9)

So:

$n =6+6-1 = 11$

$r = 6$

Selection =

$^{11}C_6 = \frac{11!}{(11-6)!6!}$

$^{11}C_6 = \frac{11!}{5!6!}$

$^{11}C_6 = 462$

Using inclusion/exclusion rule of two sets:

$Selection = At\ most\ 12 + At\ most\ 8 - (12\ red\ and\ 8\ blue)$

$Only\ 12\ red\ and\ only\ 8\ blue\ = 170544+ 657800- 462$

$Only\ 12\ red\ and\ only\ 8\ blue\ = 827882$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = Total - Only\ 12\ red\ and\ only\ 8\ blue$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 6724520 - 827882$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 5896638$

3 0

3 years ago