Well there’s 180° in one triangle. a right angle is 90°. 2x+3x+90=180. 5x=90 divide by five, x=18°
Answer:
C
Step-by-step explanation:
(2x + 3)^5 = C(5,0)2x^5*3^0 +
C(5,1)2x^4*3^1 + C(5,2)2x^3*3^2 + C(5,3)2x^2*3^3 + C(5,4)2x^1*3^4 + C(5,5)2x^0*3^5
Recall that
C(n,r) = n! / (n-r)! r!
C(5,0) = 1
C(5,1) = 5
C(5,2) = 10
C(5,3) = 10
C(5,4) = 5
C(5,5) = 1
= 1(2x^5)1 + 5(2x^4)3 + 10(2x^3)3^2 + 10(2x^2)3^3 + 5(2x^1)3^4 + 1(2x^0)3^5
= 2x^5 + 15(2x^4) + 90(2x^3) + 270(2x^2) + 405(2x) +243
= 32x^5 + 15(16x^4) + 90(8x^3) + 270(4x^2) + 810x + 243
= 32x^5 + 240x^4 + 720x^3 + 1080x^2 + 810x + 243
M2=5x+10
.................
Answer:
The cost of a school banquet is $75+30n, where n is the number of people attending.
If 53 people are attending to visit school banquet, then you have to find the value of given expression at n=53.
At n=53, the value is $(75+30·53)=$1,665.
Step-by-step explanation:
(a) From the histogram, you can see that there are 2 students with scores between 50 and 60; 3 between 60 and 70; 7 between 70 and 80; 9 between 80 and 90; and 1 between 90 and 100. So there are a total of 2 + 3 + 7 + 9 + 1 = 22 students.
(b) This is entirely up to whoever constructed the histogram to begin with... It's ambiguous as to which of the groups contains students with a score of exactly 60 - are they placed in the 50-60 group, or in the 60-70 group?
On the other hand, if a student gets a score of 100, then they would certainly be put in the 90-100 group. So for the sake of consistency, you should probably assume that the groups are assigned as follows:
50 ≤ score ≤ 60 ==> 50-60
60 < score ≤ 70 ==> 60-70
70 < score ≤ 80 ==> 70-80
80 < score ≤ 90 ==> 80-90
90 < score ≤ 100 ==> 90-100
Then a student who scored a 60 should be added to the 50-60 category.
