X=40.5
Here pls give brainliest
Answer:
Step-by-step explanation:
If we take out the extra $3, we can group the bills into one each of $5 and $1, for a value of $6. There will be 7 such groups in the remaining $42.
That means there are 7 bills of the $5 denomination, and 3 more than that (10 bills) of the $1 denomination.
There are 7 $5 bills and 10 $1 bills.
_____
If you want to write an equation, it is usually best to let a variable stand for the most-valuable contributor. Here, we can let x represent then number of $5 bills. Then the value of the cash box is ...
5x +(x+3) = 45
6x = 42 . . . . . . . . subtract 3, collect terms
x = 7 . . . . . . . . . . . there are 7 $5 bills
x+3 = 10 . . . . . . . . there are 10 $1 bills
You may notice that this working parallels the verbal description above. (After we subtract $3, x is the number of $6 groups.)
Answer:
Let p(x) = x3 + ax2 + bx +6
(x-2) is a factor of the polynomial x3 + ax2 + b x +6
p(2) = 0
p(2) = 23 + a.22 + b.2 +6 =8+4a+2b+6 =14+ 4a+ 2b = 0
7 +2 a +b = 0
b = - 7 -2a -(i)
x3 + ax2 + bx +6 when divided by (x-3) leaves remainder 3.
p(3) = 3
p(3) = 33 + a.32 + b.3 +6= 27+9a +3b +6 =33+9a+3b = 3
11+3a +b =1 => 3a+b =-10 => b= -10-3a -(ii)
Equating the value of b from (ii) and (i) , we have
(- 7 -2a) = (-10 - 3a)
a = -3
Substituting a = -3 in (i), we get
b = - 7 -2(-3) = -7 + 6 = -1
Thus the values of a and b are -3 and -1 respectively.
Step-by-step explanation:
Answer:
-1
Step-by-step explanation:
-7/10+7/15+1/-20+-9/10+11/15+11/-20
= -7/10 + -9/10 + 7/15 + 11/15 + 1/-20 + 11/-20
= -16/10 + 18/15 + 12/-20
= -8/5 + 6/5 + -3/5
= (-8+6-3)/5
= -5/5
= -1
Answer: if Gregory draws the segment with endpoints A and A’, then the midpoint will lie on the line of reflection.
Explanation:
Given that a triangle ABC is reflected in triangle A'B'C'
Here reflection is done on a line
If you imagine the line as a mirror then ABC will have image on the mirror line as A'B'C'
Recall that in a mirror the object and image would be equidistant from the mirror and also the line joining the image and object would be perpendicular to the mirror
But note that corresponding images will only be perpendicular bisector to the line
So A and A' only will be corresponding so AA' will have mid point on line
Option 1 is right
