The number of members that prefers geometry are 54members
Given the total number of people in the math club = 120 members
Let algebra + arithmetic + geometry = 100%
20 + 35 + geometry = 100
55 + geometry = 100
geometry = 100 - 55
geometry = 45%
This shows that the percentage of the member of the club that prefers geometry is 45%
Number of member that prefer geometry = 45% * 120
Number of members that prefer geometry 0.45 * 120
Number of members that prefers geometry = 54 members
Hence the number of members that prefers geometry are 54members
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In a number line, the distance between the points is the absolute value of their difference. For this given, the distance between points in negative 9 and 13 is the absolute value of -9 minus 13. Thus, the answer is letter A.
The <em><u>correct answer</u></em> is:
If she wants the same number of pieces of each kind of fruit in each bowl (same number of melons, same number of pears, and same number of apples in each bowl), then she can put 11 pieces in each of 4 bowls.
Explanation:
To answer this, we find the greatest common factor (GCF) of all 3 numbers. To do this, we find the prime factorization of 8, 12 and 24:
8 = 4(2)
4 = 2(2)
→8 = 2(2)(2)
12 = 4(3)
4 = 2(2)
→12 = 2(2)(3)
24 = 4(6)
4 = 2(2)
6 = 2(3)
→24 = 2(2)(2)(3)
The GCF is made of all of the common factors. The factors common to all 3 numbers are 2 and 2; 2(2) = 4 for the GCF.
This means we can use 4 bowls.
She has a total of 8+12+24 = 44 pieces of fruit; 44/4 = 11. She would have 11 pieces of fruit in each bowl.
Answer:
Plan A is a salary of $360 per month, plus a commission of 8% of sales. Plan B is a salary of $740 per month, plus a commission of 3% of sales.
Step-by-step explanation:
Answer:
We should attempt synthetic division to find when
x
3
−
k
x
2
+
2
k
x
−
15
x
−
3
has a remainder of
0
, which would signify that it is a factor of the polynomial.
The synthetic substitution would be set up as:
3
−
∣
∣
∣
1
−
k
2
k
−
15
−−−
−−−
−−−
−−−
−−−
|
0
Treating the synthetic division like any other synthetic division problem, we see that
3
−
∣
∣
∣
1
−
k
2
k
−
15
−−−
−−−
3
−−−−−−−−
−
3
k
+
9
−−−−−−−−
−
3
k
+
27
−−−−−−−−−
1
−
k
+
3
−
k
+
9
|
0
If the remainder equals
0
, then we know that
−
3
k
+
27
+
(
−
15
)
=
0
Solve this to see that
k
=
4
.
Thus,
(
x
−
3
)
is a factor of
x
3
−
4
x
2
+
8
x
−
15
.
