Answer:
20 dimes and 2 quarters
Step-by-step explanation:
x = number of dimes
y = number of quarters
minimum is greater than or equal to ≥
x+y ≥16
no more than is less than or equal to ≤ ,
value of dime is $ 0.10
value of quarter is $0.25 so
0.10x + 0.25y ≤ 3.00
Rewrite inequalities in terms of y:
y ≥16 - x
(0.25/0.25)y ≤ (3.00/0.25)-(0.10/0.25)x→
y≤12 - 0.4x
Graph system: see file
Answer:
As you must know, If one root of the polynomial 7-√5, the other will be 7+√5 i.e irrational root occur in pairs.
A polynomial function cannot have single unreal i.e irrational root. It always occur in pairs.
So , consider a polynomial function of any degree, if it has a root 7-√5, then it must have another root as 7+√5.
A polynomial can't have 7-√5 as a single root.
A quotient of a number 21 is 3 and 63
Hope this helps!
F(x) = 2^x; h(x) = x^3 + x + 8
Table
x f(x) = 2^x h(x) = x^3 + x + 8
0 2^0 = 1 0 + 0 + 8 = 8
1 2^1 = 2 1^3 + 1 + 8 = 10
2 2^2 = 4 2^3 + 2 + 8 = 8 + 2 + 8 = 18
3 2^3 = 8 3^3 + 3 + 8 = 27 + 3 + 8 = 38
4 2^2 = 16 4^3 + 4 + 8 = 76
10 2^10 = 1024 10^3 +10 + 8 = 1018
9 2^9 = 512 9^3 + 9 + 8 = 729 + 9 + 8 = 746
Answer: an approximate value of 10
Answer:
3×5×53
Step-by-step explanation:
You can use divisibility rules to find the small prime factors.
The number ends in 5, so is divisible by 5.
795/5 = 159
The sum of digits is 1+5+9 = 15; 1+5 = 6, a number divisible by 3, so 3 is a factor.
159/3 = 53 . . . . . a prime number,* so we're done.
795 = 3×5×53
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* If this were not prime, it would be divisible by a prime less than its square root. √53 ≈ 7.3. We know it is not divisible by 2, 3, or 5. We also know the closest multiples of 7 are 49 and 56, so it is not divisible by 7. Hence 53 is prime.
