Answer:
870 as it is the only number in the list which closest and lesser then 956.
(if you divide, then you get 10 as quotient and 86 as remainder)
we can also take 87 but the question mentions with multiples of 10.
Step-by-step explanation:
Answer:
4,68556
Step-by-step explanation:
√5+√6
<u>=</u><u> </u><u>4,68556</u>
Question:
Five times the sum of a number and 27 is greater than or equal to six times the sum of that number and 26.
What is the solution set of this problem?
Answer:

Step-by-step explanation:
Given
<em>Represent the number with x</em>
So:

Required
Determine the solution set

Open Both Brackets


Collect Like Terms


Multiply both sides by -1

Hence, the solution set is 
(2-7i)(9+5i)
18+10i-81i-35i^2
The answer is 35i^2 - 71i +18
Answer:
y = 2x² - 5x + 7
Step-by-step explanation:
General form of a quadratic is:
y = Ax² + Bx + C
Using the given points, construct equations and solve for A,B and C.
When x = 0, y = 7
7 = A(0)² + B(0) + C
C = 7
y = Ax² + Bx + 7
When x = -1, y = 14
14 = A(-1)² + B(-1) + 7
A - B = 7
A = 7 + B
When x = 1, y = 4
4 = A(1)² + B(1) + 7
A + B = -3
(7 + B) + B = -3
2B = -10
B = -5
A = 7 + (-5)
A = 2
y = 2x² - 5x + 7