Answer:
a) possible progressions are 5
b) the smallest and largest possible values of the first term are 16 and 82
Step-by-step explanation:
<u>Sum of terms:</u>
- Sₙ = n/2(a₁ + aₙ) = n/2(2a₁ + (n-1)d)
- S₂₀ = 20/2(2a₁ + 19d) = 10(2a₁ + 19d)
- 2020 = 10(2a₁ + 19d)
- 202 = 2a₁ + 19d
<u>In order a₁ to be an integer, d must be even number, so d = 2k</u>
- 202 = 2a₁ + 38k
- 101 = a₁ + 19k
<u>Possible values of k= 1,2,3,4,5</u>
- k = 1 ⇒ a₁ = 101 - 19 = 82
- k = 2 ⇒ a₁ = 101 - 38 = 63
- k = 3 ⇒ a₁ = 101 - 57 = 44
- k = 4 ⇒ a₁ = 101 - 76 = 25
- k = 5 ⇒ a₁ = 101 - 95 = 16
<u>As per above, </u>
- a) possible progressions are 5
- b) the smallest and largest possible values of the first term are 16 and 82
Answer:
i think it's y = 3x + 10
Step-by-step explanation:
hope this helps =￣ω￣=
Answer:
Option A is correct
The function 
has real zeroes at x =-10 and x =-6
Explanation:
Given: The real zeroes or roots are x = -10, and x = -6
To find the quadratic function of degree 2.
where α,β are real roots. ....[1]
Here, α= -10 and β= -6
Sum of the roots:
α+β = -10+(-6) = -10-6 = -16
Product of the roots:
αβ = (-10)(-6)= 60
Substitute these value in equation [1] we have;
Therefore, the quadratic function for the real roots at x =-10 and x =-6 ;
62-32 = 30
30* 5/9 = 16.666 rounded to 17 degree celius
