G(x)-f(x)=10x+2- (-8x+5)
=10x+2+8x-5
=18x-3
Step-by-step explanation:
<em>1st year</em>:
(1200 x 3.5 x 1) ÷ 100 = $42
<em>2nd year:</em>
(1242 x 3.5 x 1) ÷ 100= $43.47
<em>3rd year:</em>
(1285.47 x 3.5 x 1) ÷ 100= $44.99≈ $45
<em>4th year:</em>
(1330.47 x 3.5 x 1) ÷ 100= $46.56
Compound interest:
$(42 + 43.47 + 45 + 46.56)
=<u>$ 177.03</u>
Answer:
r = 1/3
= 1
Step-by-step explanation:
1 + 1/3 + 1/9 + 1/27 + 1/ 81
In this series a is the first term = 1
r is the common ratio = 2nd term/1st term = 3rd term/ 2nd term
r = 1/3 ÷ 1 = 1/9 ÷ 1/3
r = 1/3
I hope this was helpful, please mark as brainliest
Answer:
see explanation
Step-by-step explanation:
∠2 and ∠3 are same- side interior angles and are supplementary.
Answer:
Roots are not real
Step-by-step explanation:
To prove : The roots of x^2 +(1-k)x+k-3=0x
2
+(1−k)x+k−3=0 are real for all real values of k ?
Solution :
The roots are real when discriminant is greater than equal to zero.
i.e. b^2-4ac\geq 0b
2
−4ac≥0
The quadratic equation x^2 +(1-k)x+k-3=0x
2
+(1−k)x+k−3=0
Here, a=1, b=1-k and c=k-3
Substitute the values,
We find the discriminant,
D=(1-k)^2-4(1)(k-3)D=(1−k)
2
−4(1)(k−3)
D=1+k^2-2k-4k+12D=1+k
2
−2k−4k+12
D=k^2-6k+13D=k
2
−6k+13
D=(k-(3+2i))(k+(3+2i))D=(k−(3+2i))(k+(3+2i))
For roots to be real, D ≥ 0
But the roots are imaginary therefore the roots of the given equation are not real for any value of k.
