In order to graph the relationship, we will need to write the expression as the equation of a straight line as shown:
d = mt + b
d is the distance covered
t is the time taken
m is the speed
If you are skateboarding at a pace of 30 meters every 5 seconds. your friend is in-line skating at a pace of 9 meters every 2 seconds, this can be written as (5, 30) and (2, 9)
Get the slope of the line:
m = (9-30)/(2-5)
m = -21/-3
m = 7
Substitute m = 7 and the coordinate (2, 9) into the equation y = mt + b
9 = 7(2) + b
9 = 14 + b
b = -5
The required equation to plot will be expressed as y = 7t - 5
Plot the required graph
Learn more here: brainly.com/question/17003809
18= (d - 0.25d) - 0.25d
18= 0.75d - 0.25d
18= 0.50d
d = 36
the original price was $36
<h3>
Answer: 2x(x^2-2)(x+1)</h3>
=========================================================
Explanation:
First factor out the GCF 2x
2x^4+2x^3-4x^2-4x
2x*x^3+2x*x^2-2x*2x-2x*2
2x(x^3 + x^2 - 2x - 2)
Then let's factor the expression inside the parenthesis using the factor by grouping method
x^3 + x^2 - 2x - 2
(x^3 + x^2) + (- 2x - 2)
x^2(x + 1) - 2(x + 1)
(x^2 - 2)(x+1)
We see that x^3 + x^2 - 2x - 2 factors to (x^2-2)(x+1)
Overall, the original expression fully factors to 2x(x^2-2)(x+1)
length = 2x
width = x^2-2
height = x+1
The order of length, width, and height doesn't matter.
Answer:
B
Step-by-step explanation:
Given that r is inversely proportional to s then the equation relating them is
r = 
← k is the constant of proportion
To find k use the condition r = 16 when s = 3
k = rs = 16 × 3 = 48
r = 
← equation of variation → B
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.
