Answer:
HJ = 52
JK = 79
Step-by-step explanation:
Hello, I want to assume you are trying to determine either HJ or JK. Below is an illustration of how to determine either HJ or JK.
From the question given above, the following data were obtained:
HJ = 5x – 3
JK = 8x – 9
KH = 131
If we draw a straight line, we'll observe the following:
H_______J_______K
KH = HJ + JK
Next, we shall determine the value of x.
HJ = 5x – 3
JK = 8x – 9
KH = 131
KH = HJ + JK
131 = (5x – 3) + (8x – 9)
131 = 5x – 3 + 8x – 9
Collect like terms
131 + 3 + 9 = 5x + 8x
143 = 13x
Divide both side by 13
x = 143 / 13
x = 11
Finally, we shall determine HJ and JK. This can be obtained as follow:
For HJ:
HJ = 5x – 3
x = 11
HJ = 5(11) – 3
HJ = 55 – 3
HJ = 52
For JK:
JK = 8x – 9
x = 11
JK = 8x – 9
JK = 8(11) – 9
JK = 88 – 9
JK = 79
Try this option (see the attacthed file; answer is marked with red colour); note, that
answer: 2 m/h.
Answer:
f(2) = 36
Step-by-step explanation:
given f(x) = 
To evaluate f(2) substitute x = 2 into f(x)
f(2) = 6² = 6 × 6 = 36
The acceleration of the fire truck is 1.666 m/s².
<u>Step-by-step explanation:</u>
Acceleration is the measure of change in velocity with respect to time. So as the initial velocity of the fire truck is given as 0 km/hr and the final velocity is given as 66 km/hr. So at first the truck is at rest and on switching on the engine, the truck gets to the velocity of 66 km/hr in 11 seconds.
But first the units of velocity should be changed from km/hr to m/s
1 hour = 3600 seconds
1 km = 1000 m
So, 
Converted Velocity = 18.33 m/s
Then, the acceleration will be
Acceleration = 1.666 m/s²
So, the acceleration of the fire truck is 1.666 m/s².
Please don't post masses of questions all at once. 1-2 problems per post would be more realistic and fairer to others who need help.
Regarding <span>x-2/5 = 1/5
We are to solve for x. To do this, let's remove the fractions by mult. all 3 terms by 5. Then 5x - 2 = 1, or 5x = 3.
Dividing both sides by x, we get x = 3/5 (answer). Check this by subst. 3/5 into the original equation.</span>
