1. Slope intercept form is y=mx+b, where m is slope and b is y-intercept.
It gave you point (8,12) and slope of -2.
So you already know the value of m, so so far it's y = -2x + b.
Plug the coordinates as x and y: 8 for x and 12 for y.
12 = -2(8) + b
12 = -16 + b
28 = b
Adding b in, you get y = -2x + 28
2. Find the slope, then use point-slope form.
Slope is change in y over change in x, or (y2-y1)/(x2-x1).
m = (9-1)/(2-3) = 8/(-1) = -8
Point slope form: y - y1 = m(x - x1)
y - y1 = -8(x - x1)
Now plug in either of the coordinates as x1 and y1, let's do (3,1).
y - 1 = -8(x - 3)
y - 1 = -8x + 24
y = -8x + 25
3. Do it the same way as number 2. Choose two points, for example (1,1) and (2,-2). You find m = -3 and get final answer as y=-3x + 4.
4. B = 150 + 80d
B is account balance, d is number of monthly deposits.
If you rearrange the terms: B = 80d + 150, you see that that is in slope-intercept form y = mx+b, and if you remember from #1 m is the slope. In this case, the variables are changed, so d is the slope here. And the question told you that d, the slope, is the number of monthly deposits. Its coefficient is 80, so Bianca deposits $80/month.
<span>5. </span>
14a + 16b = 560
16b = 560 - 14a
b = (560 - 14a)/16
b = 560/16 - 14a/16
b = 35 - (7/8)a
<span>b = -(7/8)a + 35</span>
Answer:
"Lines s and t have slopes that are negative reciprocals of one another."
Step-by-step explanation:
First note down the relevant variables from the question.
Ua (Initial velocity a) = 320ft/s
Ub (initial velocity b) = 240ft/s
Aay (acceleration of a in the vertical axis) = Aby = -32.17ft/s/s
We want to know when they will be at the same height so should use the formula for displacement:
s = ut + 1/2 * at^2
We want to find when both firework a and firework b will be at the same height. Therefore mathematically when: say = sby (the vertical displacements of firework A and B are equal). We also know that firework B was launched 0.25s before firework A so we should either add 0.25s to the time variable for the displacement formula for firework B or subtract 0.25s for firework A.
SO:
Say = Sby
320t + 1/2*-32.17t^2 = 240(t+0.25) + 1/2 * -32.17(t+0.25)^2
320t - 16.085t^2 = 240t + 60 - 16.085(t+0.25)^2
320t - 16.085t^2 = 240t + 60 - 16.085(t^2 + 0.5t + 6.25)
320t - 16.085t^2 = 240t + 60 -16.085t^2 - 8.0425t - 100.53
320t - 240t - 8.0425t - 16.085t^2 + 16.085t^2 = 60 - 100.53
71.958t = -40.53
t = -0.56s (negative because we set t before Firework A was launched)
Now we know both fireworks explode 0.56 seconds AFTER fireworks B launches (because we added 0.25 seconds to the t variable in the equation above for the vertical displacement of Firework B).
You could continue on to find the displacement they both explode at and verify the answer by ensuring that it is equal (because the question stated they should explode at the same height by substituting the value we found for t of 0.56s into the vertical displacement formula for firework A and t+0.25s=0.81s into the same formula for Firework B
Verification:
Say = ut + 1/2at^2
Say = 320*0.56 + 1/2*-32.17*0.56^2
Say = 179.2 + -5.04
Say = 174.16ft
Sby = ut + 1/2at^2
Sby = 240*0.81 + 1/2*-32.17*0.81^2
Sby = 194.4 - 10.5
Sby = 183.9ft
While Say is close to Sby I would have expected them to be almost perfectly equal… can you please check if this matches the answer in your textbook? There could be wires due to rounding. I also usually work in SI units which use the metric system and not the imperial system although that shouldn’t make a difference. The working out and thought process is correct though and this is why trying to verify the answer is an important step to make sure it works out.
Answer: 0.56s (I think)
True
Product=multiplication
Quotient=division
Difference=subtraction
Sum=addition
Answer:
- 60/49 would be the answer