**WILL GIVE THE BRAINLIEST! & LIKE**
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The complete question in the attached figure
we know that
<span>Fred bought
</span><span>4 liters of liquid laundry detergent
</span><span>3250 mL of fabric softener------------> 3250/1000-------> 3.25 liters
</span><span>2.5 L of bleach-----------> 2.5 liters
then
</span><span>statements
4a. Fred bought 75 milliliters more fabric softener than bleach
</span>3.25-2.5-------> 0.75 liters--------> 750 mi
the statement is false
<span>4b. Fred brought 1.75 liters more laundry detergent than bleach
4-2.5---------> 1.5 liters
</span>the statement is false
4c. Fred brought 750 milliliters more fabric softener than bleach. <span>
3.25-2.5-------> 0.75 liters-------> 750 ml
</span>the statement is true
4d. Fred brought 150 milliliters more laundry detergent than bleach.
4-2.5--------> 1.5 liters---------> 1500 ml
the statement is false
<span>4e. Fred bought 0.75 liters more laundry detergent than fabric softener.
</span>4-3.25--------> 0.75 liters
the statement is true
we know that
<span>Fred bought
</span><span>4 liters of liquid laundry detergent
</span><span>3250 mL of fabric softener------------> 3250/1000-------> 3.25 liters
</span><span>2.5 L of bleach-----------> 2.5 liters
then
</span><span>statements
4a. Fred bought 75 milliliters more fabric softener than bleach
</span>3.25-2.5-------> 0.75 liters--------> 750 mi
the statement is false
<span>4b. Fred brought 1.75 liters more laundry detergent than bleach
4-2.5---------> 1.5 liters
</span>the statement is false
4c. Fred brought 750 milliliters more fabric softener than bleach. <span>
3.25-2.5-------> 0.75 liters-------> 750 ml
</span>the statement is true
4d. Fred brought 150 milliliters more laundry detergent than bleach.
4-2.5--------> 1.5 liters---------> 1500 ml
the statement is false
<span>4e. Fred bought 0.75 liters more laundry detergent than fabric softener.
</span>4-3.25--------> 0.75 liters
the statement is true
Answers:
- Part a)
- Part b) 3
- Part c) 2.24
- Part d) 1.58
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Work Shown:
Part (a)
The origin is the point (0,0) which we'll make the first point, so let (x1,y1) = (0,0)
The other point is of the form (x,y) where y = x^2-3. So the point can be stated as (x2,y2) = (x,y). We'll replace y with x^2-3
We apply the distance formula to say...
We could expand things out and combine like terms, but that's just extra unneeded work in my opinion.
Saying is the same as writing d = sqrt(x^2-(x^2-3)^2)
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Part (b)
Plug in x = 0 and you should find the following
This says that the point (x,y) = (0,3) is 3 units away from the origin (0,0).
-------------------------------------------
Part (c)
Repeat for x = 1
-------------------------------------------
Part (d)
Graph the d(x) function found back in part (a)
Use the minimum function on your graphing calculator to find the lowest point such that x > 0.
See the diagram below. I used GeoGebra to make the graph. Desmos probably has a similar feature (but I'm not entirely sure). If you have a TI83 or TI84, then your calculator has the minimum function feature.
The red point of this diagram is what we're after. That point is approximately (1.58, 1.66)
This means the smallest d can get is d = 1.66 and it happens when x = 1.58 approximately.
