Diego’s family truck holds 21 gallons of fuel. Each day the car uses 1.5 gallons of fuel. A warning light comes on when the remaining fuel is 2.5 gallons or less. 1.Starting from a full tank, can Diego’s family drive the car for 10 days without the warning light coming on? Explain or show your reasoning.

2.Starting from a full tank, can Diego’s family drive the car for 15 days without the warning light coming on? Explain or show your reasoning.

3.Diego says the expression 21 - 1.5t helps him understand this situation. In this situation, what does this expression represent, and what does the variable t stand for?

4.Write and solve an equation to determine the number of days Diego’s father can drive the car without the warning light coming on.

5.Write and solve an inequality that represents the number of days driven without the warning light coming on. Explain clearly what the solution to the inequality means in the context of this situation.

Problem 1

We'll refer to the expression in problem 3 which is a valid way to represent he amount of fuel left after t days have elapsed.

If no days go by, and t = 0, then 21-1.5*t = 21-1.5*0 = 21 gallons remain.

If 1 day elapses, then t = 1 and 21-1.5*t = 21-1.5*1 = 21-1.5 = 19.5 gallons remain.

If 2 days elapse, then t = 2 and 21-1.5*t = 21-1.5*2 = 21-3 = 18 gallons remain.

and so on.

Let's plug in t = 10 and see what we get

21-1.5*t = 21-1.5*10 = 21-15 = 6

There will be 6 gallons left after driving for 10 days. This is larger than 2.5 gallons (that triggers the warning light) so the light won't come on.

Yes, the family can drive the car 10 days without the light coming on.

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Problem 2

We'll use the same idea as before, but this time we plug in t = 15

21 - 1.5*t = 21 - 1.5*15 = 21 - 22.5 = -1.5

Since we get a negative result, this means the family cannot drive for 15 days without having to refuel.

Put another way: if they drove for 15 days, then they use up 1.5*t = 1.5*15 = 22.5 gallons which exceeds the max capacity of the tank (21 gallons).

The warning light will come on at some point during the 15 day time period, and they'll need to refuel.

 No, the light will come on at some point

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Problem 3

Refer to problems 1 and 2 for specific examples of 21 - 1.5*t being used.

The 21 is the max capacity of the tank

The 1.5 refers to the car using 1.5 gallons of fuel per day

Saying 1.5t is the amount of fuel burned over t days, where t is some positive whole number. For example, if t = 4, then 1.5*t = 1.5*4 = 6 gallons are used up over the four day period.

Therefore, 21 - 1.5t represents the amount of fuel left over when t days elapse.

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Problem 4

Set 21 - 1.5*t equal to 2.5 and solve for t

21 - 1.5*t = 2.5

-1.5t = 2.5 - 21

-1.5t = -18

t = -18/(-1.5)

t = 12.333 approximately

If we round down to t = 12, then we would find that 21 - 1.5*t is equal to

21 - 1.5*t = 21 - 1.5*12 = 3

which is over 2.5 and the light does not come on.

However, if we rounded up to t = 13, then,

21 - 1.5*t = 21 - 1.5*13 = 1.5

which is now under the 2.5 gallon threshold and the light will come on at this point. It's not entirely clear when it comes on during that 13th day, but we know it does so somewhere on this day.

12 days

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Problem 5

We want the expression 21 - 1.5*t to be greater than 2.5

If 21 - 1.5*t were equal to 2.5 or smaller, then the light would come on.

So that's why we want

Let's isolate t

The steps are pretty much identical to problem 4, except that we're dealing with an inequality sign now. The sign flips when we divide both sides by a negative number.

Since t is a whole number, the t < 12.333 can be replaced with t < 12 to ensure we're under the 12.333 marker.

This backs up the conclusion of problem 4 which said that if the family drives for 12 days or less, then the light won't come on. Driving for 13 days or more will make the light come on.

So effectively, this is just a rephrasing of problem 4.

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

answer: i realy don't get it that much can you me figer it out pleas you

step-by-step explanation: