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2 years ago

Task #3: Fuel for Thought-Student Activity Sheet Part 1

Mathematics

2 answers:

Ilia_Sergeevich [38]2 years ago

3 0

Answer:

Step-by-step explanation:

mrs_skeptik [129]2 years ago

3 0

Answer:

The method that would save the most fuel would be:

a. <u>Replacing a compact car that gets 34 miles per gallon (mpg) with a hybrid that gets 54 mpg</u>

Step-by-step explanation:

Hybrid cars are a very good way to save money on fuel and protect the environment by avoiding CO2 or CO emissions, this is because the hybrid vehicle uses both an internal combustion engine (which uses gasoline) and an electric motor (that uses electrical energy), in the example provided, <u>the more miles per gallon (mpg) a car travels, it means that its fuel will last longer, as you can see, in case A the difference when making the change is 20 mpg, while in case B it is only 10 mpg, which is why option a is chosen</u>.

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If a(x) = 3x + 1 and b (x) = StartRoot x minus 4 EndRoot, what is the domain of (b circle a) (x)?

kiruha [24]

Answer:

$[1,\infty)$

Step-by-step explanation:

$b(x)=\sqrt{x-4}$

$a(x)=3x+1$

Since we want to know the domain of $(b \circ a)(x)$ , let's first consider the domain of the inside function, that is, that of $a(x)=3x+1$ . Every polynomial function has domain all real numbers.

So we can plug anything for function $a$ and get a number back.

Now the other function is going to be worrisome because it has a square root. You cannot take square root of negative numbers if you are only considering real numbers which that is the case with most texts.

Let's find $(b \circ a)(x)$ and simplify now.

$(b \circ a)(x)$

$b(a(x))$

$b(3x+1)$

$\sqrt{(3x+1)-4}$

$\sqrt{3x+1-4}$

$\sqrt{3x-3}$

Now again we can only square root positive or zero numbers so we want $3x-3 \ge 0$ .

Let's solve this to find the domain of $(b \circ a)(x)$ .

$3x-3 \ge 0$

Add 3 on both sides:

$3x \ge 3$

Divide both sides by 3:

$x \ge 1$

So we want $x$ to be a number greater than or equal to 1.

The option that says this is $[1,\infty)$

-------------------------------

Give an example why option A fails:

A number in the given set is -2.

$a(x)=3x+1$

$b(x)=\sqrt{x-4}$

So $a(-2)=3(-2)+1=-6+1=-5$ and $b(-5)=\sqrt{-5-4}=\sqrt{-9} \text{ which is not real}$ .

Give an example why option B fails:

A number in the given set is 0.

$a(x)=3x+1$

$b(x)=\sqrt{x-4}$

So $a(0)=3(0)+1=0+1=1$ and $b(1)=\sqrt{1-4}=\sqrt{-3} \text{ which is not real}$ .

Give an example why option D fails:

While all the numbers in set D work, there are more numbers outside that range of numbers that also work.

A number not in the given set that works is 3.

$a(x)=3x+1$

$b(x)=\sqrt{x-4}$

So $a(3)=3(3)+1=9+1=10$ and $b(1)=\sqrt{10-4}=\sqrt{6} \text{ which is real}$ .

4 0

2 years ago

Read 2 more answers

Rectangle ABCD was dilated to create rectangle A'B'C'D' using point R as the center of dilation.

Slav-nsk [51]

Answer:

Step-by-step explanation:

8 0

2 years ago

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Agent Hunt transferred classified files from the CIA mainframe onto his flash drive. The drive had some files on it before the t

topjm [15]

Answer: The drive was 24.32% full when the transfer began.

Step-by-step explanation:

Given: The rate of transferring file = 4.44 megabytes per second.

After 32 seconds, the amount of data on the drive = 384 megabytes

The amount of transferred data = $32\times4.4=140.8$ megabytes

Therefore, the amount of data in the drive at the beginning is given by :-

$\text{Beginning amount=Amount after 32 seconds -Amount transferred}\\\\\Rightarrrow\ \text{Beginning amount=}384-140.8=243.2$

The drive had a maximum capacity of 1000 megabytes.

$\frac{243.2}{1000}\times100=24.32\%$

Hence, the drive was 24.32% full when the transfer began.

4 0

2 years ago

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Squaring both sides of an equation is irreversible. Is Cubing both sides of an equation reversible? Provide numerical examples t

nikitadnepr [17]

Answer:

<h2>Cubing both sides of an equation is reversible.</h2>

Step-by-step explanation:

Squaring both sides of an equation is irreversible, because the square power of negative number gives a positive result, but you can't have a negative base with a positive number, given that the square root of a negative number doesn't exist for real numbers.

In case of cubic powers, this action is reversible, because the cubic root of a negative number is also a negative number. For example

$\sqrt[3]{x} =-1$