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2 years ago

7

A bamboo shoot is 20 inches tall at the zero and gross 24 inches each day which functions show below can be used to determine he

ight F(n) of the bamboo after n days

2 answers:

ale4655 [162]2 years ago

8 0

It is definitely D because on part 1 n=24n+20 and if n equals 0 than 24 (0) +20= 20 and that is true. on part 2 if n =0 than 24 (0)+ 20 (0-1)=20 than 20 (-1)=20 -20= 20. that statement is false because-20 does not equal 20. AND for part 3 n=1 and f (1-1) + 24= 44 so f (0) + 24 =44 and keep in mind f (0) equals 20 so 44=44 because 24+20 is 44. so that is a true statement too

Alexandra [31]2 years ago

5 0

It is given bamboo shoot is 20 inches tall when time is 0. It grows at the rate of 24 inches each day .

The y intercept b=20 and m= slope = 24.

Any equation can be written in the form y=mx+b where m is slope and b is y intercept.

The equation to determine height for n number of days can be written as

f(n)=24n+20

It can also be written as 24+the growth 1 day back or

F(n)=F(n-1)+24 where f(0)=20.

Among the given options equations 1 and 111 hold true

Option D is the right answer.

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Penny and her two friends each ate 1/6 of cake . how much cake did they eat

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They ate 1/3 of the cake

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Suppose we want to choose 2 objects, without replacement, from the 5 objects pencil, eraser, desk, chair, and lamp. (a)How many

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Answer:

a) 20 ways

b) 10 ways

Step-by-step explanation:

When the order of selection/choice matters, we use Permutations to find the number of ways and if the order of selection/choice does not matter, we use Combinations to find the number of ways.

Part a)

We have to chose 2 objects from a group of 5 objects and order of choice matters. This is a problem of permutations, so we have to find 5P2

General formula of permutations of n objects taken r at time is:

$nPr=\frac{n!}{(n-r)!}$

Using the value of n=5 and r=2, we get:

$5P2=\frac{5!}{(5-2)!} =20$

Therefore, we can choose 2 objects from a group of 5 given objects if the order of choice matters.

Part b)

Order of choice does not matter in this case, so we will use combinations to find the number of ways of choosing 2 objects from a group of 5 objects which is represented by 5C2.

The general formula of combinations of n objects taken r at a time is:

$nCr=\frac{n!}{r!(n-r)!}$

Using the value of n=5 and r=2, we get:

$5C2=\frac{5!}{2!(5-2)!} =10$

Therefore, we can choose 2 objects from a group of 5 given objects if the order of choice does not matters.

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A jeweler had a fixed amount of gold to make bracelets and necklaces. The amount of gold in each bracelet is 6 grams and the amo

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Let x represent number of bracelets and y represent number of necklaces.

We have been given that a jeweler made 7 more necklaces than bracelets. This means that number of necklaces will be $x+7$ . We can represent this information in an equation as:

$y=x+7...(1)$

We have been given that the amount of gold in each bracelet is 6 grams, so amount used for x bracelets would be $6x$ grams.

We are also told that the amount of gold in each necklace is 16 grams, so amount used for y necklaces would be $16y$ grams.

Since the jeweler used 178 grams of gold, so we will equate the amount of gold used in x bracelets and y necklaces with 178 as:

$6x+16y=178...(2)$

Therefore, our required system of equations would be:

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2 years ago

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 $b(1)=\sqrt{10-4}=\sqrt{6} \text{ which is real}$ .

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