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3 years ago

Find the y-coordinate of the y-intercept of the polynomial function defined below. f(x)=2(3x+4)(x^2+2)

Mathematics

1 answer:

Salsk061 [2.6K]3 years ago

3 0

Answer:

Step-by-step explanation:

The best way to find out the answer to your question is to graph the cubic. As you can see, the y intercept is (0,16).

How was this obtained. notice that if x = 0 then

f(0) = 2(3*0+4)(0^2 + 2)

f(0) = 2(4)(2)

f(0) = 16 just as the graph tells us.

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One bar of candy A and two bars of candy B have 777 calories. Two bars of candy A and one bar of candy B contain 783 calories. F

RSB [31]

Answer:

A has 263 calories

B has 257 calories

Step-by-step explanation:

system of equations:

a + 2b = 777

2a + b = 783

I multiplied the first equation by -2 to eliminate the 'a' terms

-2a - 4b = -1554

+ <u> 2a + b = 783</u>

-3b = -771

b = 257

solve for 'a'

a + 2(257) = 777

a + 514 = 777

a = 263

4 0

3 years ago

Here please please please

Semmy [17]

Answer: 6

explanation:
set up proportions
(4+2)/9 = 4/x
6/9 = 4/x
simplify
2/3 = 4/x
cross multiply
2x = 12
divide
x = 6

8 0

3 years ago

Which of the following is independent variable?

satela [25.4K]

I think the answer is D. hours because its variation does not depend on another variable

6 0

3 years ago

7 less than three times the sum of a number and five

Evgen [1.6K]

<u>Expression Form</u>
(3-7)*n+5
n=Unknown Number

4 0

3 years ago

The numbers of teams remaining in each round of a single-elimination tennis tournament represent a geometric sequence where an i

Anit [1.1K]

Answer:

$a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}$

Step-by-step explanation:

We are given the following in the question:

The numbers of teams remaining in each round follows a geometric sequence.

Let a be the first the of the geometric sequence and r be the common ration.

The $n^{th}$ term of geometric sequence is given by:

$a_n = ar^{n-1}$

$a_4 = 16 = ar^3\\a_6 = 4 = ar^5$

Dividing the two equations, we get,

$\dfrac{16}{4} = \dfrac{ar^3}{ar^5}\\\\4}=\dfrac{1}{r^2}\\\\\Rightarrow r^2 = \dfrac{1}{4}\\\Rightarrow r = \dfrac{1}{2}$

the first term can be calculated as:

$16=a(\dfrac{1}{2})^3\\\\a = 16\times 6\\a = 128$

Thus, the required geometric sequence is

$a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}$

4 0

3 years ago