1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
bonufazy [111]
4 years ago
13

Given the function ƒ(x) = 9x2 + 27x + 2, find ƒ(2).

Mathematics
2 answers:
deff fn [24]4 years ago
8 0

Answer:  The correct option is

(C) f(2) = 92.

Step-by-step explanation:  We are given the following function :

f(x)=9x^2+27x+2~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)

We are to find the value of f(2).

To find f(2), we must substitute x = 2 in equation (i).

From equation (i), we get

f(2)\\\\=9\times2^2+27\times2+2\\\\=36+54+2\\\\=92.

Thus, the required value of f(2) is 92.

Option (C) is CORRECT.

Burka [1]4 years ago
3 0
Solution is to substitute 2 for x and u should get 92.
You might be interested in
What is the slope and y intercept of the equation 6x - 1 = 3y - 10
-BARSIC- [3]

your x value is your slope so the  slope is 6 and your y intercept is the y value so your intercept is 3

hope this helps

3 0
4 years ago
What’s the slope of -2x+8
77julia77 [94]

Answer:

<h2>The slope m = -2</h2>

Step-by-step explanation:

The slope-intercept form of the equation of line:

y=mx+b

We have

y=-2x+8

Therefore we have the slope = -2.

6 0
4 years ago
Triangle DEF was dilated according to the rule DO,(x,y) to create similar triangle D'E'F'.
lord [1]
A dilation is a transformation, with center O and a scale factor of k that is not zero, that maps O to itself and any other point P to P'. The center O is a fixed point, P' is the image of P, points O, P and P' are on the same line.

Thus, a dilation with centre O and a scale factor of \frac{1}{3} maps the original figure to the image in such a way that the distances from O to the vertices of the image are \frac{1}{3} times the distances from O to the original figure. Also the size of the image are <span>\frac{1}{3} times the size of the original figure. Also the two resulting figures (i.e. the image and the pre-image are congruent)

Thus in the dilation of triangle DEF, the following are true.</span>
<span>∠F corresponds to ∠F'.

The measure of ∠E' is the measure of ∠E.

△DEF ≈ △D'E'F'</span>
5 0
3 years ago
Read 2 more answers
Determine the value of x for the following equations.
mafiozo [28]

Answer:

I'm so sorry!!! I don't understand how you are supposed to do this problem... I hope someone can help you out. Have a good day

Step-by-step explanation:

Super sorry

4 0
3 years ago
Divide 16x3 – 12x2 + 20x – 3 by 4x + 5.
nalin [4]

Answer:

4x^2 - 8x + 15 - \frac{78}{4x+5}

Step-by-step explanation:

<em>To solve polynomial long division problems like these, it's helpful to build a long division table. Getting used to building these can make problems like this much simpler to solve.</em>

Begin by looking at the first term of the cubic polynomial.

What would we have to multiply 4x + 5 by to get an expression containing 16x^3? The answer is 4x^2, since (4x + 5) \times 4x^2 = 16x^2 + 20x.

This is the first step of our long division, and we write out the start of our long division table like this:

{ }\qquad{ }\quad{ }\quad{ }\,\,\,\,\,4x^2\\4x + 5\quad)\!\!\overline{\,\,\,16x^3 - 12x^2 + 20x - 3}\\{ }\qquad{ }\quad{ }\quad{ }\,\,16x^3 + 20x^2\\

On the left is the divisor. On top is 4x^2. In the middle is the polynomial we are dividing, and on the bottom is the result of multiplying our divisor by

The next step is to subtract the bottom expression from the middle one, like so:

{ }\qquad{ }\quad{ }\quad{ }\,\,\,\,\,4x^2\\4x + 5\quad)\!\!\overline{\,\,\,16x^3 - 12x^2 + 20x - 3}\\{ }\qquad{ }\quad{ }\quad{ }\,\,\underline{16x^3 + 20x^2}\\{ }\qquad{ }\quad{ }\quad{ }\,\,\,\,\,0x^3 - 32x^2\\

We are left with -32x^2. The next thing to do is to add the next term of the polynomial we are dividing to the bottom line, like this:

{ }\qquad{ }\quad{ }\quad{ }\,\,\,\,\,4x^2\\4x + 5\quad)\!\!\overline{\,\,\,16x^3 - 12x^2 + 20x - 3}\\{ }\qquad{ }\quad{ }\quad{ }\,\,\underline{16x^3 + 20x^2}\\{ }\qquad{ }\quad{ }\quad{ }\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 32x^2 + 20x\\

Now we return to the beginning of the instructions, and repeat the process: namely, what would we have to multiply 4x + 5 by to get an expression containing -32x^2? The answer is -8x, and we fill out our long division table like so:

{ }\qquad{ }\quad{ }\quad{ }\,\,\,\,\,4x^2 - \,\,\,\,8x\\4x + 5\quad)\!\!\overline{\,\,\,16x^3 - 12x^2 + 20x - 3}\\{ }\qquad{ }\quad{ }\quad{ }\,\,\underline{16x^3 + 20x^2}\\{ }\qquad{ }\quad{ }\quad{ }\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 32x^2 + 20x\\{ }\qquad{ }\quad{ }\quad{ }\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 32x^2 - 40x\\

Once again, we subtract the bottom expression from the one above it, and include the next term of the divisor, like so:

{ }\qquad{ }\qquad{ }\quad{ }4x^2 - \,\,\,\,8x \,+ 15\\4x + 5\quad)\!\!\overline{\,\,\,16x^3 - 12x^2 + 20x - 3}\\{ }\qquad{ }\quad{ }\quad{ }\,\,\underline{16x^3 + 20x^2}\\{ }\qquad{ }\quad{ }\quad{ }\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 32x^2 + 20x\\{ }\qquad{ }\quad{ }\quad{ }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{- 32x^2 - 40x}\\{ }\qquad{ }\qquad{ }\qquad{ }\qquad{ }\qquad{ }\,\,\,\,\,60x - 3\\

And repeat. What do we multiply 4x + 5 by to get an expression containing 60x? The answer is 15. Our completed long division table looks like this:{ }\qquad{ }\qquad{ }\quad{ }4x^2 - \,\,\,\,8x \,+15\\4x + 5\quad)\!\!\overline{\,\,\,16x^3 - 12x^2 + 20x - 3}\\{ }\qquad{ }\quad{ }\quad{ }\,\,\underline{16x^3 + 20x^2}\\{ }\qquad{ }\quad{ }\quad{ }\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 32x^2 + 20x\\{ }\qquad{ }\quad{ }\quad{ }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{- 32x^2 - 40x}\\{ }\qquad{ }\qquad{ }\qquad{ }\qquad{ }\qquad{ }\,\,\,\,\,60x - 3\\{ }\qquad{ }\hspace{3cm}\,\,\underline{60x + 75}\\{ }\hspace{4.3cm}\,\,-78

Now, the expression at the top,

4x^2 - 8x + 20x + 15

is our quotient, and the last number, -78, is our remainder.

Hence we arrive at the solution of

\frac{16x^3-12x^2+20x-3}{4x+5} =4x^2 - 8x + 15 - \frac{78}{4x+5}.

6 0
3 years ago
Other questions:
  • Find the perimeter and area of the polygon with the given vertices. Round your answers to the nearest tenth, if necessary.
    12·1 answer
  • What is the answer 25÷5(2+3)=
    9·2 answers
  • Answer this please I leave good reviews
    6·2 answers
  • During a trip Sara knits a scarf at the rate of 469 millimetre everyday if her trip lasts 8 days how much of the scar will she h
    9·1 answer
  • write the equation for the line that is perpendicular to given line that passes through the given point -3x-6y=17; (6,3)
    12·2 answers
  • What does 2 3/5 divided by 1 1/25 equal
    10·1 answer
  • *BRAINLIEST* please help!!! the option for the second blank are the same as the first one
    10·2 answers
  • Solve for x and y<br> 3x+2y=16<br> 7x+y=19
    7·1 answer
  • 1/4 (dot) 3/5 simplifily
    10·1 answer
  • Help me!<br> No links!! Please!!
    11·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!