All categories

3 years ago

F(x) = 3x^2+ 4x^2 – 8x - 2 g(x) = 3x - 5 Find (f - g)(x).

Mathematics

1 answer:

sweet [91]3 years ago

6 0

The answer is D. 3x^3+4x^2-11x+3

You might be interested in

Ms. Clark had 135 cookies. She sold x amount of cookies at the local

Mekhanik [1.2K]

Answer:

135-x=c

Step-by-step explanation:

135 is what she has, when you subtract (x) which is what she sold, it should give you the amount she has left (c)

7 0

4 years ago

Rashid can paint 2 square feet in 10 minutes. Fill in the missing parts to tell how long it will take h to paint a fence that is

KiRa [710]

First you need to find the area of the rectangle that is made by the fence to find the sq. ft. that Rashid needs to paint.
A=length *width
A=6*25
A=150 sq. ft.
each 2 is 10 mins so divide by two
75
times 10 to get in mins
total time 750 mins
now subtract 60 until you cant get a positive number
12 hrs and 30 mins

3 0

3 years ago

Which angle is supplementary to 6 ?

natulia [17]

Answer:

$< 3$

explanation:

8 0

3 years ago

Read 2 more answers

Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.

lana [24]

The series and the sigma notations are $\sum\limits^4_0 3(5)^n = 3 + 15 +75 + 375 +1875$ , $\sum\limits^4_0 4(8)^n = 4 + 32 + 256+ 2048 + 16384$ , $\sum\limits^4_0 2(3)^n = 2 + 6 + 18 + 54 + 162$ and $\sum\limits^4_0 5(3)^n = 5 + 15 + 45 + 135 + 405$

<h3>How to match each series with the equivalent series written in sigma notation?</h3>

To do this, we simply expand each sigma notation.

So, we have:

$\sum\limits^4_0 3(5)^n$

Next, we set n = 0 to 4.

So, we have:

3(5)^0 = 3

3(5)^1 = 15

3(5)^2 = 75

3(5)^3 = 375

3(5)^4 = 1875

So, we have:

$\sum\limits^4_0 3(5)^n = 3 + 15 +75 + 375 +1875$

$\sum\limits^4_0 4(8)^n$

Next, we set n = 0 to 4.

So, we have:

4(8)^0 = 4

4(8)^1 = 32

4(8)^2 = 256

4(8)^3 = 2048

4(8)^4 = 16384

So, we have:

$\sum\limits^4_0 4(8)^n = 4 + 32 + 256+ 2048 + 16384$

$\sum\limits^4_0 2(3)^n$

Next, we set n = 0 to 4.

So, we have:

2(3)^0 = 2

2(3)^1 = 6

2(3)^2 = 18

2(3)^3 = 54

2(3)^4 = 162

So, we have:

$\sum\limits^4_0 2(3)^n = 2 + 6 + 18 + 54 + 162$

$\sum\limits^4_0 5(3)^n$

Next, we set n = 0 to 4.

So, we have:

5(3)^0 = 5

5(3)^1 = 15

5(3)^2 = 45

5(3)^3 = 135

5(3)^4 = 405

$\sum\limits^4_0 5(3)^n = 5 + 15 + 45 + 135 + 405$

Hence, the series and the sigma notations are $\sum\limits^4_0 3(5)^n = 3 + 15 +75 + 375 +1875$ , $\sum\limits^4_0 4(8)^n = 4 + 32 + 256+ 2048 + 16384$ , $\sum\limits^4_0 2(3)^n = 2 + 6 + 18 + 54 + 162$ and $\sum\limits^4_0 5(3)^n = 5 + 15 + 45 + 135 + 405$