All categories

2 years ago

Find the sum and choose the correct answer. 5n 3 + 4n 2 - 9 and -2n 3 + 7n - 5

2 answers:

8 0

<h2>Answer:</h2><h2>3n^3+4n^2+7n-14</h2>

Solution,

5n^3+4n^2-9+(-2n^3+7n-5)

= 5n^3+4n^2-9-2n^3+7n-5.

Combine the like terms,

=5n^3-2n^3+4n^2+7n-9-5

=3n^3+4n^2+7n-14

hope it helps

Good luck on your assignment

valkas [14]2 years ago

7 0

Answer:

$= 3 {n}^{3} + 4 {n}^{2} + 7n - 14 \\$

Step-by-step explanation:

$5 {n}^{3} + 4{n}^{2} - 9 + ( - 2 {n}^{3} + 7n - 5) \\ 5 {n}^{3} + 4 {n}^{2} - 9 - 2 {n}^{3} + 7n - 5 \\ = 3 {n}^{3} + 4 {n}^{2} + 7n - 14$

hope this helps

brainliest appreciated

good luck! have a nice day!

You might be interested in

3 3/4 -1 3/14 in simpilist form

lakkis [162]

For this case we must simplify the following expression:

$3 \frac {3} {4} -1 \frac {3} {14} =$

To do this, we convert the mixed numbers to improper fractions:

$\frac {4 * 3 + 3} {4} - \frac {14 * 1 + 3} {14} =\\\frac {15} {4} - \frac {17} {14} =\\\frac {14 * 15-4 * 17} {4 * 14} =\\\frac {210-68} {56} =\\\frac {142} {56} =\\\frac {71} {28} =\\2 \frac {15} {28}$

Answer:

$\frac {71} {28} = 2 \frac {15} {28}$

6 0

2 years ago

Differentiate y= ln (6 – 3x)^4

dsp73

Step-by-step explanation:

$\frac{dy}{dx} = \frac{1}{ {(6 - 3x)}^{4} } \times {4(6 - 3x)}^{3} ( - 3) \\ = \frac{ - 12 {(6 - 3x)}^{3} }{ {(6 - 3x)}^{4} } \\ = - \frac{12}{6 - 3x}$

I hope I'm correct. I've never learnt differentiation for log and exponents before

3 0

2 years ago

Please help!? 5 Stars! :0 :(

zvonat [6]

OK. These problems are easy if you know the quadratic formula,
and they're impossible if you don't.

Here's the quadratic formula:

When the equation is in the form of Ax² + Bx + C = 0

then x = [ -B plus or minus √(B²-4AC) ] / 2A

I'm sure that formula is in your text or your study notes,
right before these questions. You should cut it out or
copy it, and tape it inside the cover of your notebook.
Then, you'll always have it when you need it, until
you have it memorized and can rattle it off.

The first question says 3x² + 5x + 2 = 0

Is this in the form of    Ax² + Bx + C = 0 ?

   Yes !       A=3 B=5   C=2

   so you can use the quadratic formula to solve it.

    x = [ -B plus or minus √(B²-4AC) ] / 2A

   = [ -5 plus or minus √(5² - 4·3·2) ] / 2·3

   = [ -5 plus or minus √(25 - 24) ] / 6

   = [ -5 plus or minus √1    ] / 6

   x =    -4 / 6 = -2/3
and
   x =    -6 / 6 = -1 .
_______________________________________

The second question says

   4x² + 5x - 1 = 0

Is this in the form of Ax² + Bx + C = 0 ?

   Yes it is ! A=4 B=5 C= -1

   so you can use the quadratic formula to solve it.

    x = [ -B plus or minus √(B²-4AC) ] / 2A

Now, you take it from here.

6 0

2 years ago

What is the product?

svlad2 [7]

Answer:

Third option: 2x^3-11x^2+16x-3

Step-by-step explanation:

The product to be found is:

$(x-3)(2x^2-5x+1)$