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Hitman42 [59]
3 years ago
12

6 times the sum of a number and 5

Mathematics
2 answers:
Firdavs [7]3 years ago
5 0

Answer: 6 x (x+5)

Step-by-step explanation: Since it is six times a number AND five, you want to do the addition equation first. To show this you place parentheses around a variable (x) plus 5. You add six in front of this because it comes first in the word problem.

KatRina [158]3 years ago
4 0

Answer:

6(x+5)

Step-by-step explanation:

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Answer:

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SOMEONE PLEASE JUST ANSWER THIS FOR BRAINLIEST!!!
Sloan [31]

ANSWER

P-G = 11{w}^{4} - 5 {w}^{2}  {z}^{2}   - 4z^{4}

EXPLANATION

The given polynomials are:

G =  - 3 {w}^{4}  + 2 {w}^{2}  {z}^{2}  + 5z^{4}

and

P = 8 {w}^{4}   - 3{w}^{2}  {z}^{2}  + z^{4}

We want to subtract these two polynomials.

P-G = 8 {w}^{4}   - 3{w}^{2}  {z}^{2}  + z^{4}  - (- 3 {w}^{4}  + 2 {w}^{2}  {z}^{2}  + 5z^{4} )

Expand the right hand side to obtain:

P-G = 8 {w}^{4}   - 3{w}^{2}  {z}^{2}  + z^{4}   + 3 {w}^{4}   -  2 {w}^{2}  {z}^{2}   -  5z^{4}

Group the similar terms:

P-G = 8 {w}^{4}  + 3 {w}^{4}- 3{w}^{2}  {z}^{2} -  2 {w}^{2}  {z}^{2}   -  5z^{4} + z^{4}

Combine the similar terms to get:

P-G = 11{w}^{4} - 5 {w}^{2}  {z}^{2}   - 4z^{4}

7 0
3 years ago
3. Which of the following quadratic equations has no solution? A) 0 = −2(x − 5)2 + 3 B) 0 = −2(x − 5)(x + 3) C) 0 = 2(x − 5)2 +
klemol [59]

Answer:

D) 0 = 2(x + 5)(x + 3)

Step-by-step explanation:

Which of the following quadratic equations has no solution?

We have to solve the Quadratic equation for all the options in other to get a positive value as a solution for x.

A) 0 = −2(x − 5)2 + 3

0 = -2(x - 5) × 5

0 = (-2x + 10) × 5

0 = -10x + 50

10x = 50

x = 50/10

x = 5

Option A has a solution of 5

B) 0 = −2(x − 5)(x + 3)

Take each of the factors and equate them to zero

-2 = 0

= 0

x - 5 = 0

x = 5

x + 3 = 0

x = -3

Option B has a solution by one of its factors as a positive value of 5

C) 0 = 2(x − 5)2 + 3

0 = 2(x - 5) × 5

0 = (2x -10) × 5

0 = 10x -50

-10x = -50

x = -50/-10

x = 5

Option C has a solution of 5

D) 0 = 2(x + 5)(x + 3)

Take each of the factors and equate to zero

0 = 2

= 0

x + 5 = 0

x = -5

x + 3 = 0

x = -3

For option D, all the values of x are 0, or negative values of -5 and -3.

Therefore the Quadratic Equation for option D has no solution.

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Step-by-step explanation:

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This is a polynomial with more than 2 as a degree. Using Descartes Rule of Signs: 
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Signs: + − + − − + 4 sign changes ----> 4 or 2 or 0 negative roots 
Complex roots = 0, 2, 4, or 6 
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