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3 years ago

Simplify negative 8 over 3 divided by negative 2 over 6

Mathematics

1 answer:

Art [367]3 years ago

5 0

Answer: The answer is 8

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Your answer should be a polynomial in standard form. (x + 5)(x +3)=(x+5)(x+3)

sattari [20]

Answer:

∴(x+5)(x+3) =x²+8x+15

Step-by-step explanation:

Standard form of a polynomial:

Standard of polynomial means that the terms ordered from biggest exponent of variable to lowest exponent of variable.

Example:

$a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+......$

Given,

(x+5)(x+3)

=x(x+3)+5(x+3)

=x²+3x+5x+15

=x²+8x+15

∴(x+5)(x+3) =x²+8x+15

The degree of the polynomial is 2 and is a one variable polynomial.

7 0

2 years ago

I NEED HELPPPppp!!!!

CaHeK987 [17]

Answer:

answer choice D

Step-by-step explanation:

x => 6 - 4x > 18

= 6-18 > 4x

= - 12 > 4x

= -3 > x

= x < -3

5 0

2 years ago

Help me!

vovangra [49]

Answer:

OPTION D

Step-by-step explanation: PLS MARK BRAINLIEST :D

8 0

2 years ago

Read 2 more answers

2 tan 30°<br>II<br>1 + tan- 300

shusha [124]

Question:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Answer:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

Step-by-step explanation:

Given

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Required

Simplify

In trigonometry:

$tan(30^{\circ}) = \frac{1}{\sqrt{3}}$

So, the expression becomes:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + (\frac{1}{\sqrt{3}})^2}$

Simplify the denominator

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{3+1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{4}{3}}$

Express the fraction as:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} / \frac{4}{3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} * \frac{3}{4}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{1}{\sqrt 3} * \frac{3}{2}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3}$

Rationalize

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3} * \frac{\sqrt{3}}{\sqrt{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3\sqrt{3}}{2* 3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\sqrt{3}}{2}$

In trigonometry:

$sin(60^{\circ}) = \frac{\sqrt{3}}{2}$

Hence:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

3 0

3 years ago

Please help me with the correct answer

Sergio [31]

-1,1,3,5,and 7 in that order

4 0

2 years ago