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2 years ago

Can a negative number have a positive factor?

Mathematics

2 answers:

sertanlavr [38]2 years ago

7 0

Answer:

The laws of multiplication state that when a negative number is multiplied by a positive number, the product will be negative. So, if considering a factor pair of a negative product, one of these factors must be negative and the other factor must be positive.

Gwar [14]2 years ago

5 0

One has to be positive one has to be negative

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Prove the following

fomenos

Answer:

Step-by-step explanation:

$\large\underline{\sf{Solution-}}$

<h2 /><h2><u>Consider</u></h2>

$\rm \: \cos \bigg( \dfrac{3\pi}{2} + x \bigg) \cos \: (2\pi + x) \bigg \{ \cot \bigg( \dfrac{3\pi}{2} - x \bigg) + cot(2\pi + x) \bigg \}cos(23π+x)cos(2π+x)$

$\rm \: \cos \bigg( \dfrac{3\pi}{2} + x \bigg) = sinx$

$\rm \: {cos \: (2\pi + x) }$

$\rm \: \cot \bigg( \dfrac{3\pi}{2} - x \bigg) \: = \: tanx$

$\rm \: cot(2\pi + x) \: = \: cotx$

So, on substituting all these values, we get

$\rm \: = \: sinx \: cosx \: (tanx \: + \: cotx)$

$\rm \: = \: sinx \: cosx \: \bigg(\dfrac{sinx}{cosx} + \dfrac{cosx}{sinx}$

$\rm \: = \: sinx \: cosx \: \bigg(\dfrac{ {sin}^{2}x + {cos}^{2}x}{cosx \: sinx}$

$\rm \: = \: 1=1$

<h2>Hence,</h2>

$\boxed{\tt{ \cos \bigg( \frac{3\pi}{2} + x \bigg) \cos \: (2\pi + x) \bigg \{ \cot \bigg( \frac{3\pi}{2} - x \bigg) + cot(2\pi + x) \bigg \} = 1}}$

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<h2>ADDITIONAL INFORMATION :-</h2>

Sign of Trigonometric ratios in Quadrants

sin (90°-θ) = cos θ
cos (90°-θ) = sin θ
tan (90°-θ) = cot θ
csc (90°-θ) = sec θ
sec (90°-θ) = csc θ
cot (90°-θ) = tan θ
sin (90°+θ) = cos θ
cos (90°+θ) = -sin θ
tan (90°+θ) = -cot θ
csc (90°+θ) = sec θ
sec (90°+θ) = -csc θ
cot (90°+θ) = -tan θ
sin (180°-θ) = sin θ
cos (180°-θ) = -cos θ
tan (180°-θ) = -tan θ
csc (180°-θ) = csc θ
sec (180°-θ) = -sec θ
cot (180°-θ) = -cot θ
sin (180°+θ) = -sin θ
cos (180°+θ) = -cos θ
tan (180°+θ) = tan θ
csc (180°+θ) = -csc θ
sec (180°+θ) = -sec θ
cot (180°+θ) = cot θ
sin (270°-θ) = -cos θ
cos (270°-θ) = -sin θ
tan (270°-θ) = cot θ
csc (270°-θ) = -sec θ
sec (270°-θ) = -csc θ
cot (270°-θ) = tan θ
sin (270°+θ) = -cos θ
cos (270°+θ) = sin θ
tan (270°+θ) = -cot θ
csc (270°+θ) = -sec θ
sec (270°+θ) = cos θ
cot (270°+θ) = -tan θ

7 0

2 years ago

Read 2 more answers

PLEASE HELP ME! CLICK ON THE PICTURE FOR A BETTER VIEW IF U NEED TO!

masya89 [10]

Answer:

In order: -5, 3, 5

Step-by-step explanation:

-5g + 15h - 25

-5 ( g - 3h + 5 )

Hopefully this helps!

Brainliest please?

3 0

2 years ago

Two trains leave a city at 3 P.M. One train goes east at a constant rate of 70 mph. The other goes west at a rate of 62 mph. How

andrey2020 [161]

Answer:

4 hours

Step-by-step explanation:

The general formula for distance, rate and time is:

d = rt, where distance is the product of rate multiplied by time

Given that one train is moving at a rate of 70 mph and one is moving at a rate of 62 mph in the opposite direction, the combined distance over time is equal to 528 miles. Using the variable 't' for time:

70t + 62t = 528

Combine like terms: 132t = 528

Divide both sides by 132: 132t/132 = 528/132 or t = 4 hours

8 0

2 years ago

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Solve:<br> -3/4+p=1/2<br> Thank you so much!

shepuryov [24]

-3 / 4 + p = 1/2

LCD 4 and 2 = 4

Multiply by LCD = 4

p. (4)- 3/4 .( 4 ) = 1/2.(4)

4 p - 3 = 2

Add 3 to both sides:

4 p - 3 + 3 = 2 + 3

4 p = 5

Divide both sides by 4 :

4p / 4 = 5 /4

p = 5/4

3 0

2 years ago

What is the value of 948 x 10²?<br><br>A. 9.58<br>B. 95.8<br>C. 9,580<br>D. 95,800

Rasek [7]

The answer is 94800

8 0

2 years ago

Read 2 more answers