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Delvig [45]
3 years ago
6

Two fractions have the sane denominator. Which is the greater fractionthe fraction with the greater numerator oe the less numera

tor
Mathematics
2 answers:
Semmy [17]3 years ago
6 0
If two fractions have the same denominator, the fraction with the larger numerator is the larger number.
algol [13]3 years ago
4 0
The fraction with the greater denominator is the greater.
imagine a pizza divided into 2 pieces, if you eat one, you'll have the fraction of 1/2, but if you eat both, you'll get the fraction of 2/2, from them you ate more pizzas by eating the whole ne so, you can conclude that the fraction with the greater denominator is the greater....
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Find the distance from A(-8,-3,-2) and B(-4,8,0) , find ||AB||
RUDIKE [14]

Answer:

||AB|| = 11.8743

Step-by-step explanation:

Distance between two points:

Suppose that we have two points, (x_1,y_1,z_1) and (x_2,y_2,z_2). The distance between them is given by:

D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}

Distance from A(-8,-3,-2) and B(-4,8,0)

||AB|| = \sqrt{(-4-(-8))^2+(8-(-3))^2+(0-(-2))^2}

||AB|| = \sqrt{4^2 + 11^2 + 2^2}

||AB|| = 11.8743

3 0
2 years ago
What is 1/6 divided by 3
ira [324]
1/6 divided by 3 is 1/18

1/6÷3/1 then do reciprocal so it is 1/6×1/3 equals 1/18
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3 years ago
Can’t figure this out, I keep getting it wrong
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A positive <br> B negative <br> C none<br> D multiple
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8 0
2 years ago
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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)

<h2><u>W</u><u>e</u><u> </u><u>K</u><u>n</u><u>o</u><u>w</u><u>,</u></h2>

\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}}

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

<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
3 years ago
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