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Ipatiy [6.2K]
4 years ago
15

If x is a real number, for what values of x is the equation 3x-9/3= x-3 true?

Mathematics
2 answers:
loris [4]4 years ago
8 0
Your answer would be b. Ony some
devlian [24]4 years ago
7 0
Hey there!

<span> If x is a real number, for what values of x is the equation 3x-9/3= x-3 true?

Answer: B. some values of x

Hope this helps
Have a great day (:
</span>
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Vashti and Stacy measure the width of the gym at their school. It is 17 yards wide.
Annette [7]
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3 0
3 years ago
Find the LCM of 24,60 and 54
Andrei [34K]

Answer:

1080

Step-by-step explanation:

Multiples of 24:

24, 48, 72, 96, 120, 144, 168, 192, 216, 240, 264, 288, 312, 336, 360, 384, 408, 432, 456, 480, 504, 528, 552, 576, 600, 624, 648, 672, 696, 720, 744, 768, 792, 816, 840, 864, 888, 912, 936, 960, 984, 1008, 1032, 1056, 1080, 1104, 1128

Multiples of 54:

54, 108, 162, 216, 270, 324, 378, 432, 486, 540, 594, 648, 702, 756, 810, 864, 918, 972, 1026, 1080, 1134, 1188

Multiples of 60:

60, 120, 180, 240, 300, 360, 420, 480, 540, 600, 660, 720, 780, 840, 900, 960, 1020, 1080, 1140, 1200

Therefore,

LCM(24, 54, 60) = 1080

4 0
1 year ago
Read 2 more answers
2. Are these two ratios equivalent? 8/5, 24/20
uysha [10]

Answer:  No

Step-by-step explanation:

I took a test and this answer was correct.

4 0
3 years ago
1. A right triangle LMN is given where: side MN = 8 side NL (the hypotenuse) =
inessss [21]

Step-by-step explanation:

\underline{ \underline{ \text{Given}}} :

  • Length of MN ( Base ) = 8
  • Length of NL ( Hypotenuse ) = 10

\underline{ \underline{ \text{To \: find}}} :

  • Length of LM ( Perpendicular )

\underline{ \underline{ \text{Using \: pythagoras \: theorem}}} :

\boxed{ \sf{ {Hypotenuse}^{2}  =  {Perpendicular}^{2}  +  {Base}^{2} }}

⤑ \sf{  Perpendicular =  \sqrt{ {(Hypotenuse)}^{2}  -   {(Base)}^{2}  } }

⤑ \sf{ \sqrt{ {(10)}^{2}  -   {(8)}^{2}  } }

⤑ \sf{ \sqrt{100  -  64}}

⤑ \sf{ \sqrt{36}}

⤑ \boxed{ \sf{6\: units}}

\pink{ \boxed{ \boxed{ \tt{Our \: final \: answer :  \boxed{ \underline { \tt{6 \: units}}}}}}}

Hope I helped ! ツ

Have a wonderful day / night ! ♡

▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁

5 0
3 years ago
Find the remaining trigonometric ratios of θ if csc(θ) = -6 and cos(θ) is positive
VikaD [51]
Now, the cosecant of θ is -6, or namely -6/1.

however, the cosecant is really the hypotenuse/opposite, but the hypotenuse is never negative, since is just a distance unit from the center of the circle, so in the fraction -6/1, the negative must be the 1, or 6/-1 then.

we know the cosine is positive, and we know the opposite side is -1, or negative, the only happens in the IV quadrant, so θ is in the IV quadrant, now

\bf csc(\theta)=-6\implies csc(\theta)=\cfrac{\stackrel{hypotenuse}{6}}{\stackrel{opposite}{-1}}\impliedby \textit{let's find the \underline{adjacent side}}&#10;\\\\\\&#10;\textit{using the pythagorean theorem}\\\\&#10;c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a&#10;\qquad &#10;\begin{cases}&#10;c=hypotenuse\\&#10;a=adjacent\\&#10;b=opposite\\&#10;\end{cases}&#10;\\\\\\&#10;\pm\sqrt{6^2-(-1)^2}=a\implies \pm\sqrt{35}=a\implies \stackrel{IV~quadrant}{+\sqrt{35}=a}

recall that 

\bf sin(\theta)=\cfrac{opposite}{hypotenuse}&#10;\qquad\qquad &#10;cos(\theta)=\cfrac{adjacent}{hypotenuse}&#10;\\\\\\&#10;% tangent&#10;tan(\theta)=\cfrac{opposite}{adjacent}&#10;\qquad \qquad &#10;% cotangent&#10;cot(\theta)=\cfrac{adjacent}{opposite}&#10;\\\\\\&#10;% cosecant&#10;csc(\theta)=\cfrac{hypotenuse}{opposite}&#10;\qquad \qquad &#10;% secant&#10;sec(\theta)=\cfrac{hypotenuse}{adjacent}

therefore, let's just plug that on the remaining ones,

\bf sin(\theta)=\cfrac{-1}{6}&#10;\qquad\qquad &#10;cos(\theta)=\cfrac{\sqrt{35}}{6}&#10;\\\\\\&#10;% tangent&#10;tan(\theta)=\cfrac{-1}{\sqrt{35}}&#10;\qquad \qquad &#10;% cotangent&#10;cot(\theta)=\cfrac{\sqrt{35}}{1}&#10;\\\\\\&#10;sec(\theta)=\cfrac{6}{\sqrt{35}}

now, let's rationalize the denominator on tangent and secant,

\bf tan(\theta)=\cfrac{-1}{\sqrt{35}}\implies \cfrac{-1}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{-\sqrt{35}}{(\sqrt{35})^2}\implies -\cfrac{\sqrt{35}}{35}&#10;\\\\\\&#10;sec(\theta)=\cfrac{6}{\sqrt{35}}\implies \cfrac{6}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{6\sqrt{35}}{(\sqrt{35})^2}\implies \cfrac{6\sqrt{35}}{35}
3 0
3 years ago
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