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ohaa [14]
2 years ago
14

What is the answer to number 5

Mathematics
1 answer:
True [87]2 years ago
4 0
4050 millilitres
because there are 6 people so 675 × 6
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(0, -8)

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Help solve 11 and 13 thanks mostly 11 xx
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A cycle race began at 9:40
Ksenya-84 [330]
The entire race took Henry 1hr:36min which is 96min. He has an average speed of 30km/hr means he's been traveling 0.5km/min
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6 0
2 years ago
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What is the answer to this math problem​
vesna_86 [32]

Hello from MrBillDoesMath!

Answer:

1/243

Discussion:  

3^(-3) *  3^(-2)                => as a^m * a^n = a^(m+n), the exponents add

= 3^ (-3 -2)

= 3^ (-5) =

= 1/ 3^5                           => as 3^5 = 243

= 1/243

which is approximately  .0041

Thank you,

MrB

7 0
3 years ago
Read 2 more answers
find the values of the six trigonometric functions for angle theta in standard position if a point with the coordinates (1, -8)
frutty [35]

Answer:

cosФ = \frac{1}{\sqrt{65}} , sinФ = -\frac{8}{\sqrt{65}} , tanФ = -8, secФ = \sqrt{65} , cscФ = -\frac{\sqrt{65}}{8} , cotФ = -\frac{1}{8}

Step-by-step explanation:

If a point (x, y) lies on the terminal side of angle Ф in standard position, then the six trigonometry functions are:

  1. cosФ = \frac{x}{r}
  2. sinФ = \frac{y}{r}
  3. tanФ = \frac{y}{x}
  4. secФ = \frac{r}{x}
  5. cscФ = \frac{r}{y}
  6. cotФ = \frac{x}{y}
  • Where r = \sqrt{x^{2}+y^{2} } (the length of the terminal side from the origin to point (x, y)
  • You should find the quadrant of (x, y) to adjust the sign of each function

∵ Point (1, -8) lies on the terminal side of angle Ф in standard position

∵ x is positive and y is negative

→ That means the point lies on the 4th quadrant

∴ Angle Ф is on the 4th quadrant

∵ In the 4th quadrant cosФ and secФ only have positive values

∴ sinФ, secФ, tanФ, and cotФ have negative values

→ let us find r

∵ r = \sqrt{x^{2}+y^{2} }

∵ x = 1 and y = -8

∴ r = \sqrt{x} \sqrt{(1)^{2}+(-8)^{2}}=\sqrt{1+64}=\sqrt{65}

→ Use the rules above to find the six trigonometric functions of Ф

∵ cosФ = \frac{x}{r}

∴ cosФ = \frac{1}{\sqrt{65}}

∵ sinФ = \frac{y}{r}

∴ sinФ = -\frac{8}{\sqrt{65}}

∵ tanФ = \frac{y}{x}

∴ tanФ = -\frac{8}{1} = -8

∵ secФ = \frac{r}{x}

∴ secФ = \frac{\sqrt{65}}{1} = \sqrt{65}

∵ cscФ = \frac{r}{y}

∴ cscФ = -\frac{\sqrt{65}}{8}

∵ cotФ = \frac{x}{y}

∴ cotФ = -\frac{1}{8}    

8 0
2 years ago
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