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3 years ago

5

A train leaves at 11:32 and arrives at 13:24. It travelled at an average speed of 85 km/h. How far did it travel? Give your answ

er in km to 1DP

1 answer:

Mandarinka [93]3 years ago

4 0

Hope that helps!!
Answer: 158.7 km

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Please help it’s due tonight!!

Alina [70]

Answer:

4: 1.33 ft.

5: 516.14322017669 mm

6:

7:

8: 7 cm

9: 27314.29921175 mm

10: 112 ft.

Step-by-step explanation:

not sure about 6 and 7

6 0

3 years ago

C an you simplify 5^-8 x 5^4

EastWind [94]

Answer:

1/625

Step-by-step explanation:

Exponent Rule: $(b^m)(b^n) = b^{m+n}$

Exponent Rule: $b^{-m}=\frac{1}{b^m}$

5⁻⁸ · 5⁴ = -8 + 4 = -4

5⁻⁴

1/5⁴

1/625

6 0

4 years ago

Find all possible values of α+

const2013 [10]

Answer:

$\rm\displaystyle 0,\pm\pi$

Step-by-step explanation:

please note that to find but α+β+γ in other words the sum of α,β and γ not α,β and γ individually so it's not an equation

===========================

we want to find all possible values of α+β+γ when <u>tanα+tanβ+tanγ = tanαtanβtanγ</u><u> </u>to do so we can use algebra and trigonometric skills first

cancel tanγ from both sides which yields:

$\rm\displaystyle \tan( \alpha ) + \tan( \beta ) = \tan( \alpha ) \tan( \beta ) \tan( \gamma ) - \tan( \gamma )$

factor out tanγ:

$\rm\displaystyle \tan( \alpha ) + \tan( \beta ) = \tan( \gamma ) (\tan( \alpha ) \tan( \beta ) - 1)$

divide both sides by tanαtanβ-1 and that yields:

$\rm\displaystyle \tan( \gamma ) = \frac{ \tan( \alpha ) + \tan( \beta ) }{ \tan( \alpha ) \tan( \beta ) - 1}$

multiply both numerator and denominator by-1 which yields:

$\rm\displaystyle \tan( \gamma ) = - \bigg(\frac{ \tan( \alpha ) + \tan( \beta ) }{ 1 - \tan( \alpha ) \tan( \beta ) } \bigg)$

recall angle sum indentity of tan:

$\rm\displaystyle \tan( \gamma ) = - \tan( \alpha + \beta )$

let α+β be t and transform:

$\rm\displaystyle \tan( \gamma ) = - \tan( t)$

remember that tan(t)=tan(t±kπ) so

$\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta\pm k\pi )$

therefore <u>when</u><u> </u><u>k </u><u>is </u><u>1</u> we obtain:

$\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta\pm \pi )$

remember Opposite Angle identity of tan function i.e -tan(x)=tan(-x) thus

$\rm\displaystyle \tan( \gamma ) = \tan( -\alpha -\beta\pm \pi )$

recall that if we have common trigonometric function in both sides then the angle must equal which yields:

$\rm\displaystyle \gamma = - \alpha - \beta \pm \pi$

isolate -α-β to left hand side and change its sign:

$\rm\displaystyle \alpha + \beta + \gamma = \boxed{ \pm \pi }$

<u>when</u><u> </u><u>i</u><u>s</u><u> </u><u>0</u>:

$\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta \pm 0 )$

likewise by Opposite Angle Identity we obtain:

$\rm\displaystyle \tan( \gamma ) = \tan( -\alpha -\beta\pm 0 )$

recall that if we have common trigonometric function in both sides then the angle must equal therefore:

$\rm\displaystyle \gamma = - \alpha - \beta \pm 0$

isolate -α-β to left hand side and change its sign:

$\rm\displaystyle \alpha + \beta + \gamma = \boxed{ 0 }$

and we're done!

8 0

3 years ago

Read 2 more answers

Select the graph of the solution. Click until the correct graph appears.<br> x ≥ 4

Mkey [24]

I think the correct answer from the choices listed above would be the second graph. The given inequality is given as x is greater than or equal to 4. So, the graph should include the value 4 then to the right of 4 in the number line. Hope this answers the question.

7 0

3 years ago

Read 2 more answers

Help me on this one please

shepuryov [24]

Answer:

The equation is set up wrong

Step-by-step explanation:

$9^{2} +b^{2} =15^{2}$

3 0

3 years ago