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Ostrovityanka [42]
3 years ago
12

96%

Mathematics
2 answers:
maw [93]3 years ago
4 0

Answer:

  £930  

Step-by-step explanation:

Put the numbers in the formula and do the arithmetic. The annual earnings are expected to be 12 times the monthly earnings.

  T = 0.2(12×£1375 -11850) = 0.2(£16500 -11850) = 0.2(£4650)

  T = £930

The caterer will pay £930 in income tax for the year.

Scilla [17]3 years ago
4 0

Answer:

£930  

Step-by-step explanation:

         Monthly income = £1375

     y = Yearly income  = 12 × 1375 = £16 500

     p = Personal allowance            =    11 850

y - p = Taxable income                   = <u>£ 4 650 </u>

    T = 0.2(y - p) = 0.2 × £4650       = £    930

The income tax for the year will be £930.

 

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Answer:

51.2 oz

Step-by-step explanation:

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The distance between Anubhav’s school and his house is 45 km. He started from his house for school at 7:00 am and covered 30 km
8_murik_8 [283]

Answer:

45 km/h

Step-by-step explanation:

In the first part of the trip, he already covered 30 km so we can subtract that from the total distance

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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,

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