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Tomtit [17]
2 years ago
11

Sharad went to his friend's place on a motorcycle. When he started, the reading was 19,235 km. When he reached his friend's plac

e it was 19,301 km. How many km did sharad travel ?
Mathematics
1 answer:
Liono4ka [1.6K]2 years ago
5 0

Answer:

the number of km did sharad travel is 66 km

Step-by-step explanation:

The computation of the number of km did sharad travel is as follows:

= The km at his friend place - km when he started

= 19,301 km - 19,235 km

= 66 km

Hence, the number of km did sharad travel is 66 km

So, the same is to be considered by applying the given formula so that the correct distance could come

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If t=38.5 and s=31.4 find S. Round to the nearest tenth
zubka84 [21]

<u>Answer:</u>

The correct answer option is C. S = 54.6°.

<u>Step-by-step explanation:</u>

We are given a right angled triangle with two known sides, s and t.

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For that, we will use sine.

sin S = \frac { s } { t }

sin S = \frac { 3 1 . 4 } { 3 8 . 5 }

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3 years ago
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Step-by-step explanation:

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3 years ago
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Which exponential equation is equivalent to the logarithmic equation below?
julsineya [31]

Answer:

\implies\boxed{ 10^a = 300 }

Step-by-step explanation:

We are provided logarithmic equation , which is ,

\implies log_{10}^{300}= a

  • Here we took the base as 10 , since nothing is mentioned about that in Question .

Say if we have a expoteintial equation ,

\implies a^m = n

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Similarly our required answer will be ,

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3 0
2 years ago
Verify that the function satisfies the three hypotheses of Rolle's Theorem on the given interval. Then find all numbers c that s
WINSTONCH [101]

Answer:  \bold{c=\dfrac{1+\sqrt{19}}{3}\approx1.8}

<u>Step-by-step explanation:</u>

There are 3 conditions that must be satisfied:

  1. f(x) is continuous on the given interval
  2. f(x) is differentiable
  3. f(a) = f(b)

If ALL of those conditions are satisfied, then there exists a value "c" such that c lies between a and b and f'(c) = 0.

f(x) = x³ - x² - 6x + 2     [0, 3]

1. There are no restrictions on x so the function is continuous \checkmark

2. f'(x) = 3x² - 2x - 6 so the function is differentiable \checkmark

3. f(0) =  0³ - 0² - 6(0) + 2 = 2

   f(3) =  3³ - 3² - 6(3) + 2  = 2

   f(0) = f(3) \checkmark

f'(x) = 3x² - 2x - 6 = 0

This is not factorable so you need to use the quadratic formula:

x=\dfrac{-(-2)\pm \sqrt{(-2)^2-4(3)(-6)}}{2(3)}\\\\\\.\quad =\dfrac{2\pm \sqrt{4+72}}{2(3)}\\\\\\.\quad =\dfrac{2\pm \sqrt{76}}{2(3)}\\\\\\.\quad =\dfrac{2\pm 2\sqrt{19}}{2(3)}\\\\\\.\quad =\dfrac{1\pm \sqrt{19}}{3}\\\\\\.\quad \approx\dfrac{1+4.4}{3}\quad and\quad \dfrac{1-4.4}{3}\\\\\\.\quad \approx1.8\qquad and\quad -1.1

Only one of these values (1.8) is between 0 and 3.

5 0
3 years ago
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