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wel
3 years ago
9

A car travels at a constant speed.

Mathematics
1 answer:
ad-work [718]3 years ago
4 0

Answer:

see below

Step-by-step explanation:

<h3>Given</h3>
  • Distance is 142.2 m, correct to 1 decimal place
  • Time is 7 seconds,  correct to nearest second
<h3>To find:</h3>
  • Upper bound for the speed
<h3>Solution </h3>

<em>Upper bound for the speed  = upper bound for distance/lower bound for time</em>

  • Upper bound for distance = 142.25 m  (added 0.1/5 = 0.05)
  • Lower bound for time = 6.5 seconds  (subtracted 1/2 = 0.5)

<u>Then, the speed is:</u>

  • 142.25/6.5 = 21.88 m/s
  • 21.88 = 21.9 m/s correct to 1 decimal place
  • 21.88 = 22 m/s correct to nearest m/s

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3 years ago
Identify the zeros of f(x)= (x-7)(x+4)(3x-2) Choices:
Kobotan [32]

Answer:

second option

Step-by-step explanation:

Given

f(x) = (x - 7)(x + 4)(3x - 2)

To find the zeros let f(x) = 0, that is

(x - 7)(x + 4)(3x - 2) = 0

Equate each factor to zero and solve for x

x - 7 = 0 ⇒ x = 7

x + 4 = 0 ⇒ x = - 4

3x - 2 = 0 ⇒ 3x = 2 ⇒ x = \frac{2}{3}

zeros are x = - 4, x = \frac{2}{3}, x = 7

7 0
3 years ago
Provide reasons for the proof. Given line h is parallel to line k. line j is perpendicular to line h. Prove line j is perpendicu
dusya [7]

Answer: Provided.


Step-by-step explanation: We are given two lines 'h' and 'k' which are parallel to each other. Also, there is another line 'j' that is perpendicular to line 'h'.

We are to prove that line 'j' is perpendicular to line 'k'.

Let, m, n and p be the slopes of lines 'h', 'k' and 'j' respectively.

Now, since line 'h' and 'k' are parallel, so their slopes will be equal. i.e., m = n.

Also, lines 'h' and 'j' are perpendicular, so the product of their slopes is -1. i.e.,

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Hence, we can write from the above two relations

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Proved.


8 0
4 years ago
Please calculate this limit <br>please help me​
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Answer:

We want to find:

\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n}

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Because here we are taking the limit when n tends to infinity, we can use this approximation.

Then we get.

\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n} = \lim_{n \to \infty} \frac{\sqrt[n]{\sqrt{2*\pi*n} *(\frac{n}{e} )^n} }{n} =  \lim_{n \to \infty} \frac{n}{e*n} *\sqrt[2*n]{2*\pi*n}

Now we can just simplify this, so we get:

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6 0
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