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stealth61 [152]
3 years ago
6

Matty jogs 1 km in 5 minutes. Compute Matty's speed in m/s.

Mathematics
1 answer:
kotykmax [81]3 years ago
5 0
In km/min it is 0.2km/min while on converting into m/sec..it is 3.33 m/sec........
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F(0)=2, f(2)=4, linear equation?
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Answer:

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Step-by-step explanation:

<em>for </em><em>you </em><em>to </em><em>find </em><em>x </em><em>you </em><em>first </em><em>have </em><em>to </em><em>find </em><em>the </em><em>adjacent</em><em> </em><em>of </em><em>the </em><em>4</em><em>5</em><em>°</em><em> </em><em>angle </em><em>you </em><em>can </em><em>do </em><em>that </em><em>by </em><em>using</em><em> </em><em>the </em><em>other </em><em>triangle.</em><em>u</em><em>s</em><em>i</em><em>n</em><em>g</em><em> </em><em>the </em><em>sin </em><em>ratio</em>

<em>sin60=</em><em>opposite</em><em>/</em><em>hypotenuse</em>

<em>sin60=</em><em>a/</em><em>7</em><em>√</em><em>3</em>

<em>a</em><em>=</em><em>1</em><em>0</em><em>.</em><em>5</em>

<em>then </em><em>after </em><em>you </em><em>have </em><em>found</em><em> </em><em>the </em><em>adjacent</em><em> </em><em>you </em><em>can </em><em>use </em><em>the </em><em>cos </em><em>ratio</em>

<em>cos45=</em><em>adjacent/</em><em>hypotenuse</em>

<em>cos45=</em><em>1</em><em>0</em><em>.</em><em>5</em><em>/</em><em>x</em>

<em>cos45x/</em><em>cos45=</em><em>1</em><em>0</em><em>.</em><em>5</em><em>/</em><em>cos45</em>

<em>x=</em><em>1</em><em>4</em><em>.</em><em>8</em><em>4</em><em>9</em>

<em>which </em><em>is </em><em>the </em><em>same </em><em>as </em><em>2</em><em>1</em><em>√</em><em>2</em><em> </em><em>over </em><em>2</em>

<em>I </em><em>hope</em><em> this</em><em> helps</em>

5 0
3 years ago
Read 2 more answers
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Answer:

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Step-by-step explanation:

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5 0
2 years ago
What is a reasonable estimate for the solution? O (1,-3/4) O (-3/4,1) O (-1,3/4) O (3/4,-1)​
Zina [86]

Answer:

See Explanation

Step-by-step explanation:

Your question is incomplete, as the equations or graph or table(s) were not given.

However, I'll give a general way of solving this.

Take for instance, the equations are:

y = \frac{4}{3}x - 1

y = \frac{2}{3}x - \frac{1}{2}

To do this, we start by equating both equations.

y = y

i.e.

\frac{4}{3}x - 1= \frac{2}{3}x - \frac{1}{2}

Collect Like Terms

\frac{4}{3}x - \frac{2}{3}x= 1 - \frac{1}{2}

Take LCM

\frac{4x- 2x}{3}= \frac{2 - 1}{2}

\frac{2x}{3}= \frac{1}{2}

Cross Multiply

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4x = 3

Make x the subject

x = \frac{3}{4}

Substitute 3/4 for x in y = \frac{4}{3}x - 1

y = \frac{4}{3} * \frac{3}{4} - 1

y = 1 - 1

y = 0

Hence:

(x,y) = (\frac{3}{4},0)

6 0
3 years ago
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