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3 years ago

How long would it take to travel 425 m at the rate of 50 m/sm/s ?

Mathematics

2 answers:

Shtirlitz [24]3 years ago

6 0

Answer:

8.5 s

Step-by-step explanation:

How long means time taken, so to find the time taken to travel 425 m at the rate of 50 m/s, we will simply use the formula below

V = $\frac{d}{t}$

where v is the speed or velocity, d is the distance travelled and t is the time taken.

From the question given, distance (d) = 425 m, velocity/speed(v) = 50

We can now proceed to insert our values into the equation

V = $\frac{d}{t}$

50 = $\frac{425}{t}$

Cross -multiply

50 t = 425

Divide both-side of the equation by 50

$\frac{50t}{50}$ = $\frac{425}{50}$

(On the left-hand side of the equation 50 at the numeartor will cancel-out 50 at the denominator leaving us with just x, while on the right-hand side of the equation 425 will be divided by 50)

t = 8.5 sec.

Therefore the time taken is 8.5 s

Gnom [1K]3 years ago

5 0

The answer to your questions is A 8.5 s

Hope this helps!!

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Linda takes out a loan for $4,000 she will repay the loan in 4 years Linda will pay $560 in interest how much is Linda's loan pa

Schach [20]

Answer:

$1,280 every year and $106.66 every month.

Step-by-step explanation:

You would do 4,000 by 4. Which would be 1,000. Then you would divide 560 by 4. After, you add both of your products together and you get 1,280. Obviously there are 12 months in a year, so you would divide your sum by 12. Then you would get 106.66. Basically, that is how you would get the answer.

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3 years ago

How to solve this trig

n200080 [17]

Hi there!

To find the Trigonometric Equation, we have to isolate sin, cos, tan, etc. We are also given the interval [0,2π).

First Question

What we have to do is to isolate cos first.

$\displaystyle \large{ cos \theta = - \frac{1}{2} }$

Then find the reference angle. As we know cos(π/3) equals 1/2. Therefore π/3 is our reference angle.

Since we know that cos is negative in Q2 and Q3. We will be using π + (ref. angle) for Q3. and π - (ref. angle) for Q2.

Find Q2

$\displaystyle \large{ \pi - \frac{ \pi}{3} = \frac{3 \pi}{3} - \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{2 \pi}{3} }$

Find Q3

$\displaystyle \large{ \pi + \frac{ \pi}{3} = \frac{3 \pi}{3} + \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{4 \pi}{3} }$

Both values are apart of the interval. Hence,

$\displaystyle \large \boxed{ \theta = \frac{2 \pi}{3} , \frac{4 \pi}{3} }$

Second Question

Isolate sin(4 theta).

$\displaystyle \large{sin 4 \theta = - \frac{1}{ \sqrt{2} } }$

Rationalize the denominator.

$\displaystyle \large{sin4 \theta = - \frac{ \sqrt{2} }{2} }$

The problem here is 4 beside theta. What we are going to do is to expand the interval.

$\displaystyle \large{0 \leqslant \theta < 2 \pi}$

Multiply whole by 4.

$\displaystyle \large{0 \times 4 \leqslant \theta \times 4 < 2 \pi \times 4} \\ \displaystyle \large \boxed{0 \leqslant 4 \theta < 8 \pi}$

Then find the reference angle.

We know that sin(π/4) = √2/2. Hence π/4 is our reference angle.

sin is negative in Q3 and Q4. We use π + (ref. angle) for Q3 and 2π - (ref. angle for Q4.)

Find Q3

$\displaystyle \large{ \pi + \frac{ \pi}{4} = \frac{ 4 \pi}{4} + \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{5 \pi}{4} }$

Find Q4

$\displaystyle \large{2 \pi - \frac{ \pi}{4} = \frac{8 \pi}{4} - \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{7 \pi}{4} }$

Both values are in [0,2π). However, we exceed our interval to < 8π.

We will be using these following:-

$\displaystyle \large{ \theta + 2 \pi k = \theta \: \: \: \: \: \sf{(k \: \: is \: \: integer)}}$

Hence:-

For Q3

$\displaystyle \large{ \frac{5 \pi}{4} + 2 \pi = \frac{13 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 4\pi = \frac{21 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 6\pi = \frac{29 \pi}{4} }$

We cannot use any further k-values (or k cannot be 4 or higher) because it'd be +8π and not in the interval.

For Q4

$\displaystyle \large{ \frac{ 7 \pi}{4} + 2 \pi = \frac{15 \pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 4 \pi = \frac{23\pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 6 \pi = \frac{31 \pi}{4} }$

Therefore:-

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Then we divide all these values by 4.

$\displaystyle \large \boxed{\theta = \frac{5 \pi}{16} , \frac{7 \pi}{16} , \frac{13\pi}{16} , \frac{21\pi}{16} , \frac{29\pi}{16}, \frac{15 \pi}{16} , \frac{23\pi}{16} , \frac{31\pi}{16} }$

Let me know if you have any questions!

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3 years ago

When observing patterns, it is possible that more than one true conjecture can be made.

Fudgin [204]

Yes, it is possible.
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3 years ago

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Galina-37 [17]

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3 years ago