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3 years ago

1 hour is 550 miles how many miles would 1 hour and 42 minutes be

1 answer:

5 0

935 miles. This is because it takes 1 minute to travel 9.16667 miles. There is 102 minutes in 1 hour and 42 minutes, so you take 9.16667 * 102, this gets you 935 miles. Hope this helps.

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The table shows the distance Randy drove on one day of her vacation

horsena [70]

Given the table showing the distance Randy drove on one day of her vacation as follows:

$\begin{tabular}
{|c|c|c|c|c|c|}
Time (h)&1&2&3&4&5\\[1ex]
Distance (mi)&55&110&165&220&275
\end{tabular}$

The rate at which she travels is given by

$rate= \frac{distance}{time} \\ \\ = \frac{55}{1} =55mi/h$

If Randy has driven for one more hour at the same rate, the number of hours she must have droven is 6 hrs and the total distance is given by

distance = 55 x 6 = 330 miles.

4 0

2 years ago

How do you construct the inscribed and circumscribed circle of a triangle?

suter [353]

Answer:

Draw the triangle.

Draw the perpendicular bisector to each side of the triangle. Draw the lines long enough so that you see a point of intersection of all three lines.

Draw the circle with radius at the intersection point of the bisectors that passes through one of the vertices

6 0

3 years ago

CALCULUS - Find the values of in the interval (0,2pi) where the tangent line to the graph of y = sinxcosx is

Rufina [12.5K]

Answer:

$\{\frac{\pi}{4}, \frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\}$

Step-by-step explanation:

We want to find the values between the interval (0, 2π) where the tangent line to the graph of y=sin(x)cos(x) is horizontal.

Since the tangent line is horizontal, this means that our derivative at those points are 0.

So, first, let's find the derivative of our function.

$y=\sin(x)\cos(x)$

Take the derivative of both sides with respect to x:

$\frac{d}{dx}[y]=\frac{d}{dx}[\sin(x)\cos(x)]$

We need to use the product rule:

$(uv)'=u'v+uv'$

So, differentiate:

$y'=\frac{d}{dx}[\sin(x)]\cos(x)+\sin(x)\frac{d}{dx}[\cos(x)]$

Evaluate:

$y'=(\cos(x))(\cos(x))+\sin(x)(-\sin(x))$

Simplify:

$y'=\cos^2(x)-\sin^2(x)$

Since our tangent line is horizontal, the slope is 0. So, substitute 0 for y':

$0=\cos^2(x)-\sin^2(x)$

Now, let's solve for x. First, we can use the difference of two squares to obtain:

$0=(\cos(x)-\sin(x))(\cos(x)+\sin(x))$

Zero Product Property:

$0=\cos(x)-\sin(x)\text{ or } 0=\cos(x)+\sin(x)$

Solve for each case.

Case 1:

$0=\cos(x)-\sin(x)$

Add sin(x) to both sides:

$\cos(x)=\sin(x)$

To solve this, we can use the unit circle.

Recall at what points cosine equals sine.

This only happens twice: at π/4 (45°) and at 5π/4 (225°).

At both of these points, both cosine and sine equals √2/2 and -√2/2.

And between the intervals 0 and 2π, these are the only two times that happens.

Case II:

We have:

$0=\cos(x)+\sin(x)$

Subtract sine from both sides:

$\cos(x)=-\sin(x)$

Again, we can use the unit circle. Recall when cosine is the opposite of sine.

Like the previous one, this also happens at the 45°. However, this times, it happens at 3π/4 and 7π/4.

At 3π/4, cosine is -√2/2, and sine is √2/2. If we divide by a negative, we will see that cos(x)=-sin(x).

At 7π/4, cosine is √2/2, and sine is -√2/2, thus making our equation true.

Therefore, our solution set is:

$\{\frac{\pi}{4}, \frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\}$

And we're done!

Edit: Small Mistake :)

5 0

2 years ago

If you can help 1-12 that’s great if not just help with what you can thanks

bearhunter [10]

(1. -2) (2. 0) (3. 6) (4. 3)

5 0

2 years ago

Please help mee

NemiM [27]

(2^6)^x=1
(4)^x=1
(4)^0=1
1=1
X=0

(5^0)^x=1
(1)^x=1
x=(-∞,∞)

5 0

2 years ago