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Charra [1.4K]
3 years ago
15

Equation of the parabola

Mathematics
1 answer:
castortr0y [4]3 years ago
5 0

Answer:

(x-h)2=4p (y-k)

Step-by-step explanation:

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seraphim [82]
Sorry not in high school
7 0
2 years ago
Ann is 5 feet 3 inches tall. To find the height of a lamppost, she measured her shadow to be 8 feet 9 inches and the lamppost’s
Makovka662 [10]

Answer: The lamppost is 7 feet 2 inches

Step-by-step explanation: If Ann measured her own height and her shadow, then what she used is a ratio between both measurements. If she can measure the shadow of the lamppost, then she can use the same ratio of her height and it’s shadow to derive the correct measurement of the lamppost.

If Ann’s height was measured as 5 feet 3 inches, and her shadow was 8 feet 9 inches, the ratio between them can be expressed as 3:5.

Reduce both dimensions to the same unit, that is, inches. (Remember 12 inches = 1 foot)

Ratio = 63/105

Reduce to the least fraction

Ratio = 3/5

If the height of the lamppost is H, then

H/144 = 3/5

H = (144 x3)/5

H = 86.4

Therefore the lamppost is approximately 86 inches, that is 7 feet and 2 inches tall.

5 0
2 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
What's the estimate of 608×8 ​
rusak2 [61]

Answer:

3,200

Step-by-step explanation:

you would have to multiply then estimate wich round it down an dyou get 3,200

3 0
2 years ago
8. Find an equation in standard form for the ellipse with the vertical major axis of length 16 and minor axis of length 4
MA_775_DIABLO [31]

Hello,

If center is (0,0)

x²/a²+y²/b²=1 for a half horizontal axis of a, and a half vertical axis of b

Here

a=2 ==> a²=4

b=8 ==> b²=64

\boxed{\dfrac{x^2}{4}+ \dfrac{y^2}{64}=1}\\

3 0
2 years ago
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