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2 years ago

If y = x^2 and dx/dt = 3, what is dy/dt when x = -1

1 answer:

5 0

<span>If x = y^3 - y and dy/dt = 5, then what is dx/dt when y = 2? 10. A balloon is rising vertically above a level, straight ...

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Two horses are ready to return to their barn after a long workout session at the track. The horses are at coordinates H(1,10) an

Eva8 [605]

This is hard 234669jajj

6 0

2 years ago

A snail travels at a rate of 2.36 feet per minute. a. Write a rule to describe the function. b. How far will the snail travel in

Andreyy89

Y = 2.36x....where y is the distance traveled and x is the number of minutes

how far will the snail travel in 6 minutes...so sub in 6 for x
y = 2.36(6)
y = 14.16 ft <==

6 0

3 years ago

The ratio of the perimeter of two similar polygons is 5:7. The area of the smaller polygon is 30 sq. cm. Find the area of the la

Debora [2.8K]

Answer:

42sq

Step-by-step explanation:

5:7

30 sq: x sq

30/5=6 sq for 1 of the ratio

6*7=42 sq for 7 of the ratio

4 0

2 years ago

Find the smallest integer n for an O(xn) estimate of order of the function f(x)

alexdok [17]

Answer with explanation:

$f(x)=\frac{(x^3+x^2+5)(x^4-3x^2)}{(2x^2+2x-3)(4x^2+7)}\\\\f(x)=\frac{x^3\times (x^4-3x^2)+x^2 \times (x^4-3x^2)+5 \times (x^4-3x^2)}{2x^2\times (4x^2+7)+2x(4x^2+7)-3\times (4x^2+7)}\\\\f(x)=\frac{x^7-3x^5+x^6-3x^4+5x^4-15x^2}{8x^4+14x^2+8x^3+14 x-12x^2-21}\\\\f(x)=\frac{x^7+x^6-3x^5+2x^4-15x^2}{8x^4+8x^3+2x^2+14 x-21}$

The largest degree of numerator is 7 , while the largest degree of denominator is 4.So, as we know

$\frac{x^a}{x^b}=x^{a-b}\\\\\frac{x^7}{x^4}=x^{7-4}\\\\=x^3$

→So, order of the function is the highest degree in the function raised to power.Highest degree is 3,when you will reduce the function in Expression form.

So, Degree=3

Order=1

6 0

3 years ago

If sin(x) = 5/13, and x is in quadrant 1, then tan(x/2) equals what?

Rufina [12.5K]

$x$ is in quadrant I, so $0$ , which means $0$ , so $\dfrac x2$ belongs to the same quadrant.

Now,

$\tan^2\dfrac x2=\dfrac{\sin^2\frac x2}{\cos^2\frac x2}=\dfrac{\frac{1-\cos x}2}{\frac{1+\cos x}2}=\dfrac{1-\cos x}{1+\cos x}$

Since $\sin x=\dfrac5{13}$ , it follows that

$\cos^2x=1-\sin^2x\implies \cos x=\pm\sqrt{1-\left(\dfrac5{13}\right)^2}=\pm\dfrac{12}{13}$

Since $x$ belongs to the first quadrant, you take the positive root ( $\cos x>0$ for $x$ in quadrant I). Then

$\tan\dfrac x2=\pm\sqrt{\dfrac{1-\frac{12}{13}}{1+\frac{12}{13}}}$

$\tan x$ is also positive for $x$ in quadrant I, so you take the positive root again. You're left with

$\tan\dfrac x2=\dfrac15$

4 0

2 years ago