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3 years ago

X and y are two positive numbers and y is three greater than x. the product of the numbers is 180. find the smaller number, x.

1 answer:

7 0

First positive number (x) = x
Second positive number (y) = x+3
Product: 180 = x · y
   180 = x · (x + 3)
   180 = x² + 3x
   0 = x² + 3x - 180
   0 = (x + 15)(x - 12)
   0 = x + 15, 0 = x - 12
       x = -15, x = 12
x = -15 is NOT a positive number so it can be disregarded.

Answer: 12

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Find dy/dx of y=csc(square root of x)

Vitek1552 [10]

Answer:

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Step-by-step explanation:

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3 years ago

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3 years ago

Solve the equation for all real values of x.<br> cosxtanx - 2 cos x=-1

IceJOKER [234]

Solution to equation $cosxtanx - 2 cos^2 x=-1$ for all real values of x is $x=2k\pi + \frac{\pi}{6} , x=2k\pi + \frac{5\pi}{6}$ .

<u>Step-by-step explanation:</u>

Here we have , $cosxtanx - 2 cos^2 x=-1$ . Let's solve :

⇒ $cosxtanx - 2 cos^2 x=-1$

⇒ $cosx(\frac{sinx}{cosx}) - 2 cos^2 x=-1$

⇒ $sinx = 2 cos^2 x-1$

⇒ $sinx = 2 (1-sin^2x)-1$

⇒ $sinx = 1-2sin^2x$

⇒ $2sin^2x+sinx-1=0$

By quadratic formula :

⇒ $sinx = \frac{-b \pm \sqrt{b^2-4ac} }{2a}$

⇒ $sinx = \frac{-1 \pm \sqrt{1^2-4(2)(-1)} }{2(2)}$

⇒ $sinx = \frac{-1 \pm3}{4}$

⇒ $sinx = \frac{1}{2} , sinx =-1$

⇒ $sinx = sin\frac{\pi}{6} , sinx = sin\frac{3\pi}{2}$

⇒ $x=\frac{\pi}{6} , x=\frac{3\pi}{2}$

But at $x=\frac{3\pi}{2}$ we have equation undefined as $cos\frac{3\pi}{2}=0$ . Hence only solution is :

⇒ $x=\frac{\pi}{6}$

Since , $sin(\pi -x)=sinx$

⇒ $x=\pi -\frac{\pi}{6} = \frac{5\pi}{6}$

Now , General Solution is given by :

⇒ $x=2k\pi + \frac{\pi}{6} , x=2k\pi + \frac{5\pi}{6}$

Therefore , Solution to equation $cosxtanx - 2 cos^2 x=-1$ for all real values of x is $x=2k\pi + \frac{\pi}{6} , x=2k\pi + \frac{5\pi}{6}$ .

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3 years ago

Ben and Josh went to the roof of their 40-foot tall high school to throw their math books off the edge.

krok68 [10]

Answer:

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Step-by-step explanation:

Using the given equation for v=60 and s=40, the height of Ben's book is ...

h(t) = -16t² +60t +40

We want to find t when h(t) = 0, so we're looking for the solution to ....

0 = -16t² +60t +40

Using the quadratic formula, we find the positive value of t to be ...

t = (-60 -√(60² -4(-16)(40)))/(2(-16)) = (15 +√385)/8 ≈ 4.3277

Similarly, the height of Josh's book is ...

0 = -16t² +48t +40

t = (-48 -√(48² -4(-16)(40)))/(2(-16)) = (12 +√304)/8 ≈ 3.6794

The time before Josh's book lands is shorter by ...

4.3277 -3.6794 ≈ 0.6482 . . . . . seconds

Josh's book reaches the ground first, by about 0.648 seconds.

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Step-by-step explanation:

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