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2 years ago

12

Help pleasee having a bad day,and really stressed out

2 answers:

Elanso [62]2 years ago

5 0

Answer:

anything

Step-by-step explanation:

Virty [35]2 years ago

3 0

Answer:

anything :)))

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How many degrees warmer is a temperature of 13°F than a temperature of -19°F?

Nata [24]

It'd be 32 degrees F warmer

13-(-19)
=32

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2 years ago

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HELLP anyone know the answer?!

katovenus [111]

Answer: 212998.58

Step-by-step explanation:

plz mark me brainliest

if the price increases 6% every year for 7 years you would have to multiply 7 by 6 which is 42 so you have to add 42% to 149,999 which is 212998.58.

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Xelga [282]

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this is the answer.

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2 years ago

(a) Find the equation of the normal to the hyperbola x = 4t, y = 4/t at the point (8,2).

gtnhenbr [62]

The slope of the tangent line to the curve at (8, 2) is given by the derivative $\frac{dy}{dx}$ at that point. By the chain rule,

$\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{dx} = \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}$

Differentiate the given parametric equations with respect to $t$ :

$x = 4t \implies \dfrac{dx}{dt} = 4$

$y = \dfrac4t \implies \dfrac{dy}{dt} = -\dfrac4{t^2}$

Then

$\dfrac{dy}{dx} = \dfrac{-\frac4{t^2}}4 = -\dfrac1{t^2}$

We have $x=8$ and $y=2$ when $t=2$ , so the slope at the given point is $\frac{dy}{dx} = -\frac14$ .

The normal line to the same point is perpendicular to the tangent line, so its slope is +4. Then using the point-slope formula for a line, the normal line has equation

$y - 2 = 4 (x - 8) \implies \boxed{y = 4x - 30}$

Alternatively, we can eliminate the parameter and express $y$ explicitly in terms of $x$ :

$x = 4t \implies t = \dfrac x4 \implies y = \dfrac4t = \dfrac4{\frac x4} = \dfrac{16}x$

Then the slope of the tangent line is

$\dfrac{dy}{dx} = -\dfrac{16}{x^2}$

At $x = 8$ , the slope is again $-\frac{16}{64}=-\frac14$ , so the normal has slope +4, and so on.

5 0

1 year ago

What is the exact value of arccos(−2√2) ?

Strike441 [17]

Answer:

See explanation

Step-by-step explanation:

The range of the cosine function is $-1\le \cos x\le 1$

Therefore $arc \cos (-2\sqrt{2})$ is not defined.

Assuming your question is rather $arc \cos (-\frac{\sqrt{2}}{2})$

Then $arc \cos (-\frac{\sqrt{2}}{2})=\pi-arc \cos (\frac{\sqrt{2}}{2})$ in the second quadrant.

$arc \cos (-\frac{\sqrt{2}}{2})=\pi-\frac{\pi}{4}$

$arc \cos (-\frac{\sqrt{2}}{2})=\frac{3\pi}{4}$

4 0

2 years ago