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3 years ago

At what point on the curve x = 6t2 + 6, y = t3 − 2 does the tangent line have slope 1 /2 ?

Mathematics

1 answer:

Naya [18.7K]3 years ago

6 0

Answer:

Hello,

P=(30,6)

Step-by-step explanation:

$x=6t^2+6\\y=t^3-2\\\\\dfrac{dx}{dt}= 12t\\\dfrac{dy}{dt}= 3t^2\\\\\dfrac{dy}{dx} =\dfrac{\dfrac{dy}{dt} }{\dfrac{dx}{dt} } =\dfrac{3t^2}{12t} =\dfrac{t}{4} \\\\\dfrac{t}{4} =\dfrac{1}{2} \Longrightarrow t=2\\\\\\x=6t^2+6=6*2^2+6=30\\\\y=t^3-2=2^2-2=8-2=6\\\\\\Tangence\ point=(30,6)\\$

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A pitchers arm rotates at a speed of 7 degrees per millisecond

Fantom [35]

Answer:

To solve this problem, all we have to do is to use a conversion factor to convert millisecond into seconds. We know that there are 1000 milliseconds in a second, therefore:

pitcher’s arm rotation = 7 degrees / millisecond * (1000 milliseconds / second)

pitcher’s arm rotation = 7,000 degrees / second

Step-by-step explanation:

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3 years ago

2.3+0.02(x+20)-4.8=-9<br><br> Solve for x

Naya [18.7K]

Answer:

$x=-345$

Step-by-step explanation:

$2.3+0.02\left(x+20\right)-4.8=-9$

$\mathrm{Multiply\:both\:sides\:by\:}100$

$2.3\cdot \:100+0.02\left(x+20\right)\cdot \:100-4.8\cdot \:100=-9\cdot \:100$

$230+2\left(x+20\right)-480=-900$

$\mathrm{Subtract\:}230-480\mathrm{\:from\:both\:sides}$

$230+2\left(x+20\right)-480-\left(230-480\right)=-900-\left(230-480\right)$

$2\left(x+20\right)=-650$

$\mathrm{Divide\:both\:sides\:by\:}2$

$\frac{2\left(x+20\right)}{2}=\frac{-650}{2}$

$x+20=-325$

$\mathrm{Subtract\:}20\mathrm{\:from\:both\:sides}$

$x+20-20=-325-20$

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2 years ago

A function is defined by f(x)=6x+1.5. What is f(2.5)

Nataly_w [17]

f(2.5) means x is equal to 2.5.

Replace x in the equation with 2.5 and solve:

6(2.5) + 1.5 = 15 + 1.5 = 16.5

f(2.5) = 16.5

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3 years ago

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A road has a 10% grade, meaning increasing 1 unit of rise to every 10 units of run.

fenix001 [56]

The grade is the ratio of rise to run, i.e. the slope aka the tangent.

$\tan \theta = \dfrac{1}{10}$

$\theta = \arctan 0.1 \approx 5.711^\circ$

Answer: (a) 6 degrees

For part b, the road is the hypotenuse c of a right triangle whose tangent of the small angle is 1/10. The height h or rise is the side opposite the small angle.

$\sin\theta = \dfrac h c$

$h = c \sin \theta$

We could just take the sine of the angle we got but let's get it from the tangent exactly.

$\cos^2 \theta + \sin ^2 \theta = 1$

Dividing by squared cosine

$1 + \tan ^2 \theta = 1/\cos^2 \theta = 1/(1- \sin^2 \theta)$

$(1- \sin^2 \theta) = 1/(1 + \tan^2 \theta)$

$\sin^2 \theta = 1 - 1/(1 + \tan^2 \theta)$

$\sin^2 \theta = 1 - 1/(1 + (1/10)^2) = 1-1/(101/100) = 1/101$

$h = c \sin \theta = 2 \sqrt{1/101} \approx 0.199$

Answer: (b) Rise of 0.199 km

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3 years ago

Graph: y=|x - 4| + 2

kramer

Answer:

See attached graph.

Step-by-step explanation:

We can graph this with relative ease on the calculator or computer (see attachment) but let's make sure we understand it theoretically and algebraically!

y=|x - 4| + 2

First of all, this is an absolute value function, so it'll look like a v where every input equals its positive output (2, 2) (-3,3) etc. You can't get a negative number with a graph of y=|x| unless you translate it.

Next, there's a -4 inside the abs val. According to the formula y = |x - h| + k, where h is the horizontal shift, this means we need to shift the vertex right 4.

There's a +2 outside the abs val. K is the vertical shift, so we need to shift the vertex up 2.

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3 years ago

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