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3 years ago

HELP!!!! Passes through the point (−3,4) and is parallel to the graph y=2/3x+5.

Mathematics

1 answer:

Lynna [10]3 years ago

4 0

Answer:

y = 2/3x + 6

Step-by-step explanation:

To do this, make an equation and put it in the form y = mx + b where m is the slope and b is the y-intercept

So first, when lines are parallel they have the same slope. This means that m will be 2/3.

The equation so far is y = 2/3x + b

Next, you have to find what b is, and to do this since you know that x = -3 and y = 4 is a solution you can plug those values into the equation and see what b would have to be for the equation to come out true.

4 = 2/3(-3) + b

4 = -2 + b

b = 6.

The completed equation is y = 2/3x + 6.

Hope this makes sense!

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1) If y = x3 + 2x and dx/ dt = 5, find dy/dt when x = 2. Give only the numerical answer. For example, if dy, dt = 3, type only 3

Semenov [28]

Answer:

a) 70

b) 10π ft²/ft

c) 0.24 ft/sec

Step-by-step explanation:

1) y = x³ + 2x

$\frac{dy}{dt}=\frac{d(x^3+2x)}{dt}$

$\frac{dy}{dt}$ = $3x^2\frac{dx}{dt}+2\frac{dx}{dt}$

at $\frac{dx}{dt}=5$ and x = 2

$\frac{dy}{dt}$ = $3(2)^2\times(5)+2\times(5)$

$\frac{dy}{dt}$ = 60 + 10 = 70

2) A = πr² ft²

$\frac{dA}{dr}=\frac{d(\pi r^2)}{dr}$

$\frac{dA}{dr}$ = 2(πr)

at r = 5 ft

$\frac{dA}{dr}$ = 2(π × 5) ft²/ft

$\frac{dA}{dr}$ = 10π ft²/ft

3) From Pythagoras theorem

Base² + Perpendicular² = Hypotenuse²

Thus,

B² + P² = H² .............(1)

here, H = length of the ladder

P is the height of the wall

B is the distance from the wall at bottom

B² + P² = 25² ...........(1)

at B = 20 ft

20² + P² = 25²

P² = 625 - 400

P = √225

P = 15 ft

differentiating (1) with respect to time, we get

$2B\frac{dB}{dt}+2P\frac{dP}{dt}=0$

at B = 20 ft, $\frac{dB}{dt} = 0.18$ and P = 15 ft

⇒ 2(20)(0.18) + $2(15)\frac{dP}{dt}$ = 0

$30\frac{dP}{dt}$ = - 7.2

$\frac{dP}{dt}$ = - 0.24 ft/sec (Here negative sign depicts the ladder slides down)

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