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1 year ago

A line that includes the point (4,1) has a slope of 1 7 . What is its equation in point-slope form?

Mathematics

1 answer:

Sergeeva-Olga [200]1 year ago

5 0

Answer:

y = 17x-67

Step-by-step explanation:

The one point form of line also called slope point form is given by

$(y-y_1)=m(x-x_1)$

where (x1, y1) is the point and m is slope

Given: (x1, y1) = (4,1) and m= 17

Substituting these values in above equation to find equation in point-slope form

(y-1)=17(x-4)

y-1 = 17x-68

y = 17x-67

the point -slope form of equation y = 17x-67

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I'm confused like in #1 the positive is five but what do I put for the negative

yuradex [85]

You found $5, so it's positive. You don't put anything in the negative, because you aren't losing anything.

4 0

2 years ago

The area of the shaded circle below is 78.5 square inches. The area of the large circle is 314 square inches.

umka2103 [35]

From the given description it seems that the small shaded circle lies within the larger circle as shown in image below:

The probability that a point choosen at random lies within the shaded area = Area of shaded region / Total area of the bigger cricle

So,
Probability = 78.5/314 = 0.25

Thus there is 0.25 or 25% probability that the point chosen at random will lie inside the shaded region.

6 0

2 years ago

Read 2 more answers

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Archy [21]

(a) If the particle's position (measured with some unit) at time t is given by s(t), where

$s(t) = \dfrac{5t}{t^2+11}\,\mathrm{units}$

then the velocity at time t, v(t), is given by the derivative of s(t),

$v(t) = \dfrac{\mathrm ds}{\mathrm dt} = \dfrac{5(t^2+11)-5t(2t)}{(t^2+11)^2} = \boxed{\dfrac{-5t^2+55}{(t^2+11)^2}\,\dfrac{\rm units}{\rm s}}$

(b) The velocity after 3 seconds is

$v(3) = \dfrac{-5\cdot3^2+55}{(3^2+11)^2} = \dfrac{1}{40}\dfrac{\rm units}{\rm s} = \boxed{0.025\dfrac{\rm units}{\rm s}}$

$\dfrac{-5t^2+55}{(t^2+11)^2} = 0 \implies -5t^2+55 = 0 \implies t^2 = 11 \implies t=\pm\sqrt{11}\,\mathrm s \imples t \approx \boxed{3.317\,\mathrm s}$

(d) The particle is moving in the positive direction when its position is increasing, or equivalently when its velocity is positive:

$\dfrac{-5t^2+55}{(t^2+11)^2} > 0 \implies -5t^2+55>0 \implies -5t^2>-55 \implies t^2 < 11 \implies |t|$

In interval notation, this happens for t in the interval (0, √11) or approximately (0, 3.317) s.

(e) The total distance traveled is given by the definite integral,

$\displaystyle \int_0^8 |v(t)|\,\mathrm dt$

By definition of absolute value, we have

$|v(t)| = \begin{cases}v(t) & \text{if }v(t)\ge0 \\ -v(t) & \text{if }v(t)$

In part (d), we've shown that v(t) > 0 when -√11 < t < √11, so we split up the integral at t = √11 as

$\displaystyle \int_0^8 |v(t)|\,\mathrm dt = \int_0^{\sqrt{11}}v(t)\,\mathrm dt - \int_{\sqrt{11}}^8 v(t)\,\mathrm dt$

and by the fundamental theorem of calculus, since we know v(t) is the derivative of s(t), this reduces to

$s(\sqrt{11})-s(0) - s(8) + s(\sqrt{11)) = 2s(\sqrt{11})-s(0)-s(8) = \dfrac5{\sqrt{11}}-0 - \dfrac8{15} \approx 0.974\,\mathrm{units}$

7 0

2 years ago

please hurry!! i have a D in math

3241004551 [841]

Answer:

60, and m = 30, and 1 min 20.

Step-by-step explanation:

For A, the 60 is the starting point meaning that the temperature was intitally 60 degrees

For B, if we divided both sides by 2 we get the answer m = 30, meaning every minute, it was up 30 degrees.

For C, we know that in one minute it will be 90 degrees, so 60 seconds = 30 degrees. So lets divided both sides by 3 to get 10 degrees. We get 20 seconds. It will take 1 min and 20 for it to be 100 degrees.

Hopefully thats right.

4 0

2 years ago

Cody and his friends started playing a game at 6:30 p.M. It took them it took them 47 minutes to finish the game at what time di

zaharov [31]

Answer:

7:17 PM

Step-by-step explanation:

Add another 30 minutes to 6:30 = 7:00 PM

47-30 = 17

add the remaining 17 minutes to 7:00 = 7:17 PM

7 0

2 years ago