The equation of the line is ?

Mathematics

1 answer:

larisa86 [58]2 years ago

7 0

Answer:

Y = mx + b

y = how far up

x = how far along

m = Slope or Gradient (how steep the line is)

b = value of y when x=0

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Ji-Hun and Kana are running a marathon. Due to the volume of people running, the starts occur in waves. Kana starts at 9:35 a.m.

iris [78.8K]

Answer:

23.52 miles

Step-by-step explanation:

Let us find the number of miles that they would have run.

Kana starts at 9:35 a.m. and runs at an average rate of 7 mph.

Let the time she has spent so far be t.

Speed is given as:

s = d / t

where d = distance and t = time

=> d = s * t

Therefore, for Kana:

d = 7 * t = 7t _____________(1)

Ji-Hun starts at 10:00 am (25 minutes after Hannah) and runs at an average rate of 8 mph.

25 minutes = 25/60 hour = 0.42 hour

So, his time (compared to Kana's) is t - 0.42.

therefore, for Ji-Hun:

d = 8 * (t - 0.42)

=> d = 8t - 3.36 ____________(2)

Equating (1) and (2):

7t = 8t - 3.36

=> 8t - 7t = 3.36

t = 3.36 hours

Therefore, let us find d:

d = 7 * 3.36 = 23.52 miles

So, Ji-Hun will catch up to Kana after 23.52 miles.

6 0

3 years ago

Differentiate y = 8x/ 3 − tan(x)

vagabundo [1.1K]

Answer:

$\frac{dy}{dx}=\frac{8(xsec^2(x)-tan(x)+3)}{(3-tan(x))^2}$

Step-by-step explanation:

$y=\frac{8x}{3-tan(x)}\\ \\\frac{dy}{dx}=\frac{(3-tan(x))(\frac{d}{dx}8x)-(\frac{d}{dx}(3-tan(x)))(8x)}{(3-tan(x))^2}\\ \\ \frac{dy}{dx}=\frac{(3-tan(x))(8)-(-sec^2(x))(8x)}{(3-tan(x))^2}\\ \\ \frac{dy}{dx}=\frac{24-8tan(x)+8xsec^2(x)}{(3-tan(x))^2}\\ \\ \frac{dy}{dx}=\frac{8xsec^2(x)-8tan(x)+24}{(3-tan(x))^2}\\\\ \frac{dy}{dx}=\frac{8(xsec^2(x)-tan(x)+3)}{(3-tan(x))^2}$