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2 years ago

Latrice is running a race in her neighborhood.

1 answer:

5 0

The unit rates that would be most appropriate wold be : m/s

Speed is calculated by : distance/time

Distance measured in meters while time measured in second

So, m/s will be the most appropriate measurement

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Write the equation of the line that goes through<br> the points (-3, -2) and (4, 5).

sattari [20]

Find the slope m.

Use the point-slope formula.

Solve for y.

6 0

2 years ago

I need these 2 questions answered please.

Sav [38]

The area of the circle is 28 ft square.

The area of the triangle is 30 ft square.

Step-by-step explanation:

1) The area of the circle= π*r^2

where π= 3.14(default value) and "r" is the radius of the circle.

Here, Radius of the circle(r)= 3 ft

Area = 3.14*3*3

= 28.26 ft sq.= 28 ft sq(rounded to the nearest whole number)

2) The triangle has the base= 10 feet and height= 6 feet

The Area of the triangle= 1/2(b)(h)

where "b" is the base and "h" is the height of the triangle.

Substitute b=10 and h=6 ,

Area of triangle= 1/2(10 ft)(6 ft)= 60/2= 30 feet sq.

5 0

3 years ago

Consider the curve defined by the equation y=6x2+14x. Set up an integral that represents the length of curve from the point (−2,

torisob [31]

Answer:

32.66 units

Step-by-step explanation:

We are given that

$y=6x^2+14x$

Point A=(-2,-4) and point B=(1,20)

Differentiate w.r. t x

$\frac{dy}{dx}=12x+14$

We know that length of curve

$s=\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^2}dx$

We have a=-2 and b=1

Using the formula

Length of curve= $s=\int_{-2}^{1}\sqrt{1+(12x+14)^2}dx$

Using substitution method

Substitute t=12x+14

Differentiate w.r t. x

$dt=12dx$

$dx=\frac{1}{12}dt$

Length of curve= $s=\frac{1}{12}\int_{-2}^{1}\sqrt{1+t^2}dt$

We know that

$\sqrt{x^2+a^2}dx=\frac{x\sqrt {x^2+a^2}}{2}+\frac{1}{2}\ln(x+\sqrt {x^2+a^2})+C$

By using the formula

Length of curve= $s=\frac{1}{12}[\frac{t}{2}\sqrt{1+t^2}+\frac{1}{2}ln(t+\sqrt{1+t^2})]^{1}_{-2}$

Length of curve= $s=\frac{1}{12}[\frac{12x+14}{2}\sqrt{1+(12x+14)^2}+\frac{1}{2}ln(12x+14+\sqrt{1+(12x+14)^2})]^{1}_{-2}$

Length of curve= $s=\frac{1}{12}(\frac{(12+14)\sqrt{1+(26)^2}}{2}+\frac{1}{2}ln(26+\sqrt{1+(26)^2})-\frac{12(-2)+14}{2}\sqrt{1+(-10)^2}-\frac{1}{2}ln(-10+\sqrt{1+(-10)^2})$