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2 years ago

10

Given the conversion factor 12 inches/ 1 foot, what is the line segment's length in feet?

2 answers:

serg [7]2 years ago

7 0

103.2/12 is 8.6
A. 8.6 Feet

vovangra [49]2 years ago

3 0

Answer:

8.6 feet

Step-by-step explanation:

This is a very simple way to shovel the problem

103.2/ 12=8.6 .

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A train left Podunk and traveled north at 75 km/h. Two hours later, another train left Podunk and traveled in the same direction

Andreyy89

Answer:

8 hours

Steps:

Train 1:

Hours: 1, 2, 3, 4, 5, 6, 7, 8

Miles: 75, 150, 225, 300, 375, 450, 525, 600

Train 2:

Hours: 1, 2, 3, 4, 5, 6, 7, 8

Miles: 0, 0, 100, 200, 300, 400, 500, 600

6 0

3 years ago

What is the value of t? <br> 2<br> 3<br> 6<br> 8

shepuryov [24]

Answer:

3

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

What is the point-slope form given the following slope and a point on the line? m = 3/7 , (7, -4)

GarryVolchara [31]

Answer:

The equation is y = (3/7)x - 7.

Step-by-step explanation:

Given that the point slope form of an equation is y = mx + b where m represents the gradient and b is the y-intercept. Then, you have to substitute the following values into the equation :

$y = mx + b$

$let \: m = \frac{3}{7} ,x = 7,y = - 4$

$- 4 = \frac{3}{7} (7) + b$

$- 4 = 3 + b$

$- 4 - 3 = b$

$b = - 7$

7 0

3 years ago

Mr Adams has poster paper that is 10 3/4 feet long. He wants to make sheets that are 1/4 of a foot long to make paper airplanes.

8090 [49]

Answer:

43

Step-by-step explanation:

divide 10 3/4 by 1/4

4 0

2 years ago

A plane flying horizontally at an altitude of "1" mi and a speed of "430" mi/h passes directly over a radar station. Find the ra

Anika [276]

Answer:

The rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station is 372 mi/h.

Step-by-step explanation:

Given information:

A plane flying horizontally at an altitude of "1" mi and a speed of "430" mi/h passes directly over a radar station.

$z=1$

$\frac{dx}{dt}=430$

We need to find the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station.

$y=2$

According to Pythagoras

$hypotenuse^2=base^2+perpendicular^2$

$y^2=x^2+1^2$

$y^2=x^2+1$ .... (1)

Put z=1 and y=2, to find the value of x.

$2^2=x^2+1^2$

$4=x^2+1$

$4-1=x^2$

$3=x^2$

Taking square root both sides.

$\sqrt{3}=x$

Differentiate equation (1) with respect to t.

$2y\frac{dy}{dt}=2x\frac{dx}{dt}+0$

Divide both sides by 2.

$y\frac{dy}{dt}=x\frac{dx}{dt}$

Put $x=\sqrt{3}$ , y=2, $\frac{dx}{dt}=430$ in the above equation.

$2\frac{dy}{dt}=\sqrt{3}(430)$

Divide both sides by 2.

$\frac{dy}{dt}=\frac{\sqrt{3}(430)}{2}$

$\frac{dy}{dt}=372.390923627$

$\frac{dy}{dt}\approx 372$

Therefore the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station is 372 mi/h.

6 0

3 years ago