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Oduvanchick [21]
3 years ago
15

How does understanding place value help you to subtract across zeros?

Mathematics
1 answer:
Lesechka [4]3 years ago
6 0
Zero is like a negative number. It is mathematically impossible to subtract across zeros
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Runner A is initially 6.0 km west of a flagpole and is running with a constant velocity of 3.0 km/h due east. Runner B is initia
ArbitrLikvidat [17]

Answer:

The distance of the two runners from the flagpole is 0.0882 km

Step-by-step explanation:

We are going to use the minus sign when the runner is on the west and the plus sign when the runner is on the east.

Then, Taking into account  the runner A is 6 km west and is running with a constant velocity of 3 Km/h east, the distance of the runner A from the flagpole is given by the following equation:

Xa = -6 Km + (3 Km/h)* t

Where Xa is the position of the runner A from the flagpole and t is the time in hours.

At the same way the distance of the runner B, Xb, from the flagpole is given by the following equation:

Xb = 7.4 Km - (3.8 Km/h)*t

Then, the two runners cross their path when Xa is equal to Xb, so if we solve this equation for t, we get:

              Xa = Xb

     -6 + (3*t) =  7.4 - (3.8*t)

(3.8*t) + (3*t) = 7.4 + 6

         (6.8*t) = 13.4

                  t = 13.4/6.8

                  t = 1.9706

Therefore, at time t equal to 1.9706 hours, both runners cross their path. The distance of the two runners from the flagpole can be calculated replacing the value of t in equation for Xa or in equation for Xb as:

Xa = -6 Km + (3 Km/h)* t

Xa = -6 Km + (3 Km/h)*(1.9706 h)

Xa = -0.0882 Km

That means that both runners are 0.0882 Km west of a flagpole.

4 0
3 years ago
Evaluate the given integral by changing to polar coordinates. sin(x2 y2) dA R , where R is the region in the first quadrant betw
alexira [117]

Answer:

I = 1.47001

Step-by-step explanation:

we have the function

f(x,y)=sin(x^2y^2)\\

In polar coordinates we have

x=rcos\theta\\y=rsin\theta

and dA is given by

dA=rdrd\theta

Hence, the integral that we have to solve is

I=\int \limt_2^4 \int \limit_0^{\pi /2}sin(r^4cos^2\theta sin^2\theta)rdrd\theta

This integral can be solved in a convenient program of your choice (it is very difficult to solve in an analytical way, I use Wolfram Alpha on line)

I = 1.47001

Hope this helps!!!

7 0
3 years ago
Help someone please
padilas [110]
20,30,40,50 then explain that you went up in intervals of 10
5 0
3 years ago
Read 2 more answers
Part B
Leno4ka [110]

Answer:

In y=\sqrt{x-k} , the domain depends on k . As you slide k , you move the left bound of the domain interval. In y=\sqrt{x+k}  , the domain is fixed and always starts at 0 .

Step-by-step explanation:

5 0
3 years ago
How do I find x using an equation?
zzz [600]
 you have to either divide add subtract multiply 
4 0
3 years ago
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