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ycow [4]
4 years ago
13

Show that the points (3,1) and (1,3) are the same distance from the origin

Mathematics
1 answer:
Phantasy [73]4 years ago
5 0

Let's use the distance formula to find the distance between the origin and each of the points. We can use this to show that the distances are the same. Remember that the distance formula is:

\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

  • (x_1, y_1) is one point
  • (x_2, y_2) is the other point that you are trying the distance to

First, let's find the distance between the points (3, 1) and the origin, (0, 0). We will call this distance D_1.

D_1 = \sqrt{(3 - 0)^2 + (1 - 0)^2}

D_1 = \sqrt{3^2 + 1^2}

D_1 = \sqrt{10}


Next, let's find the distance between (1, 3) and the origin, (0, 0). We will call this D_2.

D_2 = \sqrt{(1 - 0)^2 + (3 - 0)^2}

D_2 = \sqrt{1^2 + 3^2}

D_2 = \sqrt{10}


We can see that D_1 = D_2, or that the distances are the same. ✔︎

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Step-by-step explanation:

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ΔONP and ΔUTV are congruent .

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Step-by-step explanation:

6 0
3 years ago
Kevin and Randy Muise have a jar containing 73 ?coins, all of which are either quarters or nickels. The total value of the coins
inna [77]

Answer:

24 quarters and 49 nickels

Step-by-step explanation:

This situation has two unknowns - the total number of nickels and the total number of quarters. Because we have two unknowns, we will write a system of equations with two equations using the two unknowns.

  • n+q=73 is an equation representing the total number of coins
  • 0.05n+0.25q=8.45 is an equation representing the total value in money based on the number of coin. 0.05 and 0.25 come from the value of a nickel and quarter individually.

We write the first equation in terms of q by subtracting it across the equal sign to get n=73-q. We now substitute this for n in the second equation.

0.05(73-q)+0.25q=8.45

3.65-0.05q+0.25q=8.45

3.65+0.20q=8.45

After simplifying, we subtract 3.65 across and divide by the coefficient of q.

0.20q=4.8

q=24

We now know of the 73 coins that 24 are quarters. To find the number of nickels, we subtract 24 from 73 and get 49 nickels.

4 0
3 years ago
Steve likes to entertain friends at parties with "wire tricks." Suppose he takes a piece of wire 60 inches long and cuts it into
Alex_Xolod [135]

Answer:

a) the length of the wire for the circle = (\frac{60\pi }{\pi+4}) in

b)the length of the wire for the square = (\frac{240}{\pi+4}) in

c) the smallest possible area = 126.02 in² into two decimal places

Step-by-step explanation:

If one piece of wire for the square is y; and another piece of wire for circle is (60-y).

Then; we can say; let the side of the square be b

so 4(b)=y

         b=\frac{y}{4}

Area of the square which is L² can now be said to be;

A_S=(\frac{y}{4})^2 = \frac{y^2}{16}

On the otherhand; let the radius (r) of the  circle be;

2πr = 60-y

r = \frac{60-y}{2\pi }

Area of the circle which is πr² can now be;

A_C= \pi (\frac{60-y}{2\pi } )^2

     =( \frac{60-y}{4\pi } )^2

Total Area (A);

A = A_S+A_C

   = \frac{y^2}{16} +(\frac{60-y}{4\pi } )^2

For the smallest possible area; \frac{dA}{dy}=0

∴ \frac{2y}{16}+\frac{2(60-y)(-1)}{4\pi}=0

If we divide through with (2) and each entity move to the opposite side; we have:

\frac{y}{18}=\frac{(60-y)}{2\pi}

By cross multiplying; we have:

2πy = 480 - 8y

collect like terms

(2π + 8) y = 480

which can be reduced to (π + 4)y = 240 by dividing through with 2

y= \frac{240}{\pi+4}

∴ since y= \frac{240}{\pi+4}, we can determine for the length of the circle ;

60-y can now be;

= 60-\frac{240}{\pi+4}

= \frac{(\pi+4)*60-240}{\pi+40}

= \frac{60\pi+240-240}{\pi+4}

= (\frac{60\pi}{\pi+4})in

also, the length of wire for the square  (y) ; y= (\frac{240}{\pi+4})in

The smallest possible area (A) = \frac{1}{16} (\frac{240}{\pi+4})^2+(\frac{60\pi}{\pi+y})^2(\frac{1}{4\pi})

= 126.0223095 in²

≅ 126.02 in² ( to two decimal places)

4 0
4 years ago
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