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3 years ago

The length of the border between the United States and Canada centimeters

Mathematics

1 answer:

hram777 [196]3 years ago

8 0

Answer:

The length of the border between the United States and Canada in centimeters is 889,162,560 cm

Step-by-step explanation:

The length of the border between the United States and Canada is 5,525 mi

convert into centimeters and you get 889,162,560 cm

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3 years ago

Suppose there are three blue marbles and three yellow marbles in a bag and you want to remove two marbles. You do not replace th

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3 years ago

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A quadratic equation is shown below:

sweet [91]

Answer:

Part A:

( 1.8333, -0.08333)

Part B:

x = 2 or x = 5/3

Step-by-step explanation:

The quadratic equation

$3x^{2}-11x+10=0$ has been given.

Part A:

We are required to determine the vertex. The vertex is simply the turning point of the quadratic function. We shall differentiate the given quadratic function and set the result to 0 in order to obtain the co-ordinates of its vertex.

$\frac{d}{dx}(3x^{2}-11x+10)=6x-11$

Setting the derivative to 0;

6x - 11 = 0

6x = 11

x = 11/6

The corresponding y value is determined by substituting x = 11/6 into the original equation;

y = 3(11/6)^2 - 11(11/6) + 10

y = -0.08333

The vertex is thus located at the point;

( 1.8333, -0.08333)

Find the attached

Part B:

We can use the quadratic formula to solve for x as follows;

The quadratic formula is given as,

$x=\frac{-b+/-\sqrt{b^{2}-4ac } }{2a}$

From the quadratic equation given;

a = 3, b = -11, c = 10

We substitute these values into the above formula and simplify to determine the value of x;

$x=\frac{11+/-\sqrt{11^{2}-4(3)(10) } }{2(3)}=\frac{11+/-\sqrt{1} }{6}\\\\x=\frac{11+/-1}{6}\\\\x=\frac{11+1}{6}=2\\\\x=\frac{11-1}{6}=\frac{5}{3}$

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4 years ago

Solve by finding the square root 2x^2=32

Cerrena [4.2K]

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3 years ago

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The distance between port M and port N is 162m. A ship left port M with a speed of 45 mph. In 45 minutes another ship left port

matrenka [14]

Answer:

After 900 minutes

Step-by-step explanation:

when we convert the velocities of the ships into miles/ min we get

$45mph=45m/60min=0.75m/min$

$36mph=36m/60min=0.6m/min$

The equations determining the distances of the ships from the port M (set at x=0) are

$D_1=0.75t$ (for 45mph ship)

$D_2=162+0.6(t-45)$ (for 36mph ship)

The solution to these equation lie at t= 900 minutes; therefore the two ships will meet 900 minutes after the departure of the first ship.

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3 years ago

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