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Elza [17]
3 years ago
7

he total length of a road trip was 15 hours. If highway signs are posted every 0.6 hours, including one at the end of the road t

rip, how many highway signs will there be on the road trip?
Mathematics
1 answer:
lianna [129]3 years ago
5 0

Answer:

26 highway signs

Step-by-step explanation:

15 ÷ 0.6 = 25 highway signs

Since they mentioned that there was one at the end of the road trip,

25 + 1 = 26

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I'll talk you through it so you can see why it's true, and then
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Look at the two pointy triangles, hanging down like moth-wings
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-- The angle between their long side and short side ... the two angles up at 'O',
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These triangles are not congruent the way they are now, because one is
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Daily snowfall in Omaha in January<br> Mode, median or mean?
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3 years ago
A quadrilateral has vertices at $(0,1)$, $(3,4)$, $(4,3)$ and $(3,0)$. Its perimeter can be expressed in the form $a\sqrt2+b\sqr
seraphim [82]

Answer:

a + b = 12

Step-by-step explanation:

Given

Quadrilateral;

Vertices of (0,1), (3,4) (4,3) and (3,0)

Perimeter = a\sqrt{2} + b\sqrt{10}

Required

a + b

Let the vertices be represented with A,B,C,D such as

A = (0,1); B = (3,4); C = (4,3) and D = (3,0)

To calculate the actual perimeter, we need to first calculate the distance between the points;

Such that:

AB represents distance between point A and B

BC represents distance between point B and C

CD represents distance between point C and D

DA represents distance between point D and A

Calculating AB

Here, we consider A = (0,1); B = (3,4);

Distance is calculated as;

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

(x_1,y_1) = A(0,1)

(x_2,y_2) = B(3,4)

Substitute these values in the formula above

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

AB = \sqrt{(0 - 3)^2 + (1 - 4)^2}

AB = \sqrt{( - 3)^2 + (-3)^2}

AB = \sqrt{9+ 9}

AB = \sqrt{18}

AB = \sqrt{9*2}

AB = \sqrt{9}*\sqrt{2}

AB = 3\sqrt{2}

Calculating BC

Here, we consider B = (3,4); C = (4,3)

Here,

(x_1,y_1) = B (3,4)

(x_2,y_2) = C(4,3)

Substitute these values in the formula above

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

BC = \sqrt{(3 - 4)^2 + (4 - 3)^2}

BC = \sqrt{(-1)^2 + (1)^2}

BC = \sqrt{1 + 1}

BC = \sqrt{2}

Calculating CD

Here, we consider C = (4,3); D = (3,0)

Here,

(x_1,y_1) = C(4,3)

(x_2,y_2) = D (3,0)

Substitute these values in the formula above

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

CD = \sqrt{(4 - 3)^2 + (3 - 0)^2}

CD = \sqrt{(1)^2 + (3)^2}

CD = \sqrt{1 + 9}

CD = \sqrt{10}

Lastly;

Calculating DA

Here, we consider C = (4,3); D = (3,0)

Here,

(x_1,y_1) = D (3,0)

(x_2,y_2) = A (0,1)

Substitute these values in the formula above

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

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

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

DA = \sqrt{9 +  1}

DA = \sqrt{10}

The addition of the values of distances AB, BC, CD and DA gives the perimeter of the quadrilateral

Perimeter = 3\sqrt{2} + \sqrt{2} + \sqrt{10} + \sqrt{10}

Perimeter = 4\sqrt{2} + 2\sqrt{10}

Recall that

Perimeter = a\sqrt{2} + b\sqrt{10}

This implies that

a\sqrt{2} + b\sqrt{10} = 4\sqrt{2} + 2\sqrt{10}

By comparison

a\sqrt{2} = 4\sqrt{2}

Divide both sides by \sqrt{2}

a = 4

By comparison

b\sqrt{10} = 2\sqrt{10}

Divide both sides by \sqrt{10}

b = 2

Hence,

a + b = 2 + 10

a + b = 12

3 0
3 years ago
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