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Ann [662]
3 years ago
9

A town has accumulated 2 inches of​ snow, and the snow depth is increasing by 6 inches every hour. A nearby town has accumulated

10 ​inches, and the depth is increasing by 2 inches every hour. In about how many hours will the snowfall of the towns be​ equal?
Mathematics
1 answer:
aleksandrvk [35]3 years ago
8 0
A=2+6t, b=10+2t

a=b when

2+6t=10+2t  subtract 2t from both sides

2+4t=10  subtract 2 from both sides

4t=8  divide both sides by 4

t=2

So they will have the same snowfall in two hours.
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Answer:

We know that the map is of North America:

The probabilities are:

1) North America:

As the map is a map of North America, you can point at any part of the map and you will be pointing at North America, so the probability is p = 1

or 100% in percentage form.

2) New York City.

Here we can think this as:

The map of North America is an extension of area, and New Yorck City has a given area.

As larger is the area of the city, more probable to being randomly choosen, so to find the exact probability we need to find the quotient between the area of New York City and the total area of North America:

New York City = 730km^2

North America = 24,709,000 km^2

So the probability of randomly pointing at New York City is:

P = ( 730km^2)/(24,709,000 km^2) = 3x10^-5 or 0.003%

3) Europe:

As this is a map of Noth America, you can not randomly point at Europe in it (Europe is other continent).

So the probaility is 0 or 0%.

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Which function has a different range than the function f(x)=bx?
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2x+3=x+x+3<br> Show your work
Nitella [24]

Step-by-step explanation:

2x+3=x+x+3

add the X's on the right side together.

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Can someone help me with this question on Prodigy? ​
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Answer:

  8 +(3/8)√53 in² ≈ 10.73 in²

Step-by-step explanation:

Given the net of a triangular pyramid with some of the dimensions filled in, you want to find the total surface area.

<h3>Triangle base</h3>

The triangle bases identified by dashed lines will have a length equal to the hypotenuse of the right triangles with legs shown as solid lines. The legs of each of those right triangles are ...

  a = (3 in)/2 = 1.5 in

  b = 2 in . . . . . . shown as the altitude of the triangle

Then the hypotenuse is found using the Pythagorean theorem:

  c² = a² +b²

  c² = 1.5² +2² = 2.25 +4 = 6.25

  c = √6.25 = 2.5

The dashed lines are 2.5 inches long.

<h3>Triangle altitude</h3>

The altitude from the solid horizontal line to the vertex at the bottom of the figure can be found using the fact that all of the outside edge lengths of the net are the same length. That edge length is found as the length of the hypotenuse of the right triangles in the left- and right-sides of the upper portion of the net. Each of those has a leg that is (2.5 in)/2 = 1.25 in and a leg marked as 2 in.

  c² = a² +b²

  c² = 1.25² +2² = 1.5625 +4 = 5.5625

  c = (√89)/4 ≈ 2.358 . . . in

The unmarked altitude of the bottom triangle is then ...

  b² = c² -a²

  b² = 89/16 -1.5² = 53/16

  b = (√53)/4 ≈ 1.820 . . . in

<h3>Surface area</h3>

The surface area of the figure is the sum of the areas of the four triangles that make up the net. Each triangle has an area given by the formula ...

  A = 1/2bh

The left and right triangles have b=2.5, h=2, so they each have an area of ...

  A = 1/2(2.5)(2) = 2.5 . . . . in²

The center triangle has dimensions of b=3, h=2, so an area of ...

  A = 1/2(3)(2) = 3 . . . . in²

The bottom triangle has dimensions of b=3, h=(√53)/4, so an area of ...

  A = 1/2(3)(√53/4) = (3/8)√53 ≈ 2.730 . . . . in²

The total surface area is the sum of the areas of these triangles, so is ...

  A = 2.5 in² +2.5 in² +3 in² +2.73 in² = 10.73 in²

The surface area of the triangular pyramid is (64+3√53)/8 ≈ 10.73 in².

__

<em>Additional comment</em>

Often we work with pyramids that are rotationally symmetrical about a vertical line through the peak. This one is not. The altitude of the bottom triangle in the net is less than the altitude of the other triangles. This short face of the pyramid will tend to be more vertical than the other two lateral faces.

4 0
1 year ago
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