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Alex17521 [72]
3 years ago
7

56:43 Theo found the driving distance from Glacier National Park to Yellowstone Park to be 448 miles. Theo used a map that had a

ratio of StartFraction 5 centimeters over 320 miles EndFraction. How many centimeters is the distance on the map? Round to the nearest unit if necessary. 4 centimeters 7 centimeters 64 centimeters 90 centimeters
Mathematics
2 answers:
dem82 [27]3 years ago
7 0

Answer:

7 cm

Step-by-step explanation:

Convert 448 miles to the equivalent in cm on the map:

  448 mi          5 cm          2240 cm

--------------- * --------------- = --------------- = 7 cm (which matches the 2nd ans.)

       1             320 mi           320

FinnZ [79.3K]3 years ago
5 0

56:43 Theo found the driving distance from Glacier National Park to Yellowstone Park to be 448 miles. Theo used a map that had a ratio of StartFraction 5 centimeters over 320 miles EndFraction. How many centimeters is the distance on the map? Round to the nearest unit if necessary. 4 centimeters 7 centimeters 64 centimeters 90 centimeters

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Three circles with centers A, B, and C are externally tangent to each other as shown in the figure. Lines EG and DG are tangent
fiasKO [112]
My solution to the problem is as follows:

EC = 15 ... draw CF = 6 (radius) ...use Pythagorean theorem to find EF. 

EF^2 + CF^2 = EC^2 
EF^2 = 15^2 - 6^2 = 189 .... EF = sq root 189 

triangle GDE is similar to CFE ... thus proportional 
GD / ED = CF / EF 
GD / 18 = 6 / (sq root 189) 
<span>GD = 108 / (sq root 189)

I hope my answer has come to your help. God bless and have a nice day ahead!


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3 years ago
Simplify the following expression as much as you can use exponential properties. (6^-2)(3^-3)(3*6)^4
gtnhenbr [62]

Answer:

Simplifying the expression (6^{-2})(3^{-3})(3*6)^4 we get \mathbf{108}

Step-by-step explanation:

We need to simplify the expression (6^{-2})(3^{-3})(3*6)^4

Solving:

(6^{-2})(3^{-3})(3*6)^4

Applying exponent rule: a^{-m}=\frac{1}{a^m}

=\frac{1}{(6^{2})}\frac{1}{(3^{3})}(18)^4\\=\frac{(18)^4}{6^{2}\:.\:3^{3}} \\

Factors of 18=2\times 3\times 3=2\times3^2

Factors of 6=2\times 3

Replacing terms with factors

=\frac{(2\times3^2)^4}{(2\times 3)^{2}\:.\:3^{3}} \\=\frac{(2)^4\times(3^2)^4}{(2)^2\times (3)^{2}\:.\:3^{3}} \\

Using exponent rule: (a^m)^n=a^{m\times n}

=\frac{(2)^4\times(3)^8}{(2)^2\times (3)^{2}\:.\:3^{3}} \\=\frac{2^4\times 3^8}{2^2\times 3^{2}\:.\:3^{3}}

Using exponent rule: a^m.a^n=a^{m+n}

=\frac{2^4\times 3^8}{2^2\times 3^{2+3}}\\=\frac{2^4\times 3^8}{2^2\times 3^{5}}

Now using exponent rule: \frac{a^m}{a^n}=a^{m-n}

=2^{4-2}\times 3^{8-5}\\=2^{2}\times 3^{3}\\=4\times 27\\=108

So, simplifying the expression (6^{-2})(3^{-3})(3*6)^4 we get \mathbf{108}

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Step-by-step explanation:

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