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3 years ago

A hiker starts at an elevation of 65 feet and descends 30 feet to the base camp. What is the elevation of the base camp?

Mathematics

2 answers:

bezimeni [28]3 years ago

6 0

Answer:

Step-by-step explanation:

Ronch [10]3 years ago

5 0

You just subtract it,

65 - 30= 35

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The half life of a radioactive kind of cadmium is 14 years. How much will be left after 28 years, if you start with 72 grams of

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3 years ago

Ms. Thomas took her family and friend to the movies. There were a total of 16 people. Children tickets cost $6 and adult tickets

Korolek [52]

Answer:

11 adults

Step-by-step explanation:

First, set up a system of equations:

Let x = number of children and y = number of adults

There is a total of 16 people, so x+y = 16. This is your first equation.

Each child ticket costs $6, so 6x = cost of children's tickets based on the number of children present (x)

Each adult ticket costs $9, so 9y = cost of adult's tickets based on the number of adults present (y)

The total cost of tickets is $129, so 6x+9y = 129. This is your second equation.

Your system of equations is now this:

x + y = 16

6x + 9y = 129

You can solve this system using the method of elimination, where we will eliminate the variable x (number of children), since we are focused on y (number of adults).

Multiply the top equation by -6. This will give the following equation:

-6x - 6y = -96

You can now solve the system of equations by placing the new equation under the second equation like this:

6x + 9y = 129

-6x - 6y = -96

Now, add the two equations together.

6x + (-6x) = 0

9y + (-6y) = 3y

129 + (-96) = 33

After doing this, you get the following equation:

3y = 33

As you can see, the variable x has been eliminated, and you are left with y.

Solve the equation for y:

3y = 33

y = 33/3 = 11

y = 11 adults

5 0

3 years ago

Match the expressions with their equivalent simplified expressions.

Tasya [4]

Answer:

$\sqrt[4]{\frac{16x^6y^4}{81x^2y^8}}\rightarrow\frac{2x}{3y}\\\sqrt[4]{\frac{81x^2y^{10}}{81x^6y^6}} \rightarrow\frac{3y}{2x}\\\sqrt[3]{\frac{64x^8y^7}{125x^2y^{10}}}\rightarrow\frac{4x^2}{5y}\\\sqrt[5]{\frac{243x^{17}y^{16}}{32x^7y^{21}}}\rightarrow\frac{3x^2}{2y}\\\sqrt[5]{\frac{32x^{12}y^{15}}{243x^7y^{10}}} \rightarrow\frac{2xy}{3}\\\sqrt[4]{\frac{16x^{10}y^{9}}{256x^2y^{17}}}\rightarrow\frac{x}{2y}$

Step-by-step explanation:

$\sqrt[4]{\frac{16x^6y^4}{81x^2y^8}} =\sqrt[4]{\frac{(2^4)(x^{6-2})(y^{4-8})}{(3^4)}} =\sqrt[4]{\frac{2^4x^4y^{-4}}{3^4}} =\frac{2xy^{-1}}{3}=\frac{2x}{3y}$

$\sqrt[4]{\frac{81x^2y^{10}}{81x^6y^6}} =\sqrt[4]{\frac{(3^4)(x^{2-6})(y^{10-6})}{(2^4)}} =\sqrt[4]{\frac{3^4x^{-4}y^{4}}{2^4}} =\frac{3x^{-1}y^1}{3}=\frac{3y}{2x}$

$\sqrt[3]{\frac{64x^8y^7}{125x^2y^{10}}} =\sqrt[3]{\frac{(4^3)(x^{8-2})(y^{7-10})}{(5^3)}} =\sqrt[3]{\frac{4^3x^6y^{-3}}{5^3}} =\frac{4x^2y^{-1}}{5}=\frac{4x^2}{5y}$

$\sqrt[5]{\frac{243x^{17}y^{16}}{32x^7y^{21}}} =\sqrt[5]{\frac{(3^5)(x^{17-7})(y^{16-21})}{(2^5)}} =\sqrt[5]{\frac{3^5x^{10}y^{-5}}{2^5}} =\frac{3x^2y^{-1}}{2}=\frac{3x^2}{2y}$

$\sqrt[5]{\frac{32x^{12}y^{15}}{243x^7y^{10}}} =\sqrt[5]{\frac{(2^5)(x^{12-7})(y^{15-10})}{(3^5)}} =\sqrt[5]{\frac{2^5x^{5}y^{5}}{3^5}} =\frac{2x^1y^{1}}{3}=\frac{2xy}{3}$

$\sqrt[4]{\frac{16x^{10}y^{9}}{256x^2y^{17}}} =\sqrt[4]{\frac{(2^4)(x^{10-2})(y^{9-17})}{(4^4)}} =\sqrt[4]{\frac{2^4x^{8}y^{-8}}{4^4}} =\frac{2x^{1}y^{-1}}{4}=\frac{x}{2y}$

Thus,

3 0

3 years ago