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Tju [1.3M]
4 years ago
9

Mr. Smith spent 6 hours hiking along a trail and back. He walked out at the rate of 4 miles per hour and walked back at the rate

of 2 miles per hour. How long did it take Mr. Smith to get to the end of the trial?
Mathematics
1 answer:
ollegr [7]4 years ago
5 0
To get to the end of the trail it would take 2 hours because the walking back hours would be double the walking out rate, but it would still have to equal 6 hours. So walking out would be 2 hours and walking back would be 4 hours which makes the trail 8 miles long.
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6 batteries cost $7.5

Step-by-step explanation:

8 batteries = $10  

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Step: cross multiply

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If the focal point is (2,5) and the directrix is y = -7. Find the equation of the parabola.
mihalych1998 [28]

Answer:

The equation of the parabola is (x-2)^2 = 48(y-5)

Step-by-step explanation:

Mathematically, we can obtain the equation of a parabola from its focal point and the directrix given.

Firstly, we need to find the distance between the director and the vertex

That would be ;

|-7-(5)| = 12

We should kindly note that the directrix is below the vertex and thus it is a right-side parabola with a positive value of p = 12

Thus, the equation needed would be;

(x-h)^2 = 4p(y-k)

From the question, h = 2

p = 12 ( calculated) and k = 5

So the equation would be;

(x-2)^2 = 4(12)(y-5)

= (x-2)^2 = 48(y-5)

8 0
4 years ago
A triangle has two sides of length 5 and 12. what value could the length of the third side be? More then one can be correct.
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Condition\ for\ existing\ triangle\\\\
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5 0
4 years ago
Cualnes el area del rombo si sus medidas en rectangulo son 12 y 5<br>​
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Answer:

34

Step-by-step explanation:

8 0
3 years ago
Tomika heard that the diagonals of a rhombus are perpendicular to each other. Help her test her conjecture. Graph quadrilateral
Stella [2.4K]

Answer:

a. The four sides of the quadrilateral ABCD are equal, therefore, ABCD is a rhombus

b. The equation of the diagonal line AC is y = 5 - x

The equation of the diagonal line BD is y = 5 - x

c. The diagonal lines AC and BD of the quadrilateral ABCD are perpendicular to each other

Step-by-step explanation:

The vertices of the given quadrilateral are;

A(1, 4), B(6, 6), C(4, 1) and D(-1, -1)

a. The length, l, of the sides of the given quadrilateral are given as follows;

l = \sqrt{\left (y_{2}-y_{1}  \right )^{2}+\left (x_{2}-x_{1}  \right )^{2}}

The length of side AB, with A = (1, 4) and B = (6, 6) gives;

l_{AB} = \sqrt{\left (6-4  \right )^{2}+\left (6-1  \right )^{2}} = \sqrt{29}

The length of side BC, with B = (6, 6) and C = (4, 1) gives;

l_{BC} = \sqrt{\left (1-6  \right )^{2}+\left (4-6  \right )^{2}} = \sqrt{29}

The length of side CD, with C = (4, 1) and D = (-1, -1) gives;

l_{CD} = \sqrt{\left (-1-1  \right )^{2}+\left (-1-4  \right )^{2}} = \sqrt{29}

The length of side DA, with D = (-1, -1) and A = (1,4)   gives;

l_{DA} = \sqrt{\left (4-(-1)  \right )^{2}+\left (1-(-1)  \right )^{2}} = \sqrt{29}

Therefore, each of the lengths of the sides of the quadrilateral ABCD are equal to √(29), and the quadrilateral ABCD is a rhombus

b. The diagonals are AC and BD

The slope, m, of AC is given by the formula for the slope of a straight line as follows;

Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}

Therefore;

Slope, \, m_{AC} =\dfrac{1-4}{4-1} = -1

The equation of the diagonal AC in point and slope form is given as follows;

y - 4 = -1×(x - 1)

y = -x + 1 + 4

The equation of the diagonal AC is y = 5 - x

Slope, \, m_{BD} =\dfrac{-1-6}{-1-6} = 1

The equation of the diagonal BD in point and slope form is given as follows;

y - 6 = 1×(x - 6)

y = x - 6 + 6 = x

The equation of the diagonal BD is y = x

c. Comparing the lines AC and BD with equations, y = 5 - x and y = x, which are straight line equations of the form y = m·x + c, where m = the slope and c = the x intercept, we have;

The slope m for the diagonal AC = -1 and the slope m for the diagonal BD = 1, therefore, the slopes are opposite signs

The point of intersection of the two diagonals is given as follows;

5 - x = x

∴ x = 5/2 = 2.5

y = x = 2.5

The lines intersect at (2.5, 2.5), given that the slopes, m₁ = -1 and m₂ = 1 of the diagonals lines satisfy the condition for perpendicular lines m₁ = -1/m₂, therefore, the diagonals are perpendicular.

5 0
3 years ago
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