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2 years ago

How do I get the answer to this

Mathematics

2 answers:

Oxana [17]2 years ago

8 0

5x7=35. Voila. You're the 35th caller.

stepladder [879]2 years ago

4 0

You do 5 X 7 which is 35 because that means that you are both one in 7 callers and one in 5 callers. Hope this helps, have a nice day :)

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What is the final cost of a book priced at $10.00 with a tax rate of 5%?

Sophie [7]

Answer:

10.50

Step-by-step explanation:

5% of 10 is 0.5

0.5+10=10.50

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2 years ago

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Find the median of the following set of data. Round your answer to the nearest hundredth, if necessary.

Usimov [2.4K]

The correct answer is 35.2 because when you arrange them in order from least to greatest, the middle is 35.2

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3 years ago

Combine like terms, then simplify if need be. 4cx + cx – cx – cx + cx

Crank

Answer:

4cx

hope this helps

have a good day :)

Step-by-step explanation:

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3 years ago

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The length of a shadow of a tree is 9 ft. The distance from the top of the tree to the tip of the shadow is 16 ft. what is the h

Margarita [4]

So for this, you'll be using the pythagorean theorem to solve for the height.

$9^2+b^2=16^2$

Solve the exponents to get $81+b^2=256$

Subtract 81 on each side to get $b^2=175$

Square root each side, and your answer should be $b= \sqrt{175}$ or 13.23 ft

7 0

3 years ago

Find the orthocenter for the triangle described by each set of vertices.

givi [52]

Answer:

The orthocentre of the given vertices ( 2 , -3.5)

Step-by-step explanation:

Step(i):-

The orthocentre is the intersecting point for all the altitudes of the triangle.

The point where the altitudes of a triangle meet is known as the orthocentre.

Given Points are K (3.-3), L (2,1), M (4,-3)

The Altitudes are perpendicular line from one side of the triangle to the opposite vertex

The altitudes are MN , KO , LP

step(ii):-

Slope of the line

$KL = \frac{y_{2}-y_{1} }{x_{2}-x_{1} } = \frac{1-(-3)}{2-3} = -4$

The slope of MN =

The perpendicular slope of KL

= $\frac{-1}{m} = \frac{-1}{-4} = \frac{1}{4}$

The equation of the altitude

$y - y_{1} = m( x-x_{1} )$

$y - (-3) = \frac{1}{4} ( x-4 )$

4y +12 = x -4

x - 4 y -16 = 0 ...(i)

Step(iii):-

Slope of the line

$LM = \frac{y_{2}-y_{1} }{x_{2}-x_{1} } = \frac{-3-1}{4-2} = -2$

The slope of KO =

The perpendicular slope of LM

= $\frac{-1}{m} = \frac{-1}{-2} = \frac{1}{2}$

The equation of the altitude

$y - y_{1} = m( x-x_{1} )$

The equation of the line passing through the point K ( 3,-3) and slope

m = 1/2

$y - (-3) = \frac{1}{2} ( x-3 )$

2y +6 = x -3

x - 2y -9 =0 ....(ii)

Solving equation (i) and (ii) , we get

subtracting equation (i) and (ii) , we get

x - 4y -16 -( x-2y-9) =0

- 2y -7 =0

-2y = 7

y = - 3.5

Substitute y = -3.5 in equation x -4y-16=0

x - 4( -3.5) - 16 =0

x +14-16 =0

x -2 =0

x = 2

The orthocentre of the given vertices ( 2 , -3.5)

4 0

2 years ago