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3 years ago

What’s the slope to (-5,5) (5,-4)

Mathematics

1 answer:

Lena [83]3 years ago

6 0

Answer:

Step-by-step explanation:

(-5,5)...x1 = -5 and y1 = 5

(5,-4)...x2 = 5 and y2 = -4

slope = (y2 - y1) / (x2 - x1)

slope = (-4 - 5) / (5 - (-5) = -9 / (5 + 5) = -9/10 <==

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A plot of land has been surveyed for a new housing development with borders AB, BC, DC, and DA. The plot of land is a right trap

sdas [7]

Answer:

m∠ABC = 119° and DC = 105 feet

Step-by-step explanation:

For better understanding of the solution see the attached figure of the trapezium ABCD :

m∠BCD = 61°

By using the property, Sum of interior opposite angles in a trapezium is supplementary ⇒ m∠BCD + m∠ABC = 180°

⇒ m∠ABC = 180° - 61°

⇒ m∠ABC = 119°

Now, to find length of DC,

Draw a perpendicular BE on CD ⇒ CD = DE + EC

AB = DE

⇒ DE = 80 feet

In ΔBEC, By using Pythagoras theorem

EC² = BC² - BE²

⇒ EC² = 65² - 60²

⇒ EC = 25 feet

Hence, DC = 80 + 25

= 105 feet

3 0

3 years ago

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Mike got in and elevator and whent down 3 floors. he meant to go to a lower level,so he stayed on the elevator and when down 3 m

Naily [24]

It is negative six because down three is -3 so 3+3 =6 but the numbers are negative so the answer is -6

7 0

3 years ago

Find the horizontal and vertical asymptotes of f(x). f(x) equals = StartFraction 6 x Over x plus 2 EndFraction 6x x+2 Find the

miss Akunina [59]

Answer:

The horizontal asymptote can be described by the line y = 6

The vertical asymptote can be described by the line x = -2

Step-by-step explanation:

* Lets the meaning of vertical and horizontal asymptotes

- Vertical asymptotes are vertical lines which correspond to the zeroes

of the denominator of a rational function

- A horizontal asymptote is a y-value on a graph which a function

approaches but does not actually reach

- If the degree of the numerator is less than the degree of the

denominator, then there is a horizontal asymptote at y = 0

- If the degree of the numerator is greater than the degree of the

denominator, then there is no horizontal asymptote

- If the degree of the numerator is equal the degree of the denominator,

then there is a horizontal asymptote at y = leading coefficient of the

numerator ÷ leading coefficient of the denominator

* Lets solve the problem

∵ $f(x)=\frac{6x}{x+2}$

∵ The numerator is 6x

∵ The denominator is x + 2

∴ The numerator and the denominator have same degree

∵ The leading coefficient of the numerator is 6

∵ The leading coefficient of the denominator is 1

∴ There is a horizontal asymptote at y = 6/1

∴ The horizontal asymptote can be described by the line y = 6

- Put the denominator equal zero to find its zeroes

∵ The denominator is x + 2

∴ x + 2 = 0

- Subtract 2 from both sides

∴ x = -2

∴ The vertical asymptote can be described by the line x = -2

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3 years ago

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Is the answer to this problem right

Morgarella [4.7K]

Answer:

it should be option 2 and 4

Step-by-step explanation:

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3 years ago

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Triangle PQR has the coordinates P(2,4), Q(-2,4), R(0,-5).

tensa zangetsu [6.8K]

Answer:

I hope I got this correct.

Step-by-step explanation:

Multiply the coordinates by 2.

P(4,8)

Q(-4,8)

R(0,-10)

For the future if you get stuck, multiply the coordinates by the dilation number.

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2 years ago