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Juli2301 [7.4K]
2 years ago
8

Please help I dont know what to do. I have already done the graph but all I need is x,y box and equation ​

Mathematics
1 answer:
Sever21 [200]2 years ago
5 0
4 is the answer hope this helps
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Helpppp plssss!! Show your work!!<br><br>Thanks
Eduardwww [97]

Slope of a Line Passing through two points (x₁ , y₁) and (x₂ , y₂) is given by :

\heartsuit\;\;Slope(m) = \frac{y_1 - y_2}{x_1 - x_2}

Given Points are (-3 , 5) and (6 , -1)

here x₁ = -3 and x₂ = 6 and y₁ = 5 and y₂ = -1

\heartsuit\;\;Slope(m) = \frac{5 + 1}{-3 - 6} = \frac{6}{-9} = \frac{-2}{3}

Option D is the Answer

4 0
3 years ago
Can y’all help me with this one?
asambeis [7]

-44+18=-26degrees

negative 26 degrees is your answer

please mark brainliest and have a nice day

6 0
2 years ago
Read 2 more answers
2. Assume f(x) = g(x). Which of the following functions may be used to represent the equation 3x+2 = 7x + 6? A. f(x) = x + 2, g(
charle [14.2K]

Answer:

C. f(x) = 3·x + 2, g(x) = 7·x + 6

Step-by-step explanation:

The given equations relates to the property of equality of values;

The given formula for the association between f(x) and g(x) is f(x) = g(x)

The given equation of two expressions is 3·x + 2 = 7·x + 6

By transitive property of equality, the two above equations are correct when f(x) = 3·x + 2 and g(x) = 7·x + 6

Therefore, the function that may be used to represent the equation is option C; f(x) = 3·x + 2, g(x) = 7·x + 6.

7 0
2 years ago
50 POINTS IF YOU ANSWER THE FOLLOWING QUESTION PLEASE<br> https://brainly.com/question/15324380
Jet001 [13]

Answer:

I added an answer. It was -2x^4

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
From a window 20 feet above the ground, the angle of elevation to the top of a building across
Nikitich [7]

Answer: The answer is 381.85 feet.

Step-by-step explanation:  Given that a window is 20 feet above the ground. From there, the angle of elevation to the top of a building across  the street is 78°, and the angle of depression to the base of the same building is 15°. We are to calculate the height of the building across the street.

This situation is framed very nicely in the attached figure, where

BG = 20 feet, ∠AWB = 78°, ∠WAB = WBG = 15° and AH = height of the bulding across the street = ?

From the right-angled triangle WGB, we have

\dfrac{WG}{WB}=\tan 15^\circ\\\\\\\Rightarrow \dfrac{20}{b}=\tan 15^\circ\\\\\\\Rightarrow b=\dfrac{20}{\tan 15^\circ},

and from the right-angled triangle WAB, we have'

\dfrac{AB}{WB}=\tan 78^\circ\\\\\\\Rightarrow \dfrac{h}{b}=\tan 15^\circ\\\\\\\Rightarrow h=\tan 78^\circ\times\dfrac{20}{\tan 15^\circ}\\\\\\\Rightarrow h=361.85.

Therefore, AH = AB + BH = h + GB = 361.85+20 = 381.85 feet.

Thus, the height of the building across the street is 381.85 feet.

8 0
3 years ago
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