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Paha777 [63]
3 years ago
13

The lines shown below are parallel. If the green line has a slope of -3/7, what is

Mathematics
1 answer:
BartSMP [9]3 years ago
7 0

Answer: the slope of green line = -\frac37

Step-by-step explanation:

We know that,

If two lines are parallel then their slopes(rate of change of y with respect to x) are equal.

Given: The red line and the green line are parallel.

The slope of red line = -\frac37

Then, the slope of green line = -\frac37  [Slopes of two parallel lines are equal]

Hence, the slope of green line = -\frac37

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(Will give brainliest if asked for)
GarryVolchara [31]

A. x <= -12

<u>Explanation:</u>

-3x >= 36

-x >= 36/ 3

-x > = 12

x < = -12

When the negative sign is shifted then the greater than equal to sign reverses to less than equal to.

Therefore, x <= -12

7 0
3 years ago
-7-7 (2m-7) = 7 (6+m)
levacccp [35]

Answer:

m= 0

Step-by-step explanation:

this is the right answer

-7-14m+49=42+7m

42-14m=42+7m

14m+7m=42-42

21m=0

m=0

4 0
3 years ago
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Eight more than seven times a number is equal to four more than five times the number. What is the number?
Alex17521 [72]

Answer: -2

Step-by-step explanation:

convert it to a equation:

7x+8=5x+4

move x's to a side and numbers to the other

2x=-4

divide by 2

x=-2

8 0
1 year ago
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Find the value of x. Round the length to the nearest tenth.
elena-s [515]

Answer:

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4 0
2 years ago
A projectile is fired at such an angle that the vertical
svetoff [14.1K]

(a) The projectile remains in the air for 10 seconds

(b) It travels a horizontal distance of 600 meters

Step-by-step explanation:

A projectile is fired at such an angle that

1. The vertical  component of its velocity is 49 m/sec

2. The horizontal  component of its velocity is 60 m/sec

We need to find:

(a) How long the projectile remains in the air

(b) The horizontal distance it travels

∵ The vertical distance y = u_{y} t - \frac{1}{2} g t², where

u_{y} is the vertical  component of its velocity, g is the acceleration

of gravity and t is the time

∵ y = 0 ⇒ it return to the same initial height

∵ u_{y} = 49 m/s

∵ g = 9.8 m/s²

- Substitute these values in the rule above

∴ 0 = 49 t - \frac{1}{2} (9.8) t²

∴ 0 = 49 t - 4.9 t²

- Take t as a common factor

∴ 0 = t (49 - 4.9 t)

- Equate each term by 0

∴ t = 0 ⇒ at initial position

∴ 49 - 4.9 t = 0

- Add 4.9 t to both side

∴ 49 = 4.9 t

- Divide both sides by 4.9

∴ t = 10 seconds

(a) The projectile remains in the air for 10 seconds

∵ The horizontal distance = u_{x} t, where u_{x} is the

   horizontal component of its velocity

∵ u_{x} = 60 m/sec

∵ t = 10 seconds

∴ x = 60 × 10 = 600 meters

(b) It travels a horizontal distance of 600 meters

Learn more:

You can learn more about the component of velocity in brainly.com/question/4464845

#LearnwithBrainly

3 0
3 years ago
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