All categories

2 years ago

How many solutions are there to the equation below?9x - 16 = 9x - 23

Mathematics

2 answers:

barxatty [35]2 years ago

8 0

Zero.
9x-16=9x-23
9x-9x-16=-23
0-16=-23
-16=-23
The equation is invalid, so there is no solution for this equation

Stella [2.4K]2 years ago

7 0

Answer:

0 solutions are there to the given equation

Step-by-step explanation:

Given the equation:

$9x-16 = 9x -23$

Subtract 9x from both sides, we have;

$9x-16-9x = 9x -23-9x$

Simplify:

$-16 =-23$

Divide both sides by -1 we have;

$16 =23$ False.

⇒ there is no solution for the given equations.

Therefore. 0 solutions are there to the given equation.

You might be interested in

The braking distance, in feet of a car a Travling at v miles per hour is given.

irakobra [83]

The braking distance is the distance the car travels before coming to a stop after the brakes are applied

a. The braking distances are as follows;

The braking distance at 25 mph, is approximately 63.7 ft.

The braking distance at 55 mph, is approximately 298.35 ft.

The braking distance at 85 mph, is approximately 708.92 ft.

b. If the car takes 450 feet to brake, it was traveling with a speed of 98.211 ft./s

Reason:

The given function for the braking distance is D = 2.6 + v²/22

a. The braking distance if the car is going 25 mph is therefore;

25 mph = 36.66339 ft./s

$D = 2.6 + \dfrac{36.66339^2}{22} = 63.7 \ ft.$

At 25 mph, the braking distance is approximately 63.7 ft.

At 55 mph, the braking distance is given as follows;

55 mph = 80.65945 ft.s

$D = 2.6 + \dfrac{80.65945^2}{22} \approx 298.35 \ ft.$

At 55 mph, the braking distance is approximately 298.35 ft.

At 85 mph, the braking distance is given as follows;

85 mph = 124.6555 ft.s

$D = 2.6 + \dfrac{124.6555^2}{22} \approx 708.92 \ ft.$

At 85 mph, the braking distance is approximately 708.92 ft.

b. The speed of the car when the braking distance is 450 feet is given as follows;

$450 = 2.6 + \dfrac{v^2}{22}$

v² = (450 - 2.6) × 22 = 9842.8

v = √(9842.2) ≈ 98.211 ft./s

The car was moving at v ≈ 98.211 ft./s

Learn more here:

brainly.com/question/18591940

8 0

1 year ago

Esther drove to work in the morning at an average speed of 45 miles per hour. She returned home in the evening along the same ro

nydimaria [60]

The key is Esther travelled the same distance - x - in both her morning and evening commute.

45(time she took in the morning, or p) = x
30(time she took in the evening, or q) = x

Therefore 45(p) = 30(q), or divide both sides by 5 and get 9(p) = 6(q). I know you can divide it further, but these numbers are small enough and it's not worth the time.

Since the whole trip took an hour, (p + q) = 60min, and so, p = 60-q.

Therefore 9(60-q) = 6q or 540-9q = 6q. So 540 = 15q, which makes q = 36. If q = 36, then by (p+q)=60, p (the time she took in the morning) must equal 24.

45 miles per hour, her speed in the morning, times (24/60) hours, her time, makes 18 miles travelled in the morning. If you check, 30 miles per hour times (36/60) hours also makes 18 miles in the evening.

Hope that makes a little sense. And I also hope it's right

7 0

3 years ago

Read 2 more answers

How do you find the range of a linear graph

LUCKY_DIMON [66]

The range is the possible y-values of the graph. If you have a specific equation, post it.

8 0

3 years ago

describe a situation in which a formula could be used more easily if it were rearranged. Include the formula in your description

pshichka [43]

I need to find the length of something when I am given the volume, width, and height.

v=lwh
l = v/wh

Hope this helps!

5 0