Aseem and Stana are building model cars. stanas car is 5 less than 2 times the length of Aseem's car. The sum of the lengths of both cars is 20. write an equation to determine the lengths of Aseem's and Stana's cars

Answers

Answer 1
Answer:

Answer:

Aseem’s : 8.3, Stana’s : 11.6

Step-by-step explanation:

Aseem: x

Stana: 2x-5

X+2x-5=20

3x=25

X = 8.3

Answer 2
Answer:

Final answer:

The length of Aseem's car is approximately 8.33 and the length of Stana's car is approximately 11.66.

Explanation:

Let's define the length of Aseem's car as x. Stana's car is 5 less than 2 times the length of Aseem's car, so the length of Stana's car can be represented as 2x - 5. The sum of the lengths of both cars is 20, so we can set up the equation x + (2x - 5) = 20. Solving this equation will give us the lengths of Aseem's and Stana's cars.

Combining like terms, we get 3x - 5 = 20. Adding 5 to both sides of the equation, we have 3x = 25. Finally, by dividing both sides by 3, we find that x = 8.33.

The length of Aseem's car is approximately 8.33 and the length of Stana's car is approximately 2(8.33) - 5 = 11.66.

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enrollment in a photography class increased from 25 to 35 students. What was the percent of increase in enrollment?

Answers

Answer

40%

Explanation

A parentage value is a value that has been compared to 100.

Percentage = (change in the value)/(the original value) × 100%

                   = (35-25)/25 × 100%

                    = 10/25 × 100%

                   = 40%

(35-25)/25x100%
=40%
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To find the midpoint of a segment using the coordinates of its endpoints: Answers: A- Calculate the average of the x-coordinates and the average of the y-coordinates of the endpoints. B- Calculate the differences of the x-coordinates and the differences of the y-coordinates of the endpoints. C-Calculate the differences of the x-coordinates and the differences of the y-coordinates of the endpoints and divide each by 2. D-Calculate the average of the x-coordinates and the average of the y-coordinates of the endpoints and divide each by 2.

Answers

The answer here is A. "Calculate the average of the x-coordinates and the average of the y-coordinates of the endpoints."

The midpoint is basically the point located between two endpoints, where the distance is equal from the midpoint to either of the points. Since this is merely a halfway point, taking the average of the coordinates would yield us the midpoint.

What is slope of the line 15x-3y=90

Answers

Final answer:

The slope of the line 15x - 3y = 90 is 5.

Explanation:

The slope of a line can be determined using the equation in slope-intercept form, y = mx + b, where m represents the slope. To find the slope of the line 15x - 3y = 90, rearrange the equation to the form y = mx + b. Subtract 15x from both sides to isolate the -3y term, and then divide both sides by -3 to solve for y as a function of x. The resulting equation will be in the form y = mx + b, and the coefficient of xwill be the slope.

15x - 3y = 90
-3y = -15x + 90
y = 5x - 30

Therefore, the slope of the line 15x - 3y = 90 is 5.

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The line is not in slope intercept form, so you need to get y by itself

15x -3y = 90

Subtract 15x

-3y = -15x + 90

Divide by -3

y = 5x -30

So, the slope (x) is 5/1.

I hope that helps!

Is 4 kilograms is egual to 4000mgs

Answers

No.
So, there are 1000 milligrams in one gram, making the 4000 mgs equal to 4 grams.
There are 1000 grams in one kilogram, so 4 kilograms is equal to 4000 grams.

4000 =/= 4
So, 4 kilograms and 4000 milligrams are not equal.
4000Mgs is not equal to 4 kilograms. 4000 Mgs is 4 grams, which is 0.004Kilograms. Hence, 1 milogram is 1x10^-6 Kilograms.

Hy! (√2+1)²+(3-√3)²-(√5+1)²=???
Thank yours!:*

Answers

(\sqrt2+1)^2+(3-\sqrt3)^2-(\sqrt5+1)^2=\n2+2\sqrt2+9-6\sqrt3+9-(5+2\sqrt5+1)=\n20+2\sqrt2-6\sqrt3-5-2\sqrt5-1=\n14+2\sqrt2-6\sqrt3-2\sqrt5

All equation of motion with examples

Answers

Answer:

Answer is below. Hope this helps you!

Step-by-step explanation:

(i)v = u + at

(ii) s = ut + 1/2 *at²

(iii)2as = v²-u²

Examples:

A train starting from rest attains a velocity of 72km/h in 5 minutes. Assuming that the acceleration is uniform, find (i) the acceleration and  (ii)  the distance travelled by the train for attaining this velocity.

72km/h = 20m/s

(i) use v= u+ at

20 = 0 + a*300

20 = 5a

20 / 300= a

1/15 ms² = a

to find distance

use 2as= v²-u²

2 *1/15 * s = (20) - 0²

2/15s = 400

s = 400 * 15/2

s = 3000m = 3 km

a car accelerates with a uniform acceleration of 1m/s². it starts with a  velocity of 18km/h and reaches 36 km/h in 5 seconds find the distance travelled.

18 km /h = 5m/s

36km/h = 10m/s

use - s= ut + 1/2at²

s = 5 * 5 + 1/2 * 1 * (5)²

s = 25 + 25/2

s = 37.5 meters

Answer:

It is described in terms of displacement, distance, velocity, acceleration, time and speed. Jogging, driving a car, and even simply taking a walk are all everyday examples of motion.

...

The three equations are,

v = u + at.

v² = u² + 2as.

s = ut + ½at²