What change would increase the acceleration of the dog sled? (1 point) Responses run the sled up a hill run the sled up a hill remove a dog from the sled remove a dog from the sled add another dog to the sled add another dog to the sled add more mass to the sled
add another dog to the sled
are you correct?
Yes, I am correct. Adding another dog to the sled would increase the total pulling force, which in turn will increase the acceleration of the sled.
Which object will accelerate more if the student exerts an equal amount of force on each object? (1 point) Responses
10 kg box
two 10 kg boxes
The 10 kg box will accelerate more if the student exerts an equal amount of force on each object.
why
The 10 kg box will accelerate more because it has less mass compared to two 10 kg boxes. According to Newton's second law of motion (F = ma), acceleration is directly proportional to the force applied and inversely proportional to the mass of the object. Since the force is the same for both objects and the mass of the single 10 kg box is less than the combined mass of the two 10 kg boxes, the single 10 kg box will experience a greater acceleration.
Solve the system of two linear equations. −3x−4y=−23 3x−5y=5 (1 point) Responses (5, 2) (−5, 2) (5,−2) (−5,−2)
To solve the system of equations:
-3x - 4y = -23
3x - 5y = 5
We can use the method of elimination by adding the two equations together:
(-3x - 4y) + (3x - 5y) = -23 + 5
Simplifying this equation, we get:
-9y = -18
Dividing both sides of the equation by -9, we have:
y = 2
Substituting this value of y back into one of the original equations, let's use the first equation:
-3x - 4(2) = -23
-3x - 8 = -23
Adding 8 to both sides of the equation:
-3x = -15
Dividing both sides of the equation by -3:
x = 5
Therefore, the solution to the system of equations is (x, y) = (5, 2). So, the correct response is (5, 2).
are you correct
Yes, I am correct. The solution to the given system of equations is indeed (x, y) = (5, 2).
How many solutions does the system of equations have? 8x−6y=48 8x+6y=48(1 point) Responses infinite two none one
The given system of equations has only one solution.
The two equations are 8x - 6y = 48 and 8x + 6y = 48.
By adding the two equations, we eliminate the variable y and get:
16x = 96
Dividing both sides by 16, we get:
x = 6
Substituting this value of x back into one of the original equations, such as the first equation, we have:
8(6) - 6y = 48
48 - 6y = 48
Subtracting 48 from both sides, we have:
-6y = 0
Dividing both sides by -6, we get:
y = 0
Therefore, the system of equations has a single solution, which is (x, y) = (6, 0).
Solve the system of two equations using elimination with multiplication. 4x−6y=−48 −2x+4y=30(1 point) Responses (−33,−9) (3, 9) (−21,−6) (−3, 6)
To solve the system of equations using elimination with multiplication:
4x - 6y = -48
-2x + 4y = 30
We can multiply the second equation by 2 in order to make the coefficients of x in both equations equal:
2(-2x + 4y) = 2(30)
-4x + 8y = 60
Now we have:
4x - 6y = -48
-4x + 8y = 60
Adding the two equations together, the x terms cancel out:
(-4x + 8y) + (4x - 6y) = 60 + (-48)
2y = 12
y = 6
Substituting this value of y back into one of the original equations, let's use the first equation:
4x - 6(6) = -48
4x - 36 = -48
4x = -12
x = -3
Therefore, the solution to the system of equations is (x, y) = (-3, 6). The correct response is (-3, 6).
Solve the system of equations. 3x−4y=40 5x−5y=55 (1 point) Responses (14, 3) left parenthesis 14 comma 3 right parenthesis (18, 7) left parenthesis 18 comma 7 right parenthesis (8,−3) left parenthesis 8 comma negative 3 right parenthesis (4,−7)
To solve the system of equations:
3x - 4y = 40
5x - 5y = 55
We can use the method of elimination by multiplying the first equation by 5 and the second equation by 3 to make the coefficients of y equal:
(5)(3x - 4y) = (5)(40)
(3)(5x - 5y) = (3)(55)
15x - 20y = 200
15x - 15y = 165
Now, we can subtract the second equation from the first equation:
(15x - 20y) - (15x - 15y) = 200 - 165
15y = 35
y = 35 / 15
y = 7/3
Substituting this value of y back into the first equation, we have:
3x - 4(7/3) = 40
3x - 28/3 = 40
3x = 40 + 28/3
3x = 120/3 + 28/3
3x = 148/3
x = (148/3) / 3
x = 148 / (3 * 3)
x = 148 / 9
Therefore, the solution to the system of equations is (x, y) = (148/9, 7/3). In decimal form, it is approximately (16.44, 2.33). However, none of the given responses match this solution.
Solve the system of equations. 3x−4y=40 5x−5y=55 (1 point) Responses (14, 3) (18, 7) (8,−3) (4,−7)
To solve the system of equations:
3x - 4y = 40
5x - 5y = 55
We can use the method of elimination by multiplying the first equation by 5 and the second equation by 3 to make the coefficients of x equal:
(5)(3x - 4y) = (5)(40)
(3)(5x - 5y) = (3)(55)
15x - 20y = 200
15x - 15y = 165
Now, we can subtract the second equation from the first equation:
(15x - 20y) - (15x - 15y) = 200 - 165
-5y = 35
y = -7
Substituting this value of y back into the first equation, we have:
3x - 4(-7) = 40
3x + 28 = 40
3x = 40 - 28
3x = 12
x = 4
Therefore, the solution to the system of equations is (x, y) = (4, -7). The correct response is (4, -7).
How many solutions does the system have? 12x−15y=18 4x−5y=6(1 point) Responses two solutions two solutions no solution no solution infinitely many solutions infinitely many solutions one solution
The given system of equations has one solution.
To determine this, we can check the slope of the lines represented by the equations. If the slopes are different, then the lines will intersect at a single point and there will be one solution.
The equations are:
12x - 15y = 18
4x - 5y = 6
To find the slopes, we can write the equations in slope-intercept form (y = mx + c), where m is the slope:
12x - 15y = 18
-15y = -12x + 18
y = (12/15)x - (18/15)
y = (4/5)x - (6/5)
4x - 5y = 6
-5y = -4x + 6
y = (4/5)x - (6/5)
Comparing the slopes (coefficient of x), we can see that they are both equal to 4/5. This means that the lines are parallel and will not intersect, resulting in no solution. However, none of the listed responses indicate no solution.