All categories

3 years ago

Systems of equations with different slopes and different y-intercepts have more than one solution.

Mathematics

1 answer:

FrozenT [24]3 years ago

7 0

Answer: OPTION C.

Step-by-step explanation:

The systems of linear equations can have:

1. No solution: When the lines have the same slope but different y-intercept. This means that the lines are parallel and never intersect, therefore, the system of equations has no solution.

2. One solution: When the lines have different slopes and intersect at one point in the plane. The point of intersection will be the solution of the system

3. Infinitely many solutions: When the lines have the same slope and the y-intercepts are equal. This means that the equations represents the same line and there are infinite number of solution.

Therefore, based on the explained above, the conclusion is: Systems of equations with different slopes and different y-intercepts never have more than one solution.

You might be interested in

Four less than the product of one and a number x.

BaLLatris [955]

X-4 These characters are for >20 rule

3 0

3 years ago

For what value of q is the equation -13(q + 4) = 7q + 16

Leno4ka [110]

Answer:

Q = -3.4

Step-by-step explanation:

Our original equation is: -13(q+4)=7q+16

if we simplify this, we get: -13q-52=7q+16

Subtract 16 from both sides: -13q-68=7q

Add 13q to both sides to get: -68=20q

divide the equation to get this: -3.4=q

5 0

3 years ago

What number is 88% of 175?

lana66690 [7]

Answer:

154

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

If you had 36 cookies and gave 1/2 to one friend and 1/4 to another, how many could you have left?

Ann [662]

Step-by-step explanation:

$\frac{1}{2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{3}{4}$

$\frac{3}{4} 36 = 27$

6 0

2 years ago

If you have a cubic polynomial of the form y = ax^3 + bx^2 + cx + d and lets say it passes through the points (2,28), (-1, -5),

nikklg [1K]

Step-by-step explanation:

Step 1: Solve using the first point

(2, 28)

$28 = a(2)^3 + b(2)^2 + c(2) + d$

$28 = 8a + 4b + 2c + d$

Step 2: Solve using the second point

(-1, -5)

$-5 = a(-1)^3 + b(-1)^2 + c(-1) + d$

$-5 = -a + b - c + d$

Step 3: Solve using the third point

(4, 220)

$220 = a(4)^3 + b(4)^2 + c(4) + d$

$220 = 64a + 16b + 4c + d$

Step 4: Solve using the fourth point

(-2, -20)

$-20 = a(-2)^3 + b(-2)^2 + c(-2) + d$

$-20 = -8a + 4b - 2c + d$

Step 5: Combine the first and fourth equations

$28 - 20 = 8a - 8a + 4b + 4b + 2c - 2c + d + d$

$8 = 8b + 2d$

$8 - 8b = 8b - 8b + 2d$

$(8 -8b)/2 = 2d/2$

$4 - 4b = d$

Step 6: Solve for c in the second equation

$-5 + 5 = -a + b - c + d + 5$

$0 + c = -a + b - c + c + d + 5$

$c = -a + b + d + 5$

Step 7: Substitute d with the stuff we got in step 5

$c = -a + b + (4 - 4b) + 5$

$c = -a + b + 4 - 4b + 5$

$c = -a - 3b + 9$

Step 8: Substitute d and c into the first equation

$28 = 8a + 4b + 2(-a - 3b + 9) + (4 - 4b)$

$28 = 8a + 4b - 2a - 6b + 18 + 4 - 4b$

$28 - 22 = 6a - 6b + 22 - 22$

$6 / 6 = (6a - 6b) / 6$

$1 + b = a - b + b$

$1 + b = a$

Step 9: Substitute a, b, and c into the third equation

$220 = 64(1 + b) + 16b + 4(-(1 + b) - 3b + 9) + (4 - 4b)$

$220 = 64 + 64b + 16b + 4(-1 - b - 3b + 9) + 4 - 4b$

$220 - 100 = 60b + 100 - 100$

$120 / 60 = 60b / 60$

$2 = b$

Step 10: Find a using b = 2

$a = b + 1$

$a = (2) + 1$

$a = 3$

Step 11: Find c using a = 3 and b = 2

$c = -a - 3b + 9$

$c = -(3) - 3(2) + 9$

$c = -3 - 6 + 9$

$c = 0$

Step 12: Find d using b = 2

$d = 4 - 4b$

$d = 4 - 4(2)$

$d = 4 - 8$

$d = -4$

Answer: $a = 3, b = 2, c = 0,d = -4$

6 0

3 years ago

Read 2 more answers