All categories

2 years ago

What point is on both lines y=12x+3 and y=x+1 (no graph)

Mathematics

2 answers:

velikii [3]2 years ago

7 0

Answer:

(-2|11, 9|11)

Step-by-step explanation:

hope it helps

Mkey [24]2 years ago

4 0

Step-by-step explanation:

by having two equations that have the same point, you can set the two equations equal to each other

y = 12x + 3, y = x + 1

12x + 3 = x + 1

11x = -2

x = -2/11

when you find the the x value, plug it back into either one of the equations

y = x + 1

y = (-2/11) + 1

y = 11/11 - 2/11

y = 9/11

the point (solution):

(-2/11, 9/11)

You might be interested in

The county fair is in town. The cost is $10 for admission and 50 cents for each ride.

STatiana [176]

Answer:

Step-by-step explanation:

For every ride you, pay an extra 50 cents with a 10 dollars for admission

8 0

3 years ago

Read 2 more answers

4(4q + 8r) marked as BRAINLIEST:)

tatyana61 [14]

Answer:

8q+32r

Step-by-step explanation:

8 0

2 years ago

7. Use the quadratic formula to find the solution(s). x² + 2x - 8 = 0

morpeh [17]

Hey there!

Use the quadratic formula to find the solution(s). x² + 2x - 8 = 0

Answer :

x = -4 or x = 2 ✅

Explanation :

Quadratic formula : ax² + bx + c = 0 where a ≠ 0

The number of real-number solutions (roots) is determined by the discriminant (b² - 4ac) :

If b² - 4ac > 0 , There are 2 real-number solutions

If b² - 4ac = 0 , There is 1 real-number solution.

If b² - 4ac < 0 , There is no real-number solution.

The roots of the equation are determined by the following calculation:

$x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}$

Here, we have :

a = 1
b = 2
c = -8

1) Calculate the discriminant :

b² - 4ac ⇔ 2² - 4(1)(-8) ⇔ 4 - (-32) ⇔ 36

b² - 4ac = 36 > 0 ; The equation admits two real-number solutions

2) Calculate the roots of the equation:

▪️ (1)

$x_1 = \frac{ - b - \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\ x_1 = \frac{ - 2 - \sqrt{36} }{2(1) } \\ \\ x_1 = \frac{ - 2 - 6}{2} \\ \\ x_1 = \frac{ - 8}{2} \\ \\ \blue{\boxed{\red{x_1 = -4}}}$

▪️ (2)

$x_2 = \frac{ - b + \sqrt{ {b}^{2} - 4ac } }{2a} \\ \\ x_2 = \frac{ - 2 + \sqrt{36} }{2(1)} \\ \\ x_2 = \frac{ - 2 + 6}{2} \\ \\ x_2 = \frac{4}{2} \\ \\ \red{\boxed{\blue{x_2 = 2}}}$