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3 years ago

One x-intercept for a parabola is at the point

Mathematics

1 answer:

ra1l [238]3 years ago

3 0

Answer:

x=-1 or (-1,0)

Step-by-step explanation:

(-4 +- sqrt(4^2-4(5)(-1)))/2*5

(-4 +- sqrt(16+20))/10

(-4 +- sqrt(36))/10

(-4 +- 6)/10

x= (-4+6)/10 = 2/10 = 1/5 = 0.2

x= (-4-6)/10 = -10/10 = -1

x-intercepts: (0.2,0), (-1,0)

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PLEASE HELP AHHH I-. I need help lol

Pie

Answer:

264

Step-by-step explanation:

do 20 mulitplyed by 30 then subtracted it by 336

4 0

3 years ago

Read 2 more answers

Do diagonals of a rhombus bisect the vertex angles???

katrin [286]

Answer:

Yes, one of a rhombus's properties is that the diagonals bisect vertex angles.

7 0

3 years ago

Someone please help me

allsm [11]

Answer:

Solutions: $x = \frac{-3}{ 4} + i \sqrt{39}$ , $x = \frac{-3}{4} - i \sqrt{39}$

Step-by-step explanation:

Given the quadratic equation, 2x² + 3x + 6 = 0, where a =2, b = 3, and c = 6:

Use the quadratic equation and substitute the values for a, b, and c to solve for the solutions:

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

$x = \frac{-3 +/- \sqrt{3^{2} - 4(2)(6)} }{2(2)}$

$x = \frac{-3 +/- \sqrt{9- 48} }{4}$

$x = \frac{-3 +/- \sqrt{-39} }{4}$

$x = \frac{-3 + i \sqrt{39} }{4}$ , $x = \frac{-3 - i \sqrt{39} }{4}$

Therefore, the solutions to the given quadratic equation are:

$x = -\frac{3}{ 4} + i \sqrt{39}$ , $x = -\frac{3}{4} - i \sqrt{39}$

4 0

3 years ago

Given: F(x) = 2x + 3, find F(X+h)-f(x) and simplify. 01 02 (2h+6)/h

antiseptic1488 [7]

F(x+h) = 2(x+h) +3= 2x + 2h +3
f(x) = 2x + 5

f(x+h) - f(x) = 2x + 2h + 3- 2x - 3= 2h

[f(x+h) - f(x)]/h = 2h/h = 2

6 0

3 years ago

What is the value of this expression?

Zepler [3.9K]

Answer:

-3

Step-by-step explanation:

Step 1: Solve (-2+(-1))^2/3 3

1. -2+(-1) = -3

2. (-3)^2 = 9

3. 9/3 = 3

Step 2: Solve (-4)^2-17 -1

1. 3/-1

Step 3: Simplify 3/-1 = -3. I hope this helped and please don't hesitate to reach out with more questions!

5 0

3 years ago