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Alja [10]
3 years ago
15

Which of the equations below could be the equation of this parabola?

Mathematics
1 answer:
Mademuasel [1]3 years ago
7 0

Answer:

Option B: y = -\frac{1}{2} x^2

Step-by-step explanation:

The parabola has its concavity downwards, so we need a function in the model:

y = ax^2 + bx + c

With a negative value of 'a'

The vertex is (0,0), so we have that:

0 = 0a + 0b + c

c = 0

The x-coordinate of the vertex is given by the equation:

x\_vertex = -b/2a

-b/2a = 0

b = 0

So we have a function in the model:

y = ax^2

With a < 0

The only option with this format is B:

y = -\frac{1}{2} x^2

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URGENT! Describe step by step how to find the following equations.
yuradex [85]

Answer:

1. x = 45°

2. x = 330°

3. x = 105°

4. x = 30°

5. x = 60°

6. x = 30° or x = -45° (315°) or x = 45°

Step-by-step explanation:

1. For

cos x = sin x

∴ sin x/(cos x) = 1 = tan x

Hence, x = tan⁻¹1 = 45°

2. For

csc(x) + 3 = 1

∴ csc(x) = 1 - 3 = -2

Which gives sin x = -1/2

∴ x = sin⁻¹(-1/2) = -30° = 360 +(-30) = 330 °

3. For

cot(3x) = -1

∵ cot(3x) = 1/(tan(3x))

Hence, 1/(tan(3x)) = -1

∴ tan(3x) = -1

3·x = tan⁻¹(-1) = -45° = 360 + (-45) = 315°

Which gives, x = -45/3 = -15° or x = 105°

4. For

2·sin²(x) + 3·sin(x) = 2

We put sin(x) = y to get

2·y² + 3·y = 2 or 2·y² + 3·y - 2 =0

Factorizing gives

(2·y -1)(y+2) =0

∴ y = 1/2 or y = -2

That is, sin(x) = 1/2 or sin(x) = -2

Hence, x = sin⁻¹(1/2) = 30° or x = sin⁻¹(-2) = (-π/2 + 1.3·i)

∴ x = 30°

5. For

4cos²(x) = 3 we have;

cos²(x) = 3/4

cos(x) = √(3/4) = (√3)/2

∴ x = cos⁻¹((√3)/2) = 60°

6. For

4·sin³(x) + 1 = 2·sin²(x) + 2·sin(x)

We put sin(x) = y to get;

4·y³ + 1 = 2·y² + 2·y which gives;

4·y³ + 1 - (2·y² + 2·y) = 0 or 4·y³ -2·y² - 2·y + 1 = 0

Factorizing gives;

\frac{(2x-1)(2x+\sqrt{2}) )(2x-\sqrt{2})}{2} =0

Therefore, x = 1/2 or x = -(√2)/2 or (√2)/2

Therefore, sin(x) = 1/2 or -(√2)/2 or (√2)/2

That is x = sin⁻¹(1/2) = 30 or sin⁻¹(-(√2)/2) = -45 or sin⁻¹((√2)/2) = 45.

8 0
3 years ago
Determine the correct set up for solving the equation using the<br> quadratic formula 4x^2-5=3x+4
slava [35]

You need to subtract everything to the left side and set it equal to zero. Combine like terms.

Then, the coefficient of x^2 is a, the coefficient of x is b, and the constant term is c.

4x^2 - 5 = 3x + 4

4x^2 - 3x - 5 - 4 = 0

4x^2 - 3x - 9 = 0

a = 4; b = -3; c = -9

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

x = \dfrac{-(-3) \pm \sqrt{(-3)^2 - 4(4)(-9)}}{2(4)}

7 0
3 years ago
An old house has a basement stairway that has steps with 7.25 inch vertical risers and 7 inch horizontal treads. What is the slo
svlad2 [7]

Answer:

1.04 inch

Step-by-step explanation:

Given: Old house has a basement stairways with 7.25 inch vertical riser and 7 inch horizontal treads.

Remember; horizontal treads is known as run in steps.

Now, calculating the slope of the stairways.

Formula: Slope= \frac{Rise}{Run}

Rise= 7.25 inch

Run= 7 inch.

Subtituting the value in the formula

Slope= \frac{7.25}{7}= 1.0357 \approx 1.04 (∵ Round to two decimal)

∴ Slope of stairways is 1.04.

4 0
3 years ago
Flying against the wind, a jet travels 5760 miles in nine hours flying with the wind, the same jet travels 5000 miles in five ho
aalyn [17]

Answer:

Step-by-step explanation:

9h+5h(5h-5760)-2+12=y

7 0
3 years ago
Where does the normal line to the parabola y = x − x2 at the point (1, 0) intersect the parabola a second time?
Ksivusya [100]
Find the line that is normal to the parabola at the given point
remember that normal means perpendicular
perpendicular lines have slopes that multiply to -1
we can use point slope form to write the equation of the line since we are given the point (1,0)

we just need the slope

take derivitive
y'=1-2x
at x=1
y'=1-2(1)
y'=1-2
y'=-1

the slope is -1
the perpendicular of that slope is what number we can multiply to get -1
-1 times what=-1?
what=1
duh
so
point (1,0) and slope 1
y-0=1(x-1)
y=x-1 is da equation

solve for where y=x-1 and y=x-x² intersect

set equatl to each other since equal y
x-1=x-x²
x²-1=0
factor difference of 2 perfect squares
(x-1)(x+1)=0
set to zero

x-1=0
x=1
we got this point already

x+1=0
x=-1
sub back
y=-1-(-1)²
y=-1-(1)
y=-1-1
y=-2

it intersects at (-1,-2)
7 0
3 years ago
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