All categories

3 years ago

Can anyone please help,me with this?

Mathematics

1 answer:

jeyben [28]3 years ago

7 0

Remark

To find the location of the focus, you need to know something about what you are being told about the horizontal line cutting the parabola at y = 4

The general formula for this is v * y = x^2. You need to have a way of finding v

Solve

When x = 4, y = 4 (from the graph)

v * 4 = 4^2

4v = 16

v = 16/4

v = 4

Now what you have is x^2 = 4p * y

4p = 4 because the value in front of y (I called it v) = 4

p = 4/4 = 1

The focus is at h,k + p

The x value of the focus is 0

The y value of the focus is 0 + 1 = 1

The focus is located at (0,1)

The y value of the focus is 0 + 1

You might be interested in

Solve the rest for me plz

Luba_88 [7]

Lo haria si pudiera que pena contigo

5 0

2 years ago

Read 2 more answers

-2/3- (-3/4) Simplify if needed Brainliest Also, 5/6+5/12

bazaltina [42]

Answer:

$-\frac{2}{3} - (-\frac{3}{4}) = \frac{1}{12}$

$\frac{5}{6} + \frac{5}{12} = \frac{5}{3}$

Step-by-step explanation:

Given

$-\frac{2}{3} - (-\frac{3}{4})$

Required

Solve:

$-\frac{2}{3} - (-\frac{3}{4})$

Open bracket

$-\frac{2}{3} - (-\frac{3}{4}) = -\frac{2}{3} +\frac{3}{4}$

Take LCM

$-\frac{2}{3} - (-\frac{3}{4}) = \frac{-8+9}{12}$

$-\frac{2}{3} - (-\frac{3}{4}) = \frac{1}{12}$

$\frac{5}{6} + \frac{5}{12}$

Take LCM

$\frac{5}{6} + \frac{5}{12} = \frac{10+5}{12}$

$\frac{5}{6} + \frac{5}{12} = \frac{15}{12}$

Divide by 3/3

$\frac{5}{6} + \frac{5}{12} = \frac{5}{3}$

4 0

3 years ago

What's the numerator for the following rational expression? 8/y+3/y=?/y

GuDViN [60]

Ur answer would be 11/y

3 0

3 years ago

Read 2 more answers

Find the longer leg of the triangle.

Paha777 [63]

Answer:

Choice A. 3.

Step-by-step explanation:

The triangle in question is a right triangle.

The length of the hypotenuse (the side opposite to the right angle) is given.
The measure of one of the acute angle is also given.

As a result, the length of both legs can be found directly using the sine function and the cosine function.

Let $\text{Opposite}$ denotes the length of the side opposite to the $30^{\circ}$ acute angle, and $\text{Adjacent}$ be the length of the side next to this $30^{\circ}$ acute angle.

$\displaystyle \begin{aligned}\text{Opposite} &= \text{Hypotenuse} \times \sin{30^{\circ}}\\ &=2\sqrt{3}\times \frac{1}{2} \\&= \sqrt{3}\end{aligned}$ .

Similarly,

$\displaystyle \begin{aligned}\text{Adjacent} &= \text{Hypotenuse} \times \cos{30^{\circ}}\\ &=2\sqrt{3}\times \frac{\sqrt{3}}{2} \\&= 3\end{aligned}$ .

The longer leg in this case is the one adjacent to the $30^{\circ}$ acute angle. The answer will be $3$ .

There's a shortcut to the answer. Notice that $\sin{30^{\circ}} < \cos{30^{\circ}}$ . The cosine of an acute angle is directly related to the adjacent leg. In other words, the leg adjacent to the $30^{\circ}$ angle will be the longer leg. There will be no need to find the length of the opposite leg.