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3 years ago

A parabola is defined by the equation x2 = y. In which direction will the parabola open?

Mathematics

2 answers:

vlada-n [284]3 years ago

4 0

Up or to the right if it's positive down or left of its negative

Oliga [24]3 years ago

3 0

Up, just put some points in the graphic and you'll see

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What percent of 56 is 35 shirts

9966 [12]

62.5% of 56 is 35

Divide 35 and 56:
$35 \div 56 = 0.625$

Convert the decimal to a percentage:
$0.625 \times 100 = 62.5$

8 0

3 years ago

What is the value of 5^3i^9?

Yakvenalex [24]

Answer:

$\displaystyle 125^i$

Step-by-step explanation:

Extended Information on the Complex Number System

$\displaystyle \sqrt{-1} = i \\ -1 = i^2 \\ -i = i^3 \\ 1 = i^4\:[all\:multiples\:of\:four]$

I am joyous to assist you at any time.

All work is shown above. ↑

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3 years ago

SOMEBODY PLEASE HELP ASAP!!! WILL GIVE BRAINLIEST

Lana71 [14]

Answer:

a- 130, 130, and 50

b- 7 and 10

7 0

3 years ago

Read 2 more answers

Two coplanar lines that are perpendicular to the same line are parallel.

Semenov [28]

Coplanar lines are lines that lie on the same plane.

Theorem: If two coplanar lines are perpendicular to the same line, then the two lines are parallel to each other.

This theorem is true always, therefore, given statement is true always.

Answer: correct choice is A

3 0

3 years ago

Read 2 more answers

Find the area of the part of the plane 3x 2y z = 6 that lies in the first octant.

gavmur [86]

The area of the part of the plane 3x 2y z = 6 that lies in the first octant is mathematically given as

A=3 √(4) units ^2

<h3>What is the area of the part of the plane 3x 2y z = 6 that lies in the first octant.?</h3>

Generally, the equation for is mathematically given as

The Figure is the x-y plane triangle formed by the shading. The formula for the surface area of a z=f(x, y) surface is as follows:

$A=\iint_{R_{x y}} \sqrt{f_{x}^{2}+f_{y}^{2}+1} d x d y(1)$

The partial derivatives of a function are f x and f y.

$\begin{aligned}&Z=f(x)=6-3 x-2 y \\&=\frac{\partial f(x)}{\partial x}=-3 \\&=\frac{\partial f(y)}{\partial y}=-2\end{aligned}$

When these numbers are plugged into equation (1) and the integrals are given bounds, we get:

$&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{(-3)^{2}+(-2)^2+1dxdy} \\\\&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{14} d x d y \\\\&=\sqrt{14} \int_{0}^{2}[y]_{0}^{3-\frac{3}{2} x} d x d y \\\\&=\sqrt{14} \int_{0}^{2}\left[3-\frac{3}{2} x\right] d x \\\\$

$&=\sqrt{14}\left[3 x-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3.2-\frac{3}{2} \cdot \frac{1}{2} \cdot 3^{2}\right] \\\\&=3 \sqrt{14} \text { units }{ }^{2}$

In conclusion, the area is

A=3 √4 units ^2