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USPshnik [31]
3 years ago
11

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:

<u>Extended Information </u><u>on</u><u> </u><u>the</u><u> Complex Number System</u>

\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. ↑

4 0
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 <u>lines</u> that lie on the same <u>plane</u>.

Theorem: If two <u>coplanar lines</u> are <u>perpendicular</u> to the same  line, then the two lines are <u>parallel</u> 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

Read more about the plane

brainly.com/question/1962726

#SPJ4

5 0
1 year ago
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