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3 years ago

What conic section is represented by the equation show below? 9x+4y^2+18x=16

2 answers:

7 0

Conic sections are cross-sectional areas of a cone (the 2-d figure you get when you slice a cone). You can get circle, an ellipse, a parabola, or a hyperbola.

In this case it is a hyperbola because, when graphed, it is not a function (so not a parabola) and does not form a circle or ellipse/oval shape.

Serjik [45]3 years ago

4 0

I believe its <span>parabola</span>

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What is the simplest form of ^4 sqrt 81x^8y^5

lara31 [8.8K]

ANSWER

$3 {x}^{2} y \sqrt[4]{y}$

EXPLANATION

We want to simplify:

$\sqrt[4]{81 {x}^{8} {y}^{5} }$

We can split the radical sign to obtain:

$\sqrt[4]{81} \times \sqrt[4]{ {x}^{8} } \times \sqrt[4]{ {y}^{5} }$

$\sqrt[4]{81} \times \sqrt[4]{ {x}^{8} } \times \sqrt[4]{ {y}^{4} \times y}$

$\sqrt[4]{81} \times \sqrt[4]{ {x}^{8} } \times \sqrt[4]{ {y}^{4}} \times \sqrt[4]{y}$

$\sqrt[4]{ {3}^{4} } \times \sqrt[4]{ {x}^{8} } \times \sqrt[4]{ {y}^{4}} \times \sqrt[4]{y}$

Recall that:

$\sqrt[n]{ {a}^{m} } = {a}^{ \frac{m}{n} }$

${3}^{4 \times \frac{1}{4} } \times {x}^{8 \times \frac{1}{4} } \times {y}^{4 \times \frac{1}{4} }\times \sqrt[4]{y}$

$3 {x}^{2} y \sqrt[4]{y}$

6 0

4 years ago

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The local utility company averages rates to determine a standard monthly charge for area businesses. If you are charged $150 for

lisabon 2012 [21]

The average of 3 numbers is their total divided by 3.
(150 +205 +178)/3 ≈ 178.67

Rounded to the nearest dollar, the average charge for the period is $178 per month.

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4 years ago

What is slope????????

Damm [24]

Step-by-step explanation:

slope another name for gradient

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3 years ago

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1/x-2/3=3/2x find x<br>

arsen [322]

Answer:

1/x-2/3=3/2x

-2/3=3/2x-1/x

-2/3=(3-2)/2x

-2/3=1/2x

by cross multiplication

-2(2x)=1(3)

-4x=3

x= -3/4

Step-by-step explanation:

I hope this will help

plz mark as brainliest

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3 years ago

I am thinking of a number. It is increased by 7. The sun is 21. What is my number?

Semenov [28]

21-7=14
The answer is 14

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3 years ago

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