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1 year ago

Does the graph represent a function?

Mathematics

1 answer:

Alex1 year ago

5 0

keeping in mind that a function does not have any X-Repeats, that is, the 1st coordinate in every pair never repeats, Check the picture below.

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49x^2 + 42xy + 9y^2 Factor completely.

mario62 [17]

(7x-3y)^2 hope this helps

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2 years ago

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Please help. Is algebra. PLEASE HELP NO LINKS OR FILES

vovikov84 [41]

The answer is C! Sorry if u get it wrong or I’m too late

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3 years ago

Need help please!!!!

ArbitrLikvidat [17]

Given : A inequality is given to us . The inequality is 19 ≥ t + 18 ≥ 11 .

To Find : The correct option between the given ones . To write the compound inequality with integers .

Solution : The given inequality to us is 19 ≥ t + 18 ≥ 11 . Let's simplify them seperately .

$\red{\tt Case\:1}$

⇒ 19 ≥ t + 18 .

⇒ t + 18 ≤ 19 .

⇒ t ≤ 19 - 18 .

⇒ t ≤ 1 = 1 ≥ t . ..................(i)

$\red{\tt Case\:2}$

⇒ t + 18 ≥ 11 .

⇒ t ≥ 11 - 18.

⇒ t ≥ -7 . ....................(ii)

On combing (i) & (ii) .

$\Large\boxed{\green{\bf \red{\longmapsto} 1 \geqslant t \geqslant -7 }}$

This means that t is less than or equal to 1 but greater than or equal to (-7) .

4 0

3 years ago

In a certain region of space, the potential is given by v=2x−5x2y 3yz2. How much is the magnitude of the electric field at point

Masja [62]

The magnitude of the electric field at the point (-2, 2, 0) is 46.52 V/m

<h3>How to calculate the electric field?</h3>

Since in a certain region of space, the potential is given by v = 2x − 5x²y 3yz²,

The electric field, E = -grad(V) where grad(V) = (dV/dx)i + (dV/dy)j + (dV/dz)k

Since V = 2x − 5x²y + 3yz²

dV/dx = d(2x − 5x²y + 3yz²)/dx

= d2x/dx - d5x²y/dx + d3yz²/dx

= 2 - 10xy + 0

= 2 - 10xy

dV/dy = d(2x − 5x²y + 3yz²)/dy

= d2x/dy - d5x²y/dy + d3yz²/dy

= 0 - 5x² + 3z²

= - 5x² + 3z²

dV/dz = d(2x − 5x²y + 3yz²)/dz

= d2x/dz - d5x²y/dz + d3yz²/dz

= 0 - 0 + 6z

= 6z

<h3>The electric field</h3>

So, E = -grad(V)

= -[(dV/dx)i + (dV/dy)j + (dV/dz)k]

= -[(2 - 10xy)i + (- 5x² + 3z²)j + (6z)k]

So, the electric field at (-2, 2, 0) is

E = -[(2 - 10xy)i + (- 5x² + 3z²)j + (6z)k]

E = -[(2 - 10(-2)(2))i + (- 5(2)² + 3(0)²)j + (6(0))k]

E = -[(2 + 40)i + (- 5(4) + 0)j + (0)k]

E = -[42i + (-20 + 0)j + (0)k]

E = -[42i - 20j + 0k]

E = -42i + 20j - 0k

<h3>The magnitude of the electric field</h3>

So, the magnitude of E at (-2, 2, 0) is

|E| = √[(-42)² + 20² + 0²]

= √[1764 + 400 + 0]

= √2164

= 46.52 V/m

So, the magnitude of the electric field at the point (-2, 2, 0) is 46.52 V/m

Learn more about electric field here:

brainly.com/question/25751825

#SPJ12

3 0

2 years ago

If the diameter of a circle is 14cm What is the measure of its circumference ? find it

Brums [2.3K]

Answer:

$Solution\\diameter(d)=14cm\\Now, \\Circumference=\pi d\\ =3.14*14cm\\ =43.96cm$

Step-by-step explanation:

Hope it will help you a lot.

3 0

2 years ago

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