If y varies directly as x and y = 49 when x = 7, what is the value of y when x = 3?

Answers

Answer 1
Answer: Direct variation:
y=kx

y=49 when x=7.
49=7k \nk=(49)/(7) \nk=7

y=? when x=3.
y=7 * 3 \ny=21

y=21 when x=3.
Answer 2
Answer: y=21 because 3/21=7 and 7*3=21 so Y=21 

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The part of the sphere x2 + y2 + z2 = 16 that lies above the cone z = x2 + y2 . (Enter your answer as a comma-separated list of equations. Let x, y, and z be in terms of u and/or v.) where z > x2 + y2?

Answers

The required, there is no part of the sphere x² + y² + z² = 16 that lies above the cone z = x² + y², where z > x² + y².

To find the part of the sphere x² + y² + z² = 16 that lies above the cone z = x² + y², where z > x² + y², we can use spherical coordinates. In spherical coordinates, the equations for the sphere and the cone are simpler.

Spherical coordinates are represented as (ρ, θ, φ), where ρ is the radial distance, θ is the azimuthal angle (measured from the positive x-axis in the xy-plane), and φ is the polar angle (measured from the positive z-axis).

For the sphere x² + y² + z² = 16, the spherical representation is:

ρ = 4 (since ρ² = x² + y² + z² = 16)

For the cone z = x² + y², the spherical representation is:

ρ = ρ (since ρ^2 = x² + y²)

Now, to find the part of the sphere that lies above the cone (z > x² + y^2), we need to restrict the values of φ.

When z > x² + y², we have z = ρ cos(φ) > ρ².

Since ρ = 4, we get 4 cos(φ) > 4², which simplifies to cos(φ) > 4.

However, the range of φ in spherical coordinates is 0 ≤ φ ≤ π, which means that the values of φ that satisfy cos(φ) > 4 are not within the valid range.

Therefore, there is no part of the sphere x² + y² + z² = 16 that lies above the cone z = x² + y², where z > x² + y².

Learn more about Sphere here:

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Final answer:

We use the given equations of the sphere and cone and express them in spherical coordinates. The sphere lies on or above the cone when z's value in the sphere equation is greater or equal than z's value in the cone equation. One method is to use spherical coordinates and represent the radius and polar angle in terms of u and v.

Explanation:

The question involves spherical and rectangular coordinates and the relationship between the two. We are given the sphere's equation as x^2 + y^2 + z^2 = 16 and the cone's equation as z = x^2 + y^2. Here's one way to think of the part of the sphere that lies on or above the cone. If we view z=x^2 + y^2 as a function of x and y, the sphere lies above this cone when z's value in the equation of the sphere is greater or equal to the value of z in the cone's equation. To express x, y, and z in terms of u and/or v, you can use a method such as spherical coordinates.

In spherical coordinates, the relationship between spherical and rectangular coordinates can be represented as:

  • x = r sin θ cos φ
  • y = r sin θ sin φ
  • z = r cos θ

Here r, θ, and φ are the radius, polar, and azimuthal angles respectively, which we can let u and v represent. One potential assignment is to let r=u and θ=v, assuming we want only two parameters.

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If x is greater than 0 and y is greater than 0, then which quadrant holds the solutions?

Answers

If both x and y are greater than 0 then it is in quadrant 1

-2.678 + 4.5 + (-0.69)

Answers

1.132isthe correct answer of this question

Triangle ABC is translated to image A′B′C′. In this translation, A(5, 1) maps to A′(6, –2). The coordinates of B′ are (–1, 0). What are the coordinates of B? B( , )

Answers

Answer:

B' (-2, 3)

Step-by-step explanation:

A(5, 1)  to A'(6, -2)

A(6-5, -2-1)

A translation is 1, -3

consider the same translation of B'

therefore, the coordinates of B' is (-2, 3)

Answer:

B. -2, 3

Step-by-step explanation:

edge2021

A vegetable garden with an area of 200 square feet is to be fertilized. If the length of the garden is 1 foot less than three times the width, find the dimensions of the garden. Please show work.

Answers

Answer:

Dimensions of the garden are 24 feet by 8.33 feet.

Step-by-step explanation:

Let the dimensions of the vegetable garden is length = l and width = w foot

Area of the vegetable garden = 200 square feet

Since length of the garden is 1 foot less than the 3 times the width.

l = 3w - 1

Area of the garden = w × l = 200

w(3w - 1) = 200

3w² - w = 200

3w² - w - 200 = 0

By quadratic formula

w = \frac{1\pm \sqrt{1^(2)+2400}}{6}

  = (1\pm √(2401))/(6)

  = (1\pm 49)/(6)

w = -8 or 8.33 foot

Since dimensions can not be negative therefore, width of the garden will be 8.33 foot

Since l = 3w - 1

           = 3×8.33 - 1

           = 25 - 1

           = 24

Dimensions of the garden are 24 feet by 8.33 feet

Ok so I like to go in steps with these questions- first draw a picture and identify your variables.

W=width
L= 3w-1

Now we know that length times width gets us area so we plug in our variables into the area equation.

200 = w(3w-1)

When you foil that equation you end up with a quadratic : 3w^2-w-200 = 0

Either factor that or use the quadratic formula to get
w= 8.33 and w= -8

Since you can't have a negative dimension you need to use 8.33 and plug it back into your length equation.

Final answer:

w= 8.33ft
l= 23.99ft

*Now I simplified the decimals a little bit so you end up with 199.8ft^2 for the area so just add a few decimals on here and there*

What is the 217th digit after the decimal point in the repeating decimal 0.3456? How can i calculate this?

Answers

Well I think you mean 0.(3456), I will try to explain as simple as I can ;p

First I would divide 217 by 4, because we need to find 217th digit in 4 digit repeating decimal:

217/4=(217)/(4)=54(1)/(4)

54(1)/(4) it means that 3456 will be reapeted 54 times untill 217th digit, and (1)/(4) means that the 217th digit will be the first of those 4 digits in this decimal, so your answer is:
The 217th number after the decimal point in this repeating decimal will be 3