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3 years ago

A braclet is $60 and additional $10 dollars for each charm what would the equation be in y=mx+b form?? HELLLLPPP!!

Mathematics

1 answer:

disa [49]3 years ago

6 0

Y= 10x+60 is the answer i think

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A package of 25 wristbands cost $5.25. At this rate, what is the cost of 1 wristband in dollars and cents?

azamat

Answer:

$0.21

Step-by-step explanation:

We can set up a proportion. Say x is the cost of 1 wristband:

$\frac{5.25}{25} =\frac{x}{1}$

Now, we cross multiply:

25x = 5.25

Solving for x, we see that x = 0.21.

Thus, the answer is $0.21.

6 0

2 years ago

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Find the value of y in this equation 16y=164

fiasKO [112]

16y = 164
y = 164 ÷ 16
y = 10.25

Hope This Helps You!

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3 years ago

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HELP PLEASE YOU'LL GET BRAINLIEST :)

inna [77]

Step-by-step explanation:

you would divide 489 by 10 and then times that number by 15 and the divide 20 and get your answer that way

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1 year ago

If (x+4)*2=49 and x<0, what is the value of x

Nady [450]

Answer:

41 is the answer bro thanks

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3 years ago

use green's theorem to evaluate the line integral along the given positively oriented curve. c 9y3 dx − 9x3 dy, c is the circle

Rina8888 [55]

The line integral along the given positively oriented curve is -216π. Using green's theorem, the required value is calculated.

<h3>What is green's theorem?</h3>

The theorem states that,

$\int_CPdx+Qdy = \int\int_D(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y})dx dy$

Where C is the curve.

<h3>Calculation:</h3>

The given line integral is

$\int_C9y^3dx-9x^3dy$

Where curve C is a circle x² + y² = 4;

Applying green's theorem,

P = 9y³; Q = -9x³

Then,

$\frac{\partial P}{\partial y} = \frac{\partial 9y^3}{\partial y} = 27y^2$

$\frac{\partial Q}{\partial x} = \frac{\partial -9x^3}{\partial x} = 27x^2$

$\int_C9y^3dx-9x^3dy = \int\int_D(-27x^2 - 27y^2)dx dy$

⇒ $-27\int\int_D(x^2 + y^2)dx dy$

Since it is given that the curve is a circle i.e., x² + y² = 2², then changing the limits as

0 ≤ r ≤ 2; and 0 ≤ θ ≤ 2π

Then the integral becomes

$-27\int\limits^{2\pi}_0\int\limits^2_0r^2. r dr d\theta$

⇒ $-27\int\limits^{2\pi}_0\int\limits^2_0 r^3dr d\theta$

⇒ $-27\int\limits^{2\pi}_0 (r^4/4)|_0^2 d\theta$

⇒ $-27\int\limits^{2\pi}_0 (16/4) d\theta$

⇒ $-108\int\limits^{2\pi}_0 d\theta$

⇒ $-108[2\pi - 0]$

⇒ -216π

Therefore, the required value is -216π.

Learn more about green's theorem here:

brainly.com/question/23265902

#SPJ4

3 0

1 year ago