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3 years ago

What is the Slope of 4x-6y=18

Mathematics

1 answer:

Paul [167]3 years ago

8 0

Answer:

2/3

Step-by-step explanation:

To find the slope solve the equation in the form of y=mx+b.

4x-6y=18

-6y= -4x+18

y= 2/3x-3

The slope is 2/3.

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WILL GIVE BRAINLIEST!!!!

Vesna [10]

$(2^8 \times 5^{-5} \times 19^0)^{-2} \times (\dfrac{5^{-2}}{2^3})^4 \times 2^{28} =$

$= 2^{-16} \times 5^{10} \times 1 \times \dfrac{5^{-8}}{2^{12}} \times 2^{28}$

$= 2^{-16} \times 5^{10} \times 5^{-8} \times 2^{-12} \times 2^{28}$

$= 2^{-16} \times 2^{-12} \times 2^{28} \times 5^{10} \times 5^{-8}$

$= 2^{-16-12+28} \times 5^{10-8}$

$= 2^{0} \times 5^{2}$

$= 1 \times 25$

$= 25$

7 0

3 years ago

A cup of coffee contains about 100 mg of caffeine. Caffeine is metabolized and leaves the body at a continuous rate of about 12%

Mamont248 [21]

Answer:

a. $A = C_{0}(1-x)^t\\x: percentage\ of \ caffeine\ metabolized\\$

b. $\frac{dA}{dt}= -11.25 \frac{mg}{h}$

Step-by-step explanation:

First, we need tot find a general expression for the amount of caffeine remaining in the body after certain time. As the problem states that every hour x percent of caffeine leaves the body, we must substract that percentage from the initial quantity of caffeine, by each hour passing. That expression would be:

$A= C_{0}(1-x)^t\\t: time \ in \ hours\\x: percentage \ of \ caffeine\ metabolized\\$

Then, to find the amount of caffeine metabolized per hour, we need to differentiate the previous equation. Following the differentiation rules we get:

$\frac{dA}{dt} =C_{0}(1-x)^t \ln (1-x)\\\frac{dA}{dt} =100*0.88\ln(0.88)\\\frac{dA}{dt} =-11.25 \frac{mg}{h}$

The rate is negative as it represents the amount of caffeine leaving the body at certain time.

3 0

3 years ago

Find the curl of ~V<br> ~V<br> = sin(x) cos(y) tan(z) i + x^2y^2z^2 j + x^4y^4z^4 k

ch4aika [34]

Given

$\vec v = f(x,y,z)\,\vec\imath+g(x,y,z)\,\vec\jmath+h(x,y,z)\,\vec k \\\\ \vec v = \sin(x)\cos(y)\tan(z)\,\vec\imath + x^2y^2z^2\,\vec\jmath+x^4y^4z^4\,\vec k$

the curl of $\vec v$ is

$\displaystyle \nabla\times\vec v = \left(\frac{\partial h}{\partial y}-\frac{\partial g}{\partial z}\right)\,\vec\imath - \left(\frac{\partial h}{\partial x}-\frac{\partial f}{\partial z}\right)\,\vec\jmath + \left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)\,\vec k$

$\nabla\times\vec v = \left(4x^4y^3z^4-2x^2y^2z\right)\,\vec\imath \\\\ - \left(4x^3y^4z^4-\sin(x)\cos(y)\sec^2(z)\right)\,\vec\jmath \\\\ + \left(2xy^2z^2+\sin(x)\sin(y)\tan(z)\right)\,\vec k$

$\nabla\times\vec v = \left(4x^4y^3z^4-2x^2y^2z\right)\,\vec\imath \\\\ + \left(\sin(x)\cos(y)\sec^2(z)-4x^3y^4z^4\right)\,\vec\jmath \\\\ + \left(2xy^2z^2+\sin(x)\sin(y)\tan(z)\right)\,\vec k$