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3 years ago

PLZ HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

notsponge [240]3 years ago

6 0

Answer:

$72 would be the retail price

Step-by-step explanation:

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Write an equation of the line that passes through the point (8, 1) with slope 5.

Lapatulllka [165]

Answer:

Step-by-step explanation:

The slope-intercept form of an equation of a line:

$y=mx+b$

m - slope

b - y-intercept

Put the value of slope m = 5 and the coordinates of the given point (8, 1) to the equation of a line:

$1=(5)(8)+b$

$1=40+b$ subtract 40 from both sides

$-39=b\to b=-39$

Finally:

$y=5x-39$

4 0

4 years ago

5th grade pls help ik my pfp is my face but I look like a different grade

Andrew [12]

Answer:

Step-by-step explanation:

First, you divide the 10 and the 5, which leaves you with 2. Then you do 15-2+5 from left to right, and it equals 18.

(hope this helps...brainliest??)

3 0

3 years ago

Por Favor! Alguém me ajuda.

Nutka1998 [239]

Answer:

The answer is A.

Step-by-step explanation:

You have to apply :

${(a - b)}^{2} ⇒ {a}^{2} - 2ab + {b}^{2}$

So for this question :

${(3a - b)}^{2}$

$= {(3a)}^{2} - 2(3a)(b) + {(b)}^{2}$

$= 9 {a}^{2} - 6ab + {b}^{2}$

8 0

4 years ago

Which ordered pairs are on the line with equation 3x-y=2.

Juli2301 [7.4K]

The ordered pair (0 , -2) lies on the line with equation 3 x - y = 2

Step-by-step explanation:

To prove that a point lies on a line

Substitute x and y in the equation by the coordinates of the point
If the two sides of the equation equal each other, then the point lies on the line
If the two sides of the equation not equal each other then the point does't lie on the line

The equation of the line is 3 x - y = 2

a) Point (0 , -2)

∵ x = 0 and y = -2

- Substitute the values of x and y in the left hand side

∵ The left hand side is 3 x - y

∵ 3(0) - (-2) = 0 + 2 = 2

∴ The left hand side = 2

∵ The right hand side = 2

∴ The two sides of the equation are equal

∴ The ordered pair (0 , -2) lies on the line

b) Point (-3 , 4)

∵ x = -3 and y = 4

- Substitute the values of x and y in the left hand side

∵ The left hand side is 3 x - y

∵ 3(-3) - (4) = -9 - 4 = -13

∴ The left hand side = -13

∵ The right hand side = 2

∴ The two sides of the equation are not equal

∴ The ordered pair (-3 , 4) doesn't lie on the line

c) Point (1 , -5)

∵ x = 1 and y = -5

- Substitute the values of x and y in the left hand side

∵ The left hand side is 3 x - y

∵ 3(1) - (-5) = 3 + 5 = 8

∴ The left hand side = 8

∵ The right hand side = 2

∴ The two sides of the equation are not equal

∴ The ordered pair (1 , -5) doesn't lie on the line

The ordered pair (0 , -2) lies on the line with equation 3 x - y = 2

Learn more:

You can learn more about the linear equation in brainly.com/question/1979240

#LearnwithBrainly

4 0

4 years ago

Find thd <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D" id="TexFormula1" title="\frac{dy}{dx}" alt="\frac{dy}{dx}" a

NARA [144]

$x^3y^2+\sin(x\ln y)+e^{xy}=0$

Differentiate both sides, treating $y$ as a function of $x$ . Let's take it one term at a time.

Power, product and chain rules:

$\dfrac{\mathrm d(x^3y^2)}{\mathrm dx}=\dfrac{\mathrm d(x^3)}{\mathrm dx}y^2+x^3\dfrac{\mathrm d(y^2)}{\mathrm dx}$

$=3x^2y^2+x^3(2y)\dfrac{\mathrm dy}{\mathrm dx}$

$=3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}$

Product and chain rules:

$\dfrac{\mathrm d(\sin(x\ln y)}{\mathrm dx}=\cos(x\ln y)\dfrac{\mathrm d(x\ln y)}{\mathrm dx}$

$=\cos(x\ln y)\left(\dfrac{\mathrm d(x)}{\mathrm dx}\ln y+x\dfrac{\mathrm d(\ln y)}{\mathrm dx}\right)$

$=\cos(x\ln y)\left(\ln y+\dfrac1y\dfrac{\mathrm dy}{\mathrm dx}\right)$

$=\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}$

Product and chain rules:

$\dfrac{\mathrm d(e^{xy})}{\mathrm dx}=e^{xy}\dfrac{\mathrm d(xy)}{\mathrm dx}$

$=e^{xy}\left(\dfrac{\mathrm d(x)}{\mathrm dx}y+x\dfrac{\mathrm d(y)}{\mathrm dx}\right)$

$=e^{xy}\left(y+x\dfrac{\mathrm dy}{\mathrm dx}\right)$

$=ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}$

The derivative of 0 is, of course, 0. So we have, upon differentiating everything,

$3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}+\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}+ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}=0$

Isolate the derivative, and solve for it:

$\left(6x^3y+\dfrac{\cos(x\ln y)}y+xe^{xy}\right)\dfrac{\mathrm dy}{\mathrm dx}=-\left(3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}\right)$

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}}{6x^3y+\frac{\cos(x\ln y)}y+xe^{xy}}$

(See comment below; all the 6s should be 2s)

We can simplify this a bit by multiplying the numerator and denominator by $y$ to get rid of that fraction in the denominator.

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^3+y\cos(x\ln y)\ln y-y^2e^{xy}}{6x^3y^2+\cos(x\ln y)+xye^{xy}}$

3 0

3 years ago