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3 years ago

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Write an equation for the statement: A kilogram is equal to approximately 2.2 pounds. A cinder block weighs x pounds. A brick ha

s a mass of 0.6 kilogram. What is the total mass in kilograms y of the cinder block and brick?

1 answer:

navik [9.2K]3 years ago

7 0

the total mass in kilograms y of the cinder block and brick is $y=0.45x + 0.6$ .

<u>Step-by-step explanation:</u>

Here we have , A kilogram is equal to approximately 2.2 pounds. A cinder block weighs x pounds. A brick has a mass of 0.6 kilogram. We need to find What is the total mass in kilograms y of the cinder block and brick . Let's find out :

According to given data in question ,

⇒ $1kg = 2.2pounds$

⇒ $1(pounds) = \frac{1}{2.2}kg$

A cinder block weighs x pounds and a brick has a mass of 0.6 kilogram . Let total mass in kg of the cinder block and brick is y , i.e.

⇒ $y (kg)= x (pounds) + 0.6 kg$

⇒ $y (kg)= x (\frac{1}{2.2}kg) + 0.6 kg$

⇒ $y=0.45x + 0.6$

Therefore , the total mass in kilograms y of the cinder block and brick is $y=0.45x + 0.6$ .

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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.

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

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3 years ago

8+[5×6]= i need answer asap pls

Karo-lina-s [1.5K]

Answer:

38

Step-by-step explanation:

Follow orders of operations.

Slove the expression in the parenthesis or brackets whatever, same thing.

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30+8=38

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3 years ago

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Answer:

C

Step-by-step explanation:

Calculate the distance (d) using the distance formula

d = √ (x₂ - x₁ )² + (y₂ - y₁ )²

with (x₁, y₁ ) = (- 3, 9) and (x₂, y₂ ) = (3, 1)

d = $\sqrt{(3+3)^2+(1-9)^2}$

= $\sqrt{6^2+(-8)^2}$

= $\sqrt{36+64}$

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Matt earns $5 for each lawn he does. He wants to earn at least $100. Which inequality represents his situation?

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Answer:

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Step-by-step explanation:

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$r^2-10r+25=0$

Finding the values of r

$r^2-5r-5r+25=0\\\\r(r-5)-5(r-5)=0\\\\(r-5)(r-5)=0\\\\r_{1,2}=5$

We got a repeated roots. Hence, the solution of the differential equation is given by

$y(x)=c_1e^{5x}+c_2xe^{5x}...(i)$

On differentiating, we get

$y'(x)=5c_1e^{5x}+5c_2xe^{5x}+c_2e^{5x}...(ii)$

Apply the initial condition y (0)= 3 in equation (i)

$3=c_1e^{0}+0\\\\c_1=3$

Now, apply the initial condition y' (0)= 13 in equation (ii)

$13=5(3)e^{0}+0+c_2e^{0}\\\\13=15+c_2\\\\c_2=-2$

Therefore, the solution of the differential equation is

$y(x)=3e^{5x}-2xe^{5x}$

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3 years ago