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3 years ago

Someone plz help me I’ll mark as brainliest

Mathematics

2 answers:

mote1985 [20]3 years ago

8 0

Answer:

Step-by-step explanation:

The total cost is $30 and you got 20% off

30 x .20 = 6 so you saved $6

faltersainse [42]3 years ago

5 0

It is $6 hope this helps

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In need help asap-10 points

Dmitry [639]

Answer:

9.8 km

Step-by-step explanation:

7 0

3 years ago

Solve y=-2x+6 and 3y-x+3=0 using substitution

Mazyrski [523]

3(-2x+6) -x+3=0
-6x+18-x+3=0
-7x+21=0
-7x=-21
x=3

y=-2x+6
y=-2(3)+6
=6-6
=0

answer: (3,0)

7 0

3 years ago

A tank contains 100 kg of salt and 1000 L of water. A solution of a concentration 0.05 kg of salt per liter enters a tank at the

Umnica [9.8K]

Answer:

Step-by-step explanation:

Initial Quantity of salt Q(0) = 100 kg

Capacity of tank = 1000 L

Inflow of liquid = 10 l /minute and outflow = same rate thus leaving the content of volume 1000 L at any time

Assume after entering constantly mixes and drains out.

Concentration flow = inflow - outflow

= 0.05(10) - $\frac{Q(t)}{1000} *10$

i.e. $Q'(t) = 0.5-0.01Q(t)$ , Q(0) = 100

$Q'(t) = -0.01(Q(t)-50), Q(0) = 100\\\frac{dQ}{Q-50} =-0.01 dt\\ln |Q-5| = -0.01t +C\\Q-50 = Ae^{-0.01t}$

Use Q(0) = 5

-45 = A

So the equation would be

$Q-50 = -45 e^{-0.01t} \\Q(t) = 50-45 e^{-0.01t} \\$

a) Initial concentratin = $\frac{100}{1000} =0.10$

b) Use the solution fo rDE substitute t =1

Q(1) = 50-45e^(-0.01) = 5.4477 kg

c) As t becomes large Q = 50

So concentration limit = 50/1000 = 0.05

8 0

3 years ago

A ten-foot-long board is cut into three pieces. The second piece is half as long as the first. The third piece is 4 feet longer

VladimirAG [237]

Let's say the second piece is x ft long
The first piece is 2x ft long
The third piece is x+4 ft long
All of the pieces should add up to 10ft
x+2x+x+4=10
4x= 6
x=1.5ft
The first piece is 2x ft long=3 ft long

7 0

3 years ago

Differentiate with respect to x and simplify your answer. Show all the appropriate steps? 1.e^-2xlog(ln x)^3 2.e^-2x(log(ln x))^

serious [3.7K]

(1) I assume "log" on its own refers to the base-10 logarithm.

$\left(e^{-2x}\log(\ln x)^3\right)'=\left(e^{-2x}\right)'\log(\ln x)^3+e^{-2x}\left(\log(\ln x)^3\right)'$

$=-2e^{-2x}\log(\ln x)^3+\dfrac{e^{-2x}}{\ln10(\ln x)^3}\left((\ln x)^3\right)'$

$=-2e^{-2x}\log(\ln x)^3+\dfrac{3e^{-2x}(\ln x)^2}{\ln10(\ln x)^3}\left(\ln x\right)'$

$=-2e^{-2x}\log(\ln x)^3+\dfrac{3e^{-2x}(\ln x)^2}{\ln10\,x(\ln x)^3}$

$=-2e^{-2x}\log(\ln x)^3+\dfrac{3e^{-2x}}{\ln10\,x\ln x}$

Note that writing $\log(\ln x)^3=3\log(\ln x)$ is one way to avoid using the power rule.

(2)

$\left(e^{-2x}(\log(\ln x))^3\right)'=(e^{-2x})'(\log(\ln x))^3+e^{-2x}\left(\log(\ln x))^3\right)'$

$=-2e^{-2x}(\log(\ln x))^3+3e^{-2x}(\log(\ln x))^2(\log(\ln x))'$

$=-2e^{-2x}(\log(\ln x))^3+3e^{-2x}(\log(\ln x))^2\dfrac{(\ln x)'}{\ln10\,\ln x}$

$=-2e^{-2x}(\log(\ln x))^3+\dfrac{3e^{-2x}(\log(\ln x))^2}{\ln10\,x\ln x}$

(3)

$\left(\sin(xe^x)^3\right)'=\left(\sin(x^3e^{3x})\right)'=\cos(x^3e^{3x}(x^3e^{3x})'$

$=\cos(x^3e^{3x})((x^3)'e^{3x}+x^3(e^{3x})')$

$=\cos(x^3e^{3x})(3x^2e^{3x}+3x^3e^{3x})$

$=3x^2e^{3x}(1+x)\cos(x^3e^{3x})$

(4)

$\left(\sin^3(xe^x)\right)'=3\sin^2(xe^x)\left(\sin(xe^x)\right)'$

$=3\sin^2(xe^x)\cos(xe^x)(xe^x)'$

$=3\sin^2(xe^x)\cos(xe^x)(x'e^x+x(e^x)')$

$=3\sin^2(xe^x)\cos(xe^x)(e^x+xe^x)$

$=3e^x(1+x)\sin^2(xe^x)\cos(xe^x)$

(5) Use implicit differentiation here.

$(\ln(xy))'=(e^{2y})'$

$\dfrac{(xy)'}{xy}=2e^{2y}y'$

$\dfrac{x'y+xy'}{xy}=2e^{2y}y'$

$y+xy'=2xye^{2y}y'$

$y=(2xye^{2y}-x)y'$

$y'=\dfrac y{2xye^{2y}-x}$

8 0

3 years ago