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3 years ago

Solve for x: 4x(3y-2)=7x+11

Mathematics

2 answers:

kupik [55]3 years ago

7 0

Answer:

x=11(12y-15)

Step-by-step explanation:

4x(3y-2)=7x+11

12xy-8x=7x+11

12xy-8x-7x=11

12xy-15x=11

x(12y-15)=11

x(12y-15)/(12y-15)=11/(12y-15)

x=11/(12y-15)

g100num [7]3 years ago

7 0

Answer: $x=\frac{11}{12y-15}$

Step-by-step explanation:

To solve for x, we want to isolate the variable.

$4x(3y-2)=7x+11$ [distribute]

$12xy-8x=7x+11$ [subtract both sides by 7x]

$12xy-8x-7x=11$ [combine like terms]

$12xy-15x=11$ [factor out x]

$x(12y-15)=11$ [divide both sides by 12y-15]

$x=\frac{11}{12y-15}$

Now, we know that $x=\frac{11}{12y-15}$ .

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It’s cost 112 to turf a lawn of 56m square what amount would it cost to turf lawn area of 99m square

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Answer:

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Step-by-step explanation:

112/56=2

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4 years ago

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3 years ago

Give each answer as mixed number in simplest form

allochka39001 [22]

Answer:

3 7/18

Step-by-step explanation:

2 1/6 + 1 2/9 (- cancels with the - to give us +)

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Final answer: 3 7/18

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2 years ago

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Suppose an unknown radioactive substance produces 8000 counts per minute on a Geiger counter at a certain time, and only 500 cou

mariarad [96]

Answer:

The half-life of the radioactive substance is of 3.25 days.

Step-by-step explanation:

The amount of radioactive substance is proportional to the number of counts per minute:

This means that the amount is given by the following differential equation:

$\frac{dQ}{dt} = -kQ$

In which k is the decay rate.

The solution is:

$Q(t) = Q(0)e^{-kt}$

In which Q(0) is the initial amount:

8000 counts per minute on a Geiger counter at a certain time

This means that $Q(0) = 8000$

500 counts per minute 13 days later.

This means that $Q(13) = 500$ . We use this to find k.

$Q(t) = Q(0)e^{-kt}$

$500 = 8000e^{-13k}$

$e^{-13k} = \frac{500}{8000}$

$\ln{e^{-13k}} = \ln{\frac{500}{8000}}$

$-13k = \ln{\frac{500}{8000}}$

$k = -\frac{\ln{\frac{500}{8000}}}{13}$

$k = 0.2133$

$Q(t) = Q(0)e^{-0.2133t}$

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

$Q(t) = Q(0)e^{-0.2133t}$

$0.5Q(0) = Q(0)e^{-0.2133t}$

$e^{-0.2133t} = 0.5$

$\ln{e^{-0.2133t}} = \ln{0.5}$

$-0.2133t = \ln{0.5}$

$t = -\frac{\ln{0.5}}{0.2133}$

$t = 3.25$

The half-life of the radioactive substance is of 3.25 days.

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3 years ago

If the volume of the pyramid shown is 360 inches cubed, what is its height?

GenaCL600 [577]

<h3>Answer: 9 inches (choice D)</h3>

Work Shown:

V = volume of pyramid

V = (1/3)*(area of base)*(height)

V = (1/3)*(12*10)*h

V = (1/3)*120h

V = 40h

Set this equal to the given pyramid volume 360 cubic inches and solve for h

40h = 360

h = 360/40

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3 years ago

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