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3 years ago

Evaluate the given algebraic expression for x = 3 and y=3. x^2y^2

Mathematics

1 answer:

Keith_Richards [23]3 years ago

7 0

Answer:

Step-by-step explanation:

$3^{2} =9\\3^{2} =9$

9 x 9 = 81

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Finding the surface area and volume to the nearest tenth

erastovalidia [21]

The cylinder:
pi x r^2 x h
= pi x 8^2 x 20
then times it by a half
= 2009.6
The prism:
l x w
= 32 x 20
= 640
All up:
2009.6 +640
The shape equals 2649.6 cm sqaured

6 0

4 years ago

Multiplying/dividing signs with the same sign will give you what sign in your anwser

Anon25 [30]

Answer:

It will give you the opposite sign.

Step-by-step explanation:

If you multiply two negatives you will end up with a positive and vice versa with positives. The same thing applies to dividing.

5 0

3 years ago

A general formula for a parabola is y2 = 4px. What is the value of p in the equation y2 = –4x?

oee [108]

Easy

compare
y²=4px vs
y²=-4x

y²=y²
4px=-4x
divide both sides by x
4p=-4
divide both sides by 4
p=-1

p=-1

3 0

3 years ago

Read 2 more answers

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Airida [17]

Answer:

15: A

16: C

Step-by-step explanation:

I have no idea but hopefully I helped you

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3 years ago

Read 2 more answers

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.

7 0

3 years ago