PLZZ HELPP FAST WILL GIVE MEDAL!!

Mathematics

1 answer:

Afina-wow [57]3 years ago

6 0

8 5 8 5 hopefully its right i'm not good at this kind of stiff

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Element X decays radioactively with a half life of 9 minutes. If there are 760 grams of Element X, how long, to the nearest tent

ElenaW [278]

Answer:

Step-by-step explanation:

760·0.5^(t/9) = 19

0.5^(t/9) = 19/760

2^(-t/9) = 0.025

-t/9 = log₂(0.025)

t = -9log₂(0.025) = 47.9 minutes

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2 years ago

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Easy way of explaining integers?

Andre45 [30]

Answer:

Step-by-step explanation:

3 Great Ways to Teach Adding Integers

Typically, you should begin by having your students think about real world examples of positive and negative integer. ...

Then, model this example using counter chips.

Finally, have your students model addition problems using the counter chips.

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Can somebody help me with this question ;-;

Hatshy [7]

Answer:

I believe the answer is 5 units. Because it shows the line's arrow at unit 5.

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2 years ago

Consider the following equations. f(x) = − 4/ x3, y = 0, x = −2, x = −1. Sketch the region bounded by the graphs of the equation

Sindrei [870]

Answer:

1.5 unit^2

Step-by-step explanation:

Solution:-

- A graphing utility was used to plot the following equations:

$f ( x ) = - \frac{4}{x^3}\\\\y = 0 , x = -1 , x = -2$

- The plot is given in the document attached.

- We are to determine the area bounded by the above function f ( x ) subjected boundary equations ( y = 0 , x = -1 , x = - 2 ).

- We will utilize the double integral formulations to determine the area bounded by f ( x ) and boundary equations.

We will first perform integration in the y-direction ( dy ) which has a lower bounded of ( a = y = 0 ) and an upper bound of the function ( b = f ( x ) ) itself. Next we will proceed by integrating with respect to ( dx ) with lower limit defined by the boundary equation ( c = x = -2 ) and upper bound ( d = x = - 1 ).

The double integration formulation can be written as:

$A= \int\limits_c^d \int\limits_a^b {} \, dy.dx \\\\A = \int\limits_c^d { - \frac{4}{x^3} } . dx\\\\A = \frac{2}{x^2} |\limits_-_2^-^1\\\\A = \frac{2}{1} - \frac{2}{4} \\\\A = \frac{3}{2} unit^2$