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swat32
2 years ago
5

A radioactive substance decays exponentially. A scientist begins with 190 milligrams of a radioactive

Mathematics
1 answer:
worty [1.4K]2 years ago
6 0

Answer: 65.1 (mg)

Step-by-step explanation:

y=190*e^{-a*t}\\\\Calculate\ a:\\\\t=33,y=95(mg)\\\\190*e^{-a*33}=95\\\\e^{-a*33}=0.5\\\\-a*33=ln(0.5)\\\\a=\dfrac{ln(\dfrac{1}{2}) }{-33} \\\\a=0,02100... \approx{0.021}\\\\If\ t=51\ h, y=190*e^{-0.02100*51}=65.09167...\approx{65.1}

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Answer:

9.25 or 9 1/4

Step-by-step explanation:

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3 years ago
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Step-by-step explanation:

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The coordinates of polygon ABCD are A(-4,-1), B(-2,3), C(2,2), and D(4,-3). Use these coordinates to complete the sentences belo
oee [108]

Answer:

The perimeter of polygon ABCD, to the nearest thousandth units is

22.227 units

The perimeter of polygon ABCD', to the nearest thousandth, would be 20.980 units

The area of polygon ABCD' is 19.5 units²

Step-by-step explanation:

The coordinates forming the polygon are

A (-4,-1), B(-2,3), C(2,2), and D(4,-3)

The perimeter then is given by the sum of the length of the sides as follows;

Length of line between two X and Y points distance, xi and yj apart is

length XY =  \sqrt{x^2+y^2}

Therefore, the length between points AB is

length AB = \sqrt{(-4 - (-2))^2+(-1-3)^2}

= \sqrt{(-4 +2)^2+(-4)^2} = \sqrt{-2^2+-4^2}  = \sqrt{20}  = 4.472 units

Similarly, length BC is given by

Length BC =  \sqrt{(-4)^2+1^2} =  \sqrt{17} = 4.123 units

Length CD = \sqrt{(-2)^2+5^2} =  \sqrt{29} = 5.385 units

Length DA = \sqrt{(8)^2+(-2)^2} =  \sqrt{68} = 8.246 units

The perimeter is equal to;

length AB + Length BC + Length CD + Length DA

= 4.472 + 4.123 + 5.385 + 8.246 = 22.227 units

If the point D is moved up 2 units and left 1 unit we have

D' = x-1, y+2 where x and y are the coordinates of point D

D(4, -3) → D'((4-1), (-3+2)) = D'(3, -1)

The length D'A =  \sqrt{(7)^2+(0)^2} =  \sqrt{49} =7 units

The perimeter of polygon ABCD'

length AB + Length BC + Length CD + Length D'A

= 4.472 + 4.123 + 5.385 + 7 = 20.980 units.

The area is given by the determinant of the 3 by 3 matrix using Cramer's Rule as follows

 S_{\bigtriangleup} = (\frac{1}{2})|x_1y_2+x_2y_3+x_3y_1-x_1y_3-x_2y_1-x_3y_2|

= (\frac{1}{2})|x_1(y_2-y_3)+x_2(y_3-y_1) +x_3(y_1-y_2)|

Triangle ABC we have

A(-4,-1)

B(-2,3)

C(2,2)

S_{{\bigtriangleup}ABC} = (\frac{1}{2})|x_1(y_2-y_3)+x_2(y_3-y_1) +x_3(y_1-y_2)|

=(\frac{1}{2})|(-4)(3-2)+(-2)(2-(-1)) +2((-1)-3)|

= =(\frac{1}{2})|-4-6 -8|= 9 units^2

Triangle ACD'

A(-4,-1)

C(2,2)

D'(3, -1)

S_{{\bigtriangleup}ACD'} = (\frac{1}{2})|x_1(y_2-y_3)+x_2(y_3-y_1) +x_3(y_1-y_2)|

=(\frac{1}{2})|(-4)(2+1)+(2)(-1+1) +3((-1)-2)| = \frac{21}{2} = 10.5 units²

Area of polygon =   S_{{\bigtriangleup}ABCD'} = S_{{\bigtriangleup}ABC} + S_{{\bigtriangleup}ACD'} = (9 + 10.5) units²

Area of polygon ABCD' = 19.5 units².

7 0
2 years ago
Please help with math question
Naddik [55]
1- Solution using graphs:
Take a look at the attached images.
The red graph represents the first given function while the blue graph represents the second given function.
We can note that the two graphs are the same line (they overlap).
This means that any chosen point on one of them will satisfy the other.
This means that there are infinite number of solutions to these two equations.

2- Solution using substitution:
The first given equation is:
y = -5x + 3 ...........> equation I
The second given equation is:
2y + 10x = 6 ...........> equation II

Substitute with I in II and solve as follows:
2(-5x+3) + 10x = 6
-10x + 6 + 10x = 6
0 = 0
This means that there are infinitely many solutions to the given system of equations.

Hope this helps :)

3 0
2 years ago
The value of y varies directly with x. If x = 4, then y= 10. What is the value of x when
Eva8 [605]
4 10
— —
10 25

answer: 10
8 0
2 years ago
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