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LuckyWell [14K]
3 years ago
11

Situation:

Mathematics
1 answer:
FromTheMoon [43]3 years ago
4 0

Answer:

5.6 days

Step-by-step explanation:

We are given;

Initial Mass; N_o = 25 g

Mass at time(t); N_t = 25/2 = 12.5 (I divide by 2 because we are dealing with half life)

k = 0.1229

Formula is given as;

N_t = N_o•e^(-kt)

Plugging in the relevant values;

12.5 = 25 × e^(-0.1229t)

e^(-0.1229t) = 12.5/25

e^(-0.1229t) = 0.5

(-0.1229t) = In 0.5

-0.1229t = -0.6931

t = -0.6931/-0.1229

t = 5.6 days

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max2010maxim [7]

hello : <span>
<span>the discriminat of each quadratic equation : ax²+bx+c=0 ....(a ≠ 0) is :
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aleksandrvk [35]

Answer:

  • Graph B

Step-by-step explanation:

<u>According to data in the table we see pairs of points:</u>

  • (3, 75), (5, 125)

<u>This gives us the rate of change:</u>

  • (125 - 75)/(5 - 3) = 50/2 = 25

<u>The y - intercept is zero:</u>

  • 75 = 3*25 + b
  • 75 = 75 + b
  • b = 0

<u>The line is:</u>

  • y = 3x

This is the line on the graph B

<u>Note.</u> <em>Its not clear what is needed regarding the table on the bottom</em>

8 0
2 years ago
Help anyone can help me do 16 and 17 question,I will mark brainlest.The no 16 question is find the area of the shaded region​
yanalaym [24]

Answer:

Question 16 = 22

Question 17 = 20 cm²

Step-by-step explanation:

<u>Concepts:</u>

Area of Square = s²

  • s = side

Area of Triangle = bh/2

  • b = base
  • h = height

Diagonals of the square are congruent and bisect each other, which forms a right angle with 90°

Segment addition postulate states that given 2 points A and C, a third point B lies on the line segment AC if and only if the distances between the points satisfy the equation AB + BC = AC.

<u>Solve:</u>

Question # 16

<em>Step One: Find the total area of two squares</em>

Large square: 5 × 5 = 25

Small square: 2 × 2 = 4

25 + 4 = 29

<em>Step Two: Find the area of the blank triangle</em>

b = 5 + 2 = 7

h = 2

A = bh / 2

A = (7) (2) / 2

A = 14 / 2

A = 7

<em>Step Three: Subtract the area of the blank triangle from the total area</em>

Total area = 29

Area of Square = 7

29 - 7 = 22

-----------------------------------------------------------

Question # 17

<em>Step One: Find the length of PT</em>

Given:

  • PR = 4 cm
  • RT = 6 cm

PT = PR + RT [Segment addition postulate]

PT = (4) + (6)

PT = 10 cm

<em>Step Two: Find the length of S to PT perpendicularly</em>

According to the diagonal are perpendicular to each other and congruent. Therefore, the length of S to PT perpendicularly is half of the diagonal

Length of Diagonal = 4 cm

4 ÷ 2 = 2 cm

<em>Step Three: Find the area of ΔPST</em>

b = PT = 10 cm

h = S to PT = 2 cm

A = bh / 2

A = (10)(2) / 2

A = 20 / 2

A = 10 cm²

<em>Step Four: Find the length of Q to PT perpendicularly</em>

Similar to step two, Q is the endpoint of one diagonal, and by definition, diagonals are perpendicular and congruent with each other. Therefore, the length of Q to PT perpendicularly is half of the diagonal.

Length of Diagonal = 4 cm

4 ÷ 2 = 2 cm

<em>Step Five: Find the area of ΔPQT</em>

b = PT = 10 cm

h = Q to PT = 2 cm

A = bh / 2

A = (10)(2) / 2

A = 20 / 2

A = 10 cm²

<em>Step Six: Combine area of two triangles to find the total area</em>

Area of ΔPST = 10 cm²

Area of ΔPQT = 10 cm²

10 + 10 = 20 cm²

Hope this helps!! :)

Please let me know if you have any questions

6 0
2 years ago
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