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Bumek [7]
3 years ago
9

I need help!! Pleasee!!!

Mathematics
1 answer:
natita [175]3 years ago
4 0

Answer:

15/20=d/16

20d. =15×16

20d=240

d=12 cm

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Answer:

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3 years ago
A boy knows his height is 6 feet at the time of day when his shadow is 4 feet a trees shadow is 24 feet. What is the height of t
Schach [20]
The tree's height is 36 feet!

Work/explanation below:

1. The ratio of the boy's shadow to height is 4 : 6 or 1 : 1.5.

2. Plug the tree's shadow into the ratio and you have 24 : 24 X 1.5

3. 24 X 1.5 = 36.

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Hank draws a line with a zero slope through the points (-2,4) and (3.b).which value of b could represent hanks second point ?
Agata [3.3K]
B = 4



A line with a slope of zero is a perfectly horizontal line, which means the y value (the second value) stays the same for every single point on the line.

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4 0
4 years ago
Read 2 more answers
The populations of termites and spiders in a certain house are growing exponentially. The house contains 120 termites the day yo
zysi [14]

Answer:

in order to triple the inicial population of spiders, will take 50395 days

Step-by-step explanation:

we can define the termite population function as T(t) and the one for spiders as S(t) , where t represents time measured in days

since both have and exponencial growth

T(t)= a*e^(b*t)

S(t)= c*e^(d*t)

1) when the day the person moves in , t=0 and T(0)= 120 termites

T(0) = a*e^(b*0) = a = 120

2) after 4 days , t=4 and  the house contains T(4) = 210 termites

T(4)= 120*e^(b*4) = 210 → 4*b = ln (210/120) → b = (1/4)* ln(210/120)= 0.14

therefore

T(t) = 120*e^(0.14*t)

3) 3 days after moving in , t=3, there were T(3) = 120*e^(0.14*3)=182.63≈ 182 termites . The number of spiders is half of the number of termites → S(3) = T(3) * 1 spider/ 2 termites  =91.31 spiders ≈ 91 spiders

4) after 8 days of moving in , t=8, there were T(8) = 120*e^(0.14*8)=367.78≈ 368 termites . The number of spiders is 0.25 times the number of termites → S(8) = T(8) * 1 spider/ 4 termites =91.94 spiders   ≈ 92 spiders

from

S(t)= c*e^(d*t) → d = ln [S (tb)/S (ta) ] / (tb-ta)

therefore d = ln [ S(8)/S(3) ] / (8 - 3 ) = 2.18*10^-5

in order to triple the initial population

S(t3) = 3 *S(0) = 3*[c*e^(d*0)] = 3*c

S(t3) = c*e^(d*t3) = 3* c → t3 = ln(3) / d = 50395 days

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3 years ago
Have 50000 and a increase of 4% each year for 15 years
madam [21]
50,000×1.04^(15)
=90,047.18
4 0
3 years ago
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