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skad [1K]
2 years ago
5

Madison plants a 3-yard by 6-yard rectangular vegetable garden. She puts rope around the edges. The rope costs $2 per yard. What

is the total cost of the rope?
Mathematics
1 answer:
just olya [345]2 years ago
6 0

Answer:

$36

Step-by-step explanation:

You have to find the area because the rope is around the edges

3 yards x 6 yards = 18 square yards (area)

Rope is $2 every yard and there are 18 yards so you multiply.

18 yards x $2 = $36

(I deserve to be the brainliest )

You might be interested in
Suppose h(t)=-0.2t squared +2t models the height in feet of ball that is kicked into the air where t is given as time in seconds
xz_007 [3.2K]
To get the maximum height, we first determine the time at which this maximum height is attained by differentiating the given equation and equating the differential to zero. 
                                 h(t) = -0.2t² + 2t
Differentiating,
                               dh(t) = -(0.2)(2)t + 2 = 0
The value of t is equal to 5. Substituting this time to the original equation,
                                   h(t) = -0.2(5²) + 2(5) = 5 ft
Thus, the maximum height is 5 ft and since it will take 5 seconds for it to reach the maximum height, the total time for it to reach the ground is 10 seconds. 

Answers: maximum height = 5 ft 
                  time it will reach the ground = 10 s
4 0
3 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

5 0
3 years ago
System of Equations<br><br> y=-x-1<br> y+x=3<br><br> Solve?
ozzi
Solve by substitution

\sf{y=-x-1}
\sf{y+x=3}

Just plug in y=-x-1 into y+x=3

So we get:

\sf{(-x-1)+x=3}

Simplify that:

\sf{-1=3}

And since that is not true, that means this system of equations has no solutions!
6 0
2 years ago
Please help with this question
Vera_Pavlovna [14]
This would be the equation:

Y= 5/2x+5

I hope I've helped!
7 0
3 years ago
Read 2 more answers
Use the properties of exponents to rewrite y=e^-0.75t in the form of a(1-r)^t
Ivahew [28]

Answer: y = (1 - 0.527)^t

Step-by-step explanation:

y=e^(-0.75t)


y=(e^-0.75)^t


y= 0.47236655^t


1 - 0.47236655 = 0.52763345

y = (1 - 0.527)^t

5 0
2 years ago
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