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sergeinik [125]
2 years ago
7

It is recommended that you use 3.75 pound of fertilizer to cover 1,200 square feet. How much space can you cover with a 1-pound

bag of fertilizer?
Mathematics
1 answer:
uysha [10]2 years ago
4 0

Answer:

320 square feet

Step-by-step explanation:

To find the space you can cover with 1-pound bag of fertilizer, we need to divide 1,200 by 3.75. It makes things easier to think of how 3.75 ÷ 3.75 = 1. Anyways, 1,200 ÷ 3.75 = 320. Therefore, you can cover 320 square feet with a 1-pound bag of fertilizer.

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If x=(2a-3b),y=(2a+3b)and z=9bsquare -4absquare,show that xy+z=0​
kaheart [24]

Answer:

<h2>Hope it helps you........</h2>

5 0
3 years ago
Indicate a general rule for the nth term of this sequence. 6a, 3a, 0, -3a, -6a, . . .
Lesechka [4]
6a-(n-1)(3a) is the nth term
Expanding we get 6a-3an+3a=9a-3an=3a(3-n)
SOLUTION: nth term is 3a(3-n)
8 0
4 years ago
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The length of the shadow of a pole on level ground increases by 90m when the angle of elevation of the sun changes from 58° to 3
siniylev [52]
<h3>Answer: Approximately 119.76 meters</h3>

===================================================

Work Shown:

x = starting length of the shadow

y = height of the pole

tan(angle) = opposite/adjacent

tan(58) = y/x

1.6003345 = y/x

1.6003345x = y

x = y/1.6003345

x = (1/1.6003345)y

x = 0.62486936y

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

When the angle changes, the adjacent side gets 90 meters longer

tan(angle) = opposite/adjacent

tan(36) = y/(x+90)

0.72654253 = y/(0.62486936y+90)

0.72654253(0.62486936y+90) = y

0.453994166y + 65.3888277 = y

65.3888277 = y-0.453994166y

65.3888277 = 0.546005834y

0.546005834y = 65.3888277

y = 65.3888277/0.546005834

y = 119.758478075162

y = 119.76

The height of the pole is about 119.76 meters.

8 0
3 years ago
A metal beam was brought from the outside cold into a machine shop where the temperature was held at 65degreesF. After 5 ​min, t
ivolga24 [154]

Answer:

The beam initial temperature is 5 °F.

Step-by-step explanation:

If T(t) is the temperature of the beam after t minutes, then we know, by Newton’s Law of Cooling, that

T(t)=T_a+(T_0-T_a)e^{-kt}

where T_a is the ambient temperature, T_0 is the initial temperature, t is the time and k is a constant yet to be determined.

The goal is to determine the initial temperature of the beam, which is to say T_0

We know that the ambient temperature is T_a=65, so

T(t)=65+(T_0-65)e^{-kt}

We also know that when t=5 \:min the temperature is T(5)=35 and when t=10 \:min the temperature is T(10)=50 which gives:

T(5)=65+(T_0-65)e^{k5}\\35=65+(T_0-65)e^{-k5}

T(10)=65+(T_0-65)e^{k10}\\50=65+(T_0-65)e^{-k10}

Rearranging,

35=65+(T_0-65)e^{-k5}\\35-65=(T_0-65)e^{-k5}\\-30=(T_0-65)e^{-k5}

50=65+(T_0-65)e^{-k10}\\50-65=(T_0-65)e^{-k10}\\-15=(T_0-65)e^{-k10}

If we divide these two equations we get

\frac{-30}{-15}=\frac{(T_0-65)e^{-k5}}{(T_0-65)e^{-k10}}

\frac{-30}{-15}=\frac{e^{-k5}}{e^{-k10}}\\2=e^{5k}\\\ln \left(2\right)=\ln \left(e^{5k}\right)\\\ln \left(2\right)=5k\ln \left(e\right)\\\ln \left(2\right)=5k\\k=\frac{\ln \left(2\right)}{5}

Now, that we know the value of k we can use it to find the initial temperature of the beam,

35=65+(T_0-65)e^{-(\frac{\ln \left(2\right)}{5})5}\\\\65+\left(T_0-65\right)e^{-\left(\frac{\ln \left(2\right)}{5}\right)\cdot \:5}=35\\\\65+\frac{T_0-65}{e^{\ln \left(2\right)}}=35\\\\\frac{T_0-65}{e^{\ln \left(2\right)}}=-30\\\\\frac{\left(T_0-65\right)e^{\ln \left(2\right)}}{e^{\ln \left(2\right)}}=\left(-30\right)e^{\ln \left(2\right)}\\\\T_0=5

so the beam started out at 5 °F.

6 0
3 years ago
2 + 6x equals 1 - x​
melomori [17]

Answer:

x = -1/7

Step-by-step explanation:

simplify by bringing all the x to the left and all the constants to the right:

6x + x = 1 - 2

7x = -1  (divide by 7)

x = -1/7

5 0
3 years ago
Read 2 more answers
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