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3 years ago

14

If 6 lbs of nails are needed for each thousand lath, how many pounds are required for 4250 lath? Round your answer to the neares

t tenth.

1 answer:

Snezhnost [94]3 years ago

8 0

Answer:

Step-by-step explanation:

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Find the length of the hypotenuse. Round your answer to the nearest hundredth

laiz [17]

Answer:

c^2=a^2+b^2

=7^2+3^2

=49+9

=58

c^2=58

c=

$\sqrt{58 = 7.6}$

the answer is C=7.62..

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2 years ago

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Two lines, E and F, are represented by the equations given below.

shusha [124]

Although you have not included the statements about the solution to the set of the given equations, I can explain you what happen with that system.

You can arrive to the same conclusion by any method of solution of systems that you like.

The quickest in this case, is to multiply the equation of the lIne F by 5.

That leads to: 5 (x + y) = 5(8)

⇒ 5x + 5y = 40.

Now you can realize that the two representations (equations) correspond to the same line.

That means that there are infinite solutions, this is infinity values of x and y meet both equations.

If you graph line E and line F they overlap in the entire domain, so the graphing method also tells you that you cannot find one solution but infinite soluttions.

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3 years ago

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Letter grades earned by a particular student in history class.

Lorico [155]

<span>Letter grades are not numerical, but can be ordered in a meaningful way. Because the responses are not numerical, they are considered qualitative. Because they can be ordered in a meaningful way, they are also considered ordinal. The best answer choice is B.</span>

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3 years ago

Y = 4x <br> y = x^4<br><br> solve without graphing

jasenka [17]

The answers to the questions

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2 years ago

Consider a population whose growth over a given time period can be described by the exponential model: dN/dt = rN. Select the co

pantera1 [17]

Answer:

b and c

Step-by-step explanation:

We are given that a population whose growth over a given time period can be described by the exponential model

$\frac{dN}{dt}=r N$

Let initial population = $N_0$ when time t=0

$\int\frac{dN}{N}=r\int_0^tdt$

After integrating

We get ln N=rt +C

Where C is integration constant

When t=0 then N= $N_0$

$ln N_0=C$

Substitute the value of C then we get

$ln N=rt +ln N_0$

$ln N-ln N_0=rt$

$ln\frac{N}{N_0}=rt$

$\frac{N}{N_0}=e^{rt}$

$N=N_0e^{rt}$

When r=0.1 then we get

$N=N_0e^{0.1t}$

Hence, the population increase not decrease.

When r= 0

Then we get

$N=N_0e^{0}$

$N=N_0$

Hence, the population do not increase or decrease.

So, a population with r of 0 will have no births or deaths during the time period under consideration.

If we take a positive value of r then the population will increase exponentially .

Hence, option b and c are both correct.

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2 years ago