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Inga [223]
4 years ago
8

One pound of ground beef makes four hamburger patties. Solve an equation to determine how many pounds of beef are needed to make

36 hamburgers?
Mathematics
1 answer:
OLga [1]4 years ago
3 0
9 pimds of beef are needed to make 36 hamburgers.
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Leanne's yearly income is 15,600 and she works twenty hours each week <br> What is her hourly rate
BartSMP [9]

Answer:

Leanne's hourly rate is $15 per hour

Step-by-step explanation:

Assuming a year is 52 weeks, divide $15,600/52 to find the $/week ($300). Then, divide $300/20 to find the $/hrs ($15)

8 0
3 years ago
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Yt3=1/2(x+4) in standard form
son4ous [18]
Do you mean 3y? If so, 3y = 1/2x + 2

3y - 1/2x = 2 is standard form
8 0
4 years ago
Can a triangle have side lengths 4, 4, 7? <br> 1.no <br> 2. Yes <br> 3. not enough information
Marina CMI [18]
Yes! The lengths of each side must be less than the sum of the other two lengths. It looks like this:
4+4>7
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8 0
3 years ago
A population of 6000 animals decrease at an annual rate of 0.7%. What is the expected population in 12 years? Round your answer
solmaris [256]

Answer:

The population is expected to be 5515 animals in 12 years.

Step-by-step explanation:

This problem can be solved by using a compounded interest formula with a negative interest rate, since the number of animals are decreasing. We have:

final population = initial population * (1 - r)^(t)

Where r is rate of decline and t is the time elapsed. Applying the data from the question we have:

final population = 6000*(1 - 0.007)^(12)

final population = 6000*(0.993)^12 = 5514.9582 animals

The population is expected to be 5515 animals in 12 years.

6 0
4 years ago
Chlorofluorocarbons (CFCs) were used as propellants in spray cans until their buildup in the atmosphere started destroying the o
melamori03 [73]

Answer:

a) C(2000)=1915

C(2014)=1915

b) C'(2000)=-\frac{275}{14}

C'(2014)=-\frac{275}{14}

c) C(t)=-\frac{275}{14}t+\frac{288405}{7}

d) t=2021

e) too early

Step-by-step explanation:

a)

Since C(t) is the concentration of CFCs in ppt in year t, all we need to do to solve this part is determine what the concentration of CFCs is in years 2000 and 2014. Luckily for us, the problem already gives us those values, so:

C(2000)=1915    concentration in year 2000

C(2014)=1915      concentration in year 2014

b)

By definition, the derivative of a function at a given point is interpreted as the slope of the tangent line to the point of interest, so in order to find this answer, we need to find the slope of the line. Since the problem specifies that the behavior is linear, this means that the slope will always be the same no matter the year, so we get:

m=C'(t)=\frac{C'(t_{2})-C'(t_{1})}{t_2-t_1}

so:

C'(t)=\frac{1640-1915}{2014-2000}=-\frac{275}{14}\approx -19.64

therefore:

C'(2000)=-\frac{275}{14}

C'(2014)=-\frac{275}{14}

c)

Since the behavior is linear, we can calculate it with the point-slope form of the line which is:

y-y_{1}=m(x-x_{1})

in this case:

C(t)-C(t_1)=C'(t)(t-t_{1})

so we get:

C(t)-1915=-\frac{275}{14}(t-2000)

and we can now solve for C(t) so we get:

C(t)=-\frac{275}{14}t+\frac{275000}{t}+1915

for a final answer of:

C(t)=-\frac{275}{14}t+\frac{288405}{7}

d)

So next we solve the equation for C(t)=1500 so we get:

1500=-\frac{275}{14}t+\frac{288405}{7}

1500-\frac{288405}{7}=-\frac{275}{14}t

-\frac{277905}{7}=-\frac{275}{14}t

t=2021 \frac{7}{55}

so we will reach that concentration level at the year 2021 approximately.

e)

Since the second derivative of the concentration function is greater than zero, this means that the original function might be a function in the form: .

This means that the decrease of the concentration levels is slower than that of a linear equation. So the projected date will be too early than the real date.

8 0
3 years ago
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