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WITCHER [35]
3 years ago
13

What is the angle to this below

Mathematics
1 answer:
fgiga [73]3 years ago
7 0

Answer:

Acute angle

Step-by-step explanation:

180° - 121° = 59°

less than 90° is acute

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The initial outlay or cost is $1,000,000 for a four-year project. The respective future cash inflows for years 1, 2, 3 and 4 are
trasher [3.6K]

Answer:

The Payback Period from non discounted cash flows  is 2 Years and 8 months

Step-by-step explanation:

With an initial outlay of $1,000,000

and Cash inflows for 4 years consecutively of $500,000 + $300,000 + $300,000 + $300,000

Pay back Period = the Period where Initial Outlay minus Cash Inflows equal Zero

= 1,000,000 - 500,000 (yr 1) - 300,000 (yr 2) - 200,000 (2/3 of Year 3) = 0

Pay back Period therefore is equal to 2 years & 8 months.

3 0
4 years ago
The product written as an improper fraction in lowest terms is
Anna007 [38]
I think the answer is 7/5.
3 0
4 years ago
A tank contains 1080 L of pure water. Solution that contains 0.07 kg of sugar per liter enters the tank at the rate 7 L/min, and
allsm [11]

(a) Let A(t) denote the amount of sugar in the tank at time t. The tank starts with only pure water, so \boxed{A(0)=0}.

(b) Sugar flows in at a rate of

(0.07 kg/L) * (7 L/min) = 0.49 kg/min = 49/100 kg/min

and flows out at a rate of

(<em>A(t)</em>/1080 kg/L) * (7 L/min) = 7<em>A(t)</em>/1080 kg/min

so that the net rate of change of A(t) is governed by the ODE,

\dfrac{\mathrm dA(t)}[\mathrm dt}=\dfrac{49}{100}-\dfrac{7A(t)}{1080}

or

A'(t)+\dfrac7{1080}A(t)=\dfrac{49}{100}

Multiply both sides by the integrating factor e^{7t/1080} to condense the left side into the derivative of a product:

e^{\frac{7t}{1080}}A'(t)+\dfrac7{1080}e^{\frac{7t}{1080}}A(t)=\dfrac{49}{100}e^{\frac{7t}{1080}}

\left(e^{\frac{7t}{1080}}A(t)\right)'=\dfrac{49}{100}e^{\frac{7t}{1080}}

Integrate both sides:

e^{\frac{7t}{1080}}A(t)=\displaystyle\frac{49}{100}\int e^{\frac{7t}{1080}}\,\mathrm dt

e^{\frac{7t}{1080}}A(t)=\dfrac{378}5e^{\frac{7t}{1080}}+C

Solve for A(t):

A(t)=\dfrac{378}5+Ce^{-\frac{7t}{1080}}

Given that A(0)=0, we find

0=\dfrac{378}5+C\implies C=-\dfrac{378}5

so that the amount of sugar at any time t is

\boxed{A(t)=\dfrac{378}5\left(1-e^{-\frac{7t}{1080}}\right)}

(c) As t\to\infty, the exponential term converges to 0 and we're left with

\displaystyle\lim_{t\to\infty}A(t)=\frac{378}5

or 75.6 kg of sugar.

7 0
3 years ago
I need help plss :)!
OLEGan [10]

3wx^2

The third choice is the answer.

Answer :C

7 0
3 years ago
Read 2 more answers
The square root of which of the following integers is between 7 and 8 ?
Angelina_Jolie [31]
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8 0
3 years ago
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