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Tanya [424]
3 years ago
12

Two angles of a triangle measure 12º and 40°.

Mathematics
2 answers:
Sloan [31]3 years ago
8 0

Answer:

12+40= 52

180-52=128

Step-by-step explanation:

Angles in a triangle add up to 180

so if two sides are given , they must be added and subtracted from 180.

which gives us 180-52=128

tatiyna3 years ago
6 0

Answer:

the angle is 128°

the right answer is C

Step-by-step explanation:

the angles of a triangle are always 180°

thus

the angle of the the triangle = 180° - (12°+40°)

= 180- 52° = 128°

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Nataly [62]

6V + 2(V+12) = 488
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V+12=70 mph for 2 hours.


 check: 58*6 = 348

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7 0
3 years ago
The least total cost method (LTC) lot-sizing technique calculates the order quantity by comparing the carrying cost and the setu
rewona [7]

Answer:

True

Step-by-step explanation:

The least total cost method is the method in which the total cost of the ordering cost and the total carrying cost is equal among various lot size available.

The order quantity should be choose when the total ordering cost and the total carrying cost equal to each other

The formula to compute the economic order quantity is shown below:

a. The computation of the economic order quantity is shown below:

= \sqrt{\frac{2\times \text{Annual demand}\times \text{Ordering cost}}{\text{Carrying cost}}}

It is always be expressed in units

The formula to compute the ordering cost is

= \frac{Annual\ demand}{Economic\ order\ quantity}\times ordering\ cost\ per\ order

And, the formula to compute the carrying cost is

= \frac{Economic\ order\ quanity}{2}\times carrying\ cost\ per\ unit

Hence, the given statement is true

8 0
4 years ago
Please calculate this limit <br>please help me​
Tasya [4]

Answer:

We want to find:

\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n}

Here we can use Stirling's approximation, which says that for large values of n, we get:

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Because here we are taking the limit when n tends to infinity, we can use this approximation.

Then we get.

\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n} = \lim_{n \to \infty} \frac{\sqrt[n]{\sqrt{2*\pi*n} *(\frac{n}{e} )^n} }{n} =  \lim_{n \to \infty} \frac{n}{e*n} *\sqrt[2*n]{2*\pi*n}

Now we can just simplify this, so we get:

\lim_{n \to \infty} \frac{1}{e} *\sqrt[2*n]{2*\pi*n} \\

And we can rewrite it as:

\lim_{n \to \infty} \frac{1}{e} *(2*\pi*n)^{1/2n}

The important part here is the exponent, as n tends to infinite, the exponent tends to zero.

Thus:

\lim_{n \to \infty} \frac{1}{e} *(2*\pi*n)^{1/2n} = \frac{1}{e}*1 = \frac{1}{e}

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