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leonid [27]
4 years ago
14

How does the "E" in STEM work with the other letters

Engineering
1 answer:
KIM [24]4 years ago
4 0

Answer:

if you are speaking of the acronym then Engineering uses science and mathematics to solve everyday problems in society

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Determine the depreciation expense for 2018 and 2019 using the following​ methods: (a)​ Straight-line (SL),​ (b) Units of produc
prohojiy [21]

Answer:

Check Explanation.

Explanation:

(1). The straight-line method: the general clue with this method is that in the two years, depreciation is the same. The formula for Calculating depreciation is given below;

straight-line method = (cost - Residual value)/ useful life in years.

From the question we know that the cost of acquisition is $30,000,000, the residual value of the asset is $4,000,000 and useful life is 7 years. Therefore;

straight-line method = ($30,000,000 - $4,000,000)/ 7.

= $3, 714,285.71 Per year.

That is $3, 714,285.71 for 2018 and 2019.

(2).Units of production​ (UOP) = (cost - Residual value)/ useful life in units.

= ($30,000,000 - $4,000,000)/ 4,375, 000.

Units of production​ (UOP) = $6 per mile.

Hence, the depreciation in 2018 = Depreciation per unit × 2018 year usage.

= 6 × 1,100,000 mile.

= $6,600,000.

depreciation in 2019 = Depreciation per unit × 2019 year usage.

= 6 × 1,200,000.

= $7,200,000.

Double-declining-balance​ (DDB)= (cost - accumulated depreciation) × 2 × 1/(useful life years).

Double-declining-balance​ (DDB) = (30,000,000 - 0)× 2 × (1/7).

= $8,571,428.57 depreciation in 2018.

= $8,571,428.6 depreciation in 2018

Double-declining-balance​ (DDB) = (30,000,000 - 8,571,428.57) × 2 × 1/7.

= $6,122,449.00 depreciation in 2019.

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

Total depreciation for straight-line method(2018 and 2019) = $7,428,571.42.

Total depreciation for Units of production​ (UOP)(2018 and 2019) = $13,800,000.

Total depreciation for Double-declining-balance (DDB)= $ 14,693,877.6.

5 0
3 years ago
Please dimension this, was due yesterday.
lorasvet [3.4K]
Love the image isn’t showing up. Try reposting it!
6 0
3 years ago
The 240-ft structure is used to provide various support services to launch vehicles prior to liftoff. In a test, a 12-ton weight
alina1380 [7]

Answer:

hello your question lacks the required question attached below is the missing diagram

Forces in GJ = -4.4444 i.e. 4.4444 tons

Forces in IG = 15.382 tons ( T )

Explanation:

Forces in GJ = -4.4444 i.e. 4.4444 tons

Forces in IG = 15.382 tons ( T )

attached below is the detailed solution

3 0
3 years ago
Steel grade AISI 1045 contains 4.5% carbon. a)-True b)- False
yarga [219]

Answer:

False

Explanation:

According to the AISI (SAE) statements the last two numbers of the designation indicates the porcentage of carbon in the steel. And the firs numbers indicate the most important alloy alements.

In the case of the AISI 1045 indicates that the composition of this steel is 0,45% of carbon.

4 0
3 years ago
Determine the voltage input to the inverting terminal of an op amp when −40 μV is applied to the non-inverting terminal and the
Ad libitum [116K]

Answer:

The voltage input to the inverting terminal is 60μV

Explanation:

Given;

open-loop gain, A = 150,000

output voltage, V₀ = 15 V

voltage at the inverting input, V_n = −40 μV = 40 x 10⁻⁶ V

The relationship between output voltage and voltage at the inverting input is given as;

V_o = A(V_p -V_n)\\\\15 = 150,000(V_p -(-40*10^{-6}))\\\\15 = 150,000 (V_p +40*10^{-6}) \\\\V_p +40*10^{-6} = \frac{15}{150,000} \\\\V_p + 40*10^{-6} = 1 *10^{-4}\\\\V_p + 40*10^{-6} = 100 *10^{-6}\\\\V_p  =  100 *10^{-6} - 40*10^{-6}\\\\V_p = 60 *10^{-6}\\\\V_p = 60 \ \mu V

Therefore, the voltage input to the inverting terminal is 60μV

3 0
3 years ago
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