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lilavasa [31]
4 years ago
5

Use the relationship between the

Mathematics
1 answer:
motikmotik4 years ago
4 0

The equation x° = 180° - (37°+ 53°) can be used to find the value of x.

Step-by-step explanation:

Step 1; There are three lines in the given diagram. There are a baseline and two other lines. Out of the other two lines, one extends above and below the baseline whereas the other extends only above. Since the baseline is horizontal and the others are at angles we have the sum of all the three angles as 180° i.e. 37°, x°, and 53°.

37° + x° + 53° = 180°.

Step 2; To solve the value of x, we keep the unknown value at the left-hand side whereas all the known values are taken to the right side of the equation.

x° = 180° - (37°+ 53°).

So the fifth option can be used to determine the value of x.

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Use a net to find the surface area of the cylinder. Use 3.14 for .
notka56 [123]

Answer:

Formula = 2nr

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3 0
2 years ago
13/25 as a percentage
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4 years ago
Read 2 more answers
On October 1, 2014, Hess Company places a new asset into service. The cost of the asset is $80,000 with an estimated 5-year life
Ede4ka [16]

Answer:

$43,200

Step-by-step explanation:

Data provided in the question:

Cost of the asset = $80,000

Useful life of the machine = 5 years

Salvage value at the end of useful life = $20,000

Now,

Using the double declining method of depreciation

Annual depreciation rate = 2 ×  [1 ÷ useful life ]

= 2 × [ 1 ÷ 5 ]

= 2 × 0.2

= 0.4 or 40%

thus,

The depreciation from October 1, 2014 to December 31, 2014

= Annual Depreciation rate × duration × Book value for 2014

= 0.4 × 3 months × $80,000

= 0.4 × 0.25 year × $80,000

= $8,000

Book value for 2015

= Cost - depreciation till December 31, 2014

= $80,000 - $8,000

= $72,000

Therefore,

Depreciation for the year 2015

= Annual Depreciation rate  × Book value for 2015

= 0.4 × $72,000

= $28,800

Therefore,

the  book value of the plant asset on the December 31, 2015

= Book value for 2015 - Depreciation for the year 2015

= $72,000 - $28,800

= $43,200

4 0
3 years ago
Determine whether each function is continuous at x=5. Justify your answer using the continuity test.
irakobra [83]

Answer:

2. f(x)=\frac{x^2}{x+5}

Step-by-step explanation:

When you replace x=5 into those functions you obtain:

1. f(x)=\sqrt{x^2-36} = \sqrt{5^2-36} = \sqrt{-11} \\2. f(x)=\frac{x^2}{x+5}=\frac{5^2}{5+5}=\frac{25}{10}

We can see that the function number 1, when the x=5 is replaced, falls into the imaginary numbers or complex domain, so the function is not continuous in the real domain.

On the other hand, the second function falls into the real domain whe x=5 is replaced, so the function is continuous in the real domain!

I hope you understand!

6 0
4 years ago
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