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Varvara68 [4.7K]
3 years ago
10

I need help asap .....

Mathematics
1 answer:
NeX [460]3 years ago
5 0
C is the correct answer
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Please help me out solve it and tell me how you got that answer so I can understand please!
NikAS [45]

The solution for s in the given equation is s = C-nD

The question seems to be incomplete

Here is the complete question:

Solve for s in this equation D= \frac{C-s}{n}   Depreciation.

To solve for s, that means we should make s the subject of the equation

From the given equation,

D= \frac{C-s}{n}

To solve for s, first multiply both sides by n to clear the fraction

We get

n\times D= n \times \frac{C-s}{n}

Then,

nD = C - s

Now, add s to both sides

nD + s= C - s+s

nD+s = C

Then, subtract nD from both sides

nD-nD+s = C-nD

∴ s = C-nD

Hence, the solution for s in the given equation is s = C-nD

Learn more here: brainly.com/question/21406377

6 0
3 years ago
A circle with area 16pi has a sector with a central angle of 8/5pi radians. What is the area of the sector?
lianna [129]

Answer:

The area of the sector is 12.8\pi\ units^2

Step-by-step explanation:

we know that

The area of a complete circle (16π units^2) subtends a central angle of 2π radians

so

using proportion

Find out the area of a sector , if the central angle is equal to 8π/5 radians

\frac{16\pi}{2\pi}=\frac{x}{(8\pi/5)}\\\\x=8(8\pi/5)\\\\x= 12.8\pi\ units^2

4 0
4 years ago
Read 2 more answers
Polynomial of degree 4 has 1 positive real root that is bouncer and 1 negative real root that is a bouncer. How many imaginary r
Rainbow [258]

Answer:

<h3>The given polynomial of degree 4 has atleast one imaginary root</h3>

Step-by-step explanation:

Given that " Polynomial of degree 4 has 1 positive real root that is bouncer and 1 negative real root that is a bouncer:

<h3>To find how many imaginary roots does the polynomial have :</h3>
  • Since the degree of given polynomial is 4
  • Therefore it must have four roots.
  • Already given that the given polynomial has 1 positive real root and 1 negative real root .
  • Every polynomial with degree greater than 1  has atleast one imaginary root.
<h3>Hence the given polynomial of degree 4 has atleast one imaginary root</h3><h3> </h3>

8 0
3 years ago
9s - 5g = -4y, for s
Oksi-84 [34.3K]
S = (5g -4y)/9

hope this helps
4 0
4 years ago
What is the LEast common multiple?
belka [17]
The least common multiple (LCM) of a set of numbers is the lowest value that all of the numbers can be divided by and have a result of a whole number.

In this case, the least common multiple is 21 because 21/7=3 and 21/21=1.
4 0
3 years ago
Read 2 more answers
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