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rodikova [14]
3 years ago
6

Please help asap 28 pts

Mathematics
2 answers:
Nat2105 [25]3 years ago
7 0

f(x) = -x+7

f(15) = -15+7

f(15) = -8

Mkey [24]3 years ago
3 0
F(15) is equal to -x + 105
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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
Ik it’s pretty easy but u knowh am not really sure if am right or not
gladu [14]

2,6,18

r = 6/2 = 3

18 *3 = 54

54*3 = 162

162*3=486

Answer: 54,162,486

8 0
4 years ago
Read 2 more answers
Find the unknown value .<br> 45%=54/⬛️
Vitek1552 [10]

Answer:

12

Step-by-step explanation:

54/12=4.5 then move the decimal once

4 0
3 years ago
If an assembler on average makes one error for 125 products assembled how many hours can he be expected to make after assembling
Roman55 [17]
I can't explain it with words so here's the equation, hopefully you understand:
875/125=7
6 0
3 years ago
Two automobiles leave the same city simultaneously and both head towards another city. The speed of one is 10 km/hour greater th
Veronika [31]

Answer:

5091 Km/hr and 505 km/hr

Step-by-step explanation:

Speed = Distance / Time

Let the speed of first automobile be 'x' and that of the second be 'y'

Since speed of one is 10 times greater than the other. therefore;

⇒ x = 10 y

also let time for faster automobile be 'T' and time for slower auto mobile be 't'

Since first arrive one hour earlier than second, therefore;

⇒ t = T + 1

⇒ For first automobile; x X T = 560 ; substituting for 'x' and 'T'. Therefore;

⇒ 10y (t+1) = 560

⇒ For Second automobile; y X t = 560

⇒ y = \frac{560}{t}

⇒ 10(\frac{560}{t})(t + 1) = 560

⇒ 5600 + \frac{5600}{t} = 560

⇒ 5600 - 560 =  -  \frac{5600}{t}

⇒ t = 1.11 hr

also ; T = 1.11 - 1 = 0.11 hr

Speed of 1st auto  = 560/0.11 = 5091 km /hr

Speed of 2nd auto = 560/1.11 = 505 km/hr

8 0
3 years ago
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