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WINSTONCH [101]
4 years ago
9

What is the input value?

Mathematics
1 answer:
zloy xaker [14]4 years ago
3 0

Answer:

x = 5

Step-by-step explanation:

Locate - 3 on the y- axis, move across to the right to where the graph is.

Go vertically up from this point to meet the x- axis at point x = 5

Thus

f(x) = - 3 ⇒ x = 5

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A $1,200 computer is losing value at a rate of 22% per year; 2 years​
zalisa [80]

Answer:

$730.08

Step-by-step explanation:

22% = 0.22

1 - 0.22 = 0.78

Value after 2 years

v = 1200(0.78)²

v = 730.08

$730.08

6 0
3 years ago
Andrea enjoys running. She keeps a daily journal to record how far she runs. She wears a pedometer to calculate the exact distan
Lana71 [14]

Again this is a simple addition problem I think.

Add 3.6+4.705+5.92=

14.225 kilometers that Andrea ran over the three days.

Hope this helps, have a good day. c;

5 0
3 years ago
Read 2 more answers
Water whose temperature is at 100∘C is left to cool in a room where the temperature is 60∘C. After 3 minutes, the water temperat
Tju [1.3M]

Answer:

21.68 minutes ≈ 21.7 minutes

Step-by-step explanation:

Given:

T=60+40e^{kt}

Initial temperature

T = 100°C

Final temperature = 60°C

Temperature after (t = 3 minutes) = 90°C

Now,

using the given equation

T=60+40e^{kt}

at T = 90°C and  t = 3 minutes

90=60+40e^{k(3)}

30=40e^{3k}

or

e^{3k}=\frac{3}{4}

taking the natural log both sides, we get

3k = \ln(\frac{3}{4})

or

3k = -0.2876

or

k = -0.09589

Therefore,

substituting k in 1 for time at temperature, T = 65°C

65=60+40e^{( -0.09589)t}

or

5=40e^{( -0.09589)t}

or

e^{( -0.09589)t}=\frac{5}{40}

or

e^{( -0.09589)t}=0.125

taking the natural log both the sides, we get

( -0.09589)t = ln(0.125)

or

( -0.09589)t = -2.0794

or

t = 21.68 minutes ≈ 21.7 minutes

6 0
3 years ago
M=110 what is m, the measure in degree s of third angle <br><br> m=40<br><br> m=70<br><br> m-30
Ymorist [56]
M=70 cause 180-110=70
8 0
3 years ago
PLEASE PLEASE PLEASE help me
Maru [420]

Answer:

Step-by-step explanation:

i dont know

8 0
3 years ago
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