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Allisa [31]
3 years ago
11

Which number will come in between the integers 0, 10 and (-10)?​

Mathematics
1 answer:
weqwewe [10]3 years ago
4 0

Answer:

Hey mate....

Step-by-step explanation:

This is ur answer....

<h3>-9, -8 ,-7 ,-6, -5, -4, -3, -2,-1, 0 ,1, 2, 3 ,4, 5,6, 7 ,8, 9</h3><h3 />

Hope it helps!

Mark me brainliest pls....

FOLLOW ME! :)

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UNO [17]

Answer:

160°

Step-by-step explanation:

Take notice that m∠1 and m∠2 are supplementary angles, meaning their angle measures will add up to 180°.

Since we know m∠2, we can find m∠1 easily.

180-20=160°

Hope this helped!

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Answer:

1st box would be 8/20 and the second 5/20. the final box would be >.

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Answer:

A=1.5 units²

Step-by-step explanation:

a²+b²=c²

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Chlorofluorocarbons (CFCs) were used as propellants in spray cans until their buildup in the atmosphere started destroying the o
melamori03 [73]

Answer:

a) C(2000)=1915

C(2014)=1915

b) C'(2000)=-\frac{275}{14}

C'(2014)=-\frac{275}{14}

c) C(t)=-\frac{275}{14}t+\frac{288405}{7}

d) t=2021

e) too early

Step-by-step explanation:

a)

Since C(t) is the concentration of CFCs in ppt in year t, all we need to do to solve this part is determine what the concentration of CFCs is in years 2000 and 2014. Luckily for us, the problem already gives us those values, so:

C(2000)=1915    concentration in year 2000

C(2014)=1915      concentration in year 2014

b)

By definition, the derivative of a function at a given point is interpreted as the slope of the tangent line to the point of interest, so in order to find this answer, we need to find the slope of the line. Since the problem specifies that the behavior is linear, this means that the slope will always be the same no matter the year, so we get:

m=C'(t)=\frac{C'(t_{2})-C'(t_{1})}{t_2-t_1}

so:

C'(t)=\frac{1640-1915}{2014-2000}=-\frac{275}{14}\approx -19.64

therefore:

C'(2000)=-\frac{275}{14}

C'(2014)=-\frac{275}{14}

c)

Since the behavior is linear, we can calculate it with the point-slope form of the line which is:

y-y_{1}=m(x-x_{1})

in this case:

C(t)-C(t_1)=C'(t)(t-t_{1})

so we get:

C(t)-1915=-\frac{275}{14}(t-2000)

and we can now solve for C(t) so we get:

C(t)=-\frac{275}{14}t+\frac{275000}{t}+1915

for a final answer of:

C(t)=-\frac{275}{14}t+\frac{288405}{7}

d)

So next we solve the equation for C(t)=1500 so we get:

1500=-\frac{275}{14}t+\frac{288405}{7}

1500-\frac{288405}{7}=-\frac{275}{14}t

-\frac{277905}{7}=-\frac{275}{14}t

t=2021 \frac{7}{55}

so we will reach that concentration level at the year 2021 approximately.

e)

Since the second derivative of the concentration function is greater than zero, this means that the original function might be a function in the form: .

This means that the decrease of the concentration levels is slower than that of a linear equation. So the projected date will be too early than the real date.

8 0
3 years ago
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