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m_a_m_a [10]
3 years ago
8

N is an integer and - 2<n+3<4, write down the possible values of n from smallest to largest​

Mathematics
1 answer:
Korolek [52]3 years ago
4 0

Answer:

- 4, - 3, - 2, - 1, 0

Step-by-step explanation:

Given

- 2 < n + 3 < 4 ( subtract 3 from each interval )

- 5 < n < 1

This indicates that n ≠ - 5, 1 but all values in between, that is

n = - 4, - 3, - 2, - 1, 0 ← possible values of n

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The answers are A and C because you need to be able to convert simple decimals like those
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3 years ago
Find the missing side to the triangle in the attached image.
kari74 [83]

Answer:

x=34

Step-by-step explanation:

1. formula for Pythagorean Theorem for right trangles is <u>a^2+b^2=c^2</u>

2. a=30 and b=16, so (30)^2+(16)^2=c^2

3. (30)^2=900 and (16)^2=256

4. 900+256=1156

5. 1156=c^2

6. \sqrt{1156} =\sqrt{c^2}

7. the square root of 1156 is 34, so c=34

8. The missing side of the triangle, labeled as "x<em>"</em> is 34

8 0
3 years ago
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Which of the following is equivalent to
aleksandr82 [10.1K]

Answer:

Its A i just took the test

Step-by-step explanation:

5 0
3 years ago
20 POINTS! + BRAINLIEST!
g100num [7]

Answer:

= x^4 -x^3 -4x^2 -3

Step-by-step explanation:

f(x) = x^4 -x^2 +9

g(x) = x^3 +3x^2 +12

We want to find f-g

(f-g) (x) = x^4 -x^2 +9 - (x^3 +3x^2 +12)

Distribute the minus sign

(f-g) (x) = x^4 -x^2 +9 - x^3 -3x^2 -12

            = x^4 -x^3 -4x^2 -3

3 0
3 years ago
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A model for tumor growth is given by the Gompertz equation dV dt = a (In b-, In V) V where a &gt;0 and b &gt;0 are constants and
Xelga [282]

Answer:

V(t) = b^{(1-e^{-at})}

Step-by-step explanation:

We are given the following information in the question:

\displaystyle\frac{dV}{dt} = a (In b - In V) V

where a > 0 and b > 0.

\displaystyle\frac{dV}{dt} = a (In b-In V) V\\\\\displaystyle\frac{dV}{dt} = -aV(ln\frac{V}{b})\\\\\frac{dV}{V(ln\frac{V}{b})} = (-a)dt\\\\\text{Put } ln\frac{V}{b} = z\\\\\text{Integrating both sides}\\\\\int \frac{dV}{V(ln\frac{V}{b})} = \int (-a)dt\\\\\text{We get}\\\\\int \frac{dz}{z} = \int (-a)dt\\\\\\\text{where C is the constant of integration}

V(0) = 1~ mm^3\\V(t) = b.e^{e^{-at+C}}\\\text{Putting t =0, V(0) = 1}\\\\V(0) = 1 = b.e^{e^{C}}\\\\V(t) = b^{(1-e^{-at})}

where v(t) is the required tumor volume as a function of time that has an initial tumor volume of V(0) = 1 cubic mm.

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