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DiKsa [7]
2 years ago
5

What’s the inequality of N<-1

Mathematics
1 answer:
Juliette [100K]2 years ago
7 0

Answer:

ℎℎ ℎ

Step-by-step explanation:

ℎℎℎℎ ℎ <em /><em /><em /><em /><em /><em /><em /><em><u /></em><em><u /></em><em><u /></em><em><u /></em><em><u /></em><em><u> </u></em><em><u /></em><em><u /></em><em><u /></em><em><u /></em><em><u /></em><em><u> </u></em><em><u>ℎ</u></em><em><u /></em><em><u /></em><em><u /></em><em><u /></em>

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Can someone explain to me how to solve this problem I’m really confused. I don’t even know where to start.
kaheart [24]
20. Hope this helped
6 0
3 years ago
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The function f(x)= x^2 -4 is translated 3 units to the right to create function g(x). What is g(9)?
algol [13]

Answer:

g(9) = 32

Step-by-step explanation:

Given f(x) then f(x + a) is a horizontal translation of f(x)

• If a > 0 then shift to the left of a units

• If a < 0 then shift to the right of a units

Here f(x) is translated 3 units right, thus

g(x) = (x - 3)² - 4, hence

g(9) = (9 - 3)² - 4 = 6² - 4 = 36 - 4 = 32

7 0
2 years ago
Find all local extrema for f(x, y) = 2y3 + 12x2 − 24xy. (if an answer does not exist, enter dne.) local minimum (x, y) = local m
OlgaM077 [116]
f(x,y)=2y^3+12x^2-24xy
Find the first derivatives:
f'_x=24x-24y \\ f'_y=6y^2-24x.
Solve the system \left \{ {{f'_x=0} \atop {f'_y=0}} \right.:
\left \{ {{24x-24y=0} \atop {6y^2-24x=0}} \right. \rightarrow \left \{ {{x=y} \atop {6y^2-24y=0}}. The second equation has solutions y_1=0, \\ y_2=4 and then x_1=0, \\ x_2=4 and you have two points A_1(0,0), \\ A_2(4,4).

Find the first derivatives:
f^{''}_{xx}=24 \\ f^{''}_{xy}=f^{''}_{yx}=-24 \\ f^{''}_{yy}=12y and calculate \Delta=\left|  \left[\begin{array}{cc}24&-24\\-24&12y\end{array}\right]\right |=24\cdot 12y-(-24)^2=288y-576.
Since \Delta(A_1)=-576 and f^{''}_{xx}\ \textgreater \ 0, A_1 is a point of maximum and f(0,0)=0.
Since \Delta(A_2)=576 and f^{''}_{xx}\ \textgreater \ 0, A_1 is a point of minimum and f(4,4)=-64.




4 0
3 years ago
(a) consider the initial-value problem a=ka,a(0)=ao as t as the model for the decay of a radioactive substance. show that, in ge
mina [271]

The initial-value problem dA/dt = kA, A(0) = A₀ is used to represent the radioactive decay. The radioactive substance having half-life T= -(ln 2)/k,will take 2.5 T for the substance to decay from A₀ to A₀ / 6 .

a.) To solve this, we have the following differential equation:

dA/dt = kA

With the initial condition A(0) = A₀

Rewriting the differential equation like this:

dA/A = kdt

And if we integrate both sides we get:

ln |A| = kt + c₁

Where  is a constant. If we apply exponential for both sides we get:

A=e^{kt}e^{c} = C e^{kt}

Using the initial condition A(0) = A₀ we get ,

A₀ = C

So the solution for the differential equation is given by:

A(t) = A_{0}e^{kt}

For the half life we know that we need to find the value of t for where we have,

A(t) = 1/2 (A₀)

Using this we get ,

\frac{1}{2}A_{0} = A_{0}e^{kt

Cancel A₀ on both sides and applying log on both sides we get ,

ln(1/2) = kt

t = {ln(1/2) / k } ----(1)

And using the fact that  ln(1/2) = -ln(2)  we get,

t = - { ln(2) / k }

b.) To solve this we consider ,

A(t) = A_{0} e^{kt}

Replacing k with value obtained from 1 we get,

k = - {ln(2) / T}

A(t) = A_{0}e^{\frac{ln(2)}{T}t

Cancel the exponential with the natural log, we get,

A(t) = A_{0} 2^{-\frac{1}{T} }

c.)For this case we find the value of t when we have remaining A₀/6

So we can use the following equation:

A₀/6 =A_{0}2^{-\frac{t}{T} }

Simplifying we got:

1/6 = 2^{-\frac{t}{T} }

We can apply natural log on both sides and we got:

ln (1/6) = -t/T{ln(2)}

And if we solve for t we got:

t = T { ln(6) / ln(2) }

We can rewrite this expression like this:

t = T { ln(2²°⁵) / ln(2) }

Using properties of natural logs we got:

t = 2.5T { ln(2) / ln(2) }

t = 2.5T

Thus it will take 2.5 T for the substance to decay from A₀ to A₀/6 

Solve more problems on Half-life at :brainly.com/question/16439717

#SPJ4

5 0
1 year ago
If a student sees a graph with an axis similar to the one in the photo, the student can assume:
mario62 [17]

Answer:

the scale is linear ,the scale is logarithmic

5 0
2 years ago
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