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Keith_Richards [23]

3 years ago

10

Explain your reasoning as you compare the values of a^n and a^-n when n<0 brainly

1 answer:

mojhsa [17]3 years ago

4 0

n < 0, is another way to say "n is negative", so let's check

$\bf ~~~~~~~~~~~~\textit{negative exponents}
\\\\
a^{-n} \implies \cfrac{1}{a^n}
\qquad \qquad
\cfrac{1}{a^n}\implies a^{-n}
\qquad \qquad
a^n\implies \cfrac{1}{a^{-n}}
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
a^n~\hspace{10.5em}\stackrel{n = -n}{a^{-n}}\implies \cfrac{1}{a^n}
\\\\\\
a^{-n}~\hspace{10em}\stackrel{n=-n}{a^{-(-n)}}\implies a^{+n}\implies a^n$

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5(12 + m) = 100 find m

Llana [10]

Answer:

8

Step-by-step explanation:

1. multiply 5 and 12 to get 60

2. multiply 5 and m to get 5m

3. the equation would be 60 + 5m = 100

4. next, subtract 60 on both sides

5. your equation so far is 5m = 40

6. then divide 5 on both sides

7. then you have your answer m = 8

3 0

3 years ago

I need help please :)) I’m not very good at math .. no links okey ......

Gelneren [198K]

You answer is the third option

5 0

3 years ago

Read 2 more answers

The dye dilution method is used to measure cardiac output with 3 mg of dye. The dye concentrations, in mg/L, are modeled by c(t)

Lemur [1.5K]

Answer:

Cardiac output: $F=0.055 L\s$

Step-by-step explanation:

Given : The dye dilution method is used to measure cardiac output with 3 mg of dye.

To Find : Find the cardiac output.

Solution:

Formula of cardiac output: $F=\frac{A}{\int\limits^T_0 {c(t)} \, dt}$ ---1

A = 3 mg

$\int\limits^T_0 {c(t)} \, dt =\int\limits^{10}_0 {20te^{-0.06t}} \, dt$

Do, integration by parts

$[\int{20te^{-0.6t}} \, dt]^{10}_0=[20t\int{e^{-0.6t} \,dt}-\int[\frac{d[20t]}{dt}\int {e^{-0.6t} \, dt]dt]^{10}_0$

$[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20}{0.6}\int {e^{-0.6t} \,dt]^{10}_0$

$[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20e^{-0.6t}}{(0.6)^2}]^{10}_{0}$

$[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-200e^{-6}}{0.6}+\frac{20e^{-6}}{(0.6)^2}]+\frac{20}{(0.60^2}$

$[\int{20te^{-0.6t}} \, dt]^{10}_0=\frac{20(1-e^{-6}}{(0.6)^2}-\frac{200e^{-6}}{0.6}$

$[\int{20te^{-0.6t}} \, dt]^{10}_0\sim {54.49}$

Substitute the value in 1

Cardiac output: $F=\frac{3}{54.49}$

Cardiac output: $F=0.055 L\s$

Hence Cardiac output: $F=0.055 L\s$

4 0

3 years ago

I need help please help

Vedmedyk [2.9K]

Answer:

$m(x)=x^4-x^3+8x-1\\\\n(x)=|x+6|$

Are not one-to-one functions

Step-by-step explanation:

As m(x) has symmetry at the line x=0, it is not one-to-one

As n(x) has symmetry at the line x=-6, it is not one-to-one

8 0

2 years ago

Solve for w.<br><br> −(1/4w 8) 8=3<br><br> w=52w=52<br> w=134w=134<br> w=−34w=−34<br> w=−12

prohojiy [21]

-(1/4 w + 8) + 8 = 3
-1/4 w - 8 = 3 - 8 = -5
1/4w = -8 + 5 = -3
w = 4 x -3 = -12
w = -12

8 0

3 years ago