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3 years ago

What is the value of the expression 3^−4?

Mathematics

2 answers:

Wewaii [24]3 years ago

7 0

$3^{-4}= \cfrac{1}{3^4} =\cfrac{1}{81}$

hjlf3 years ago

4 0

Answer:

Answer is $\frac{1}{81}$

Step-by-step explanation:

Here we can apply the property of exponents with negative exponents

in order to make exponent positive we can take the term in denominator which makes the expression $\frac{1}{3^4}$ and $3^4 means$ 81

therefore answer is $\frac{1}{81}$

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An art history professor assigns letter grades on a test according to the following scheme.A: Top 5% of scoresB: Scores below th

adoni [48]

Answer:

Grade B score:

$76 \leq x \leq 89$

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 73.3

Standard Deviation, σ = 9.7

We are given that the distribution of score on test is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

B: Scores below the top 5% and above the bottom 62%

We have to find the value of x such that the probability is 0.62

$P( X > x) = P( z > \displaystyle\frac{x - 73.3}{9.7})=0.62$

$= 1 -P( z \leq \displaystyle\frac{x - 73.3}{9.7})=0.62$

$=P( z \leq \displaystyle\frac{x - 73.3}{9.7})=0.38$

Calculation the value from standard normal z table, we have,

$\displaystyle\frac{x - 73.3}{9.7} = 0.305\\\\x = 76.26$

We have to find the value of x such that the probability is 0.05

$P(X < 0.95) = \\\\P( X < x) = P( z < \displaystyle\frac{x - 73.3}{9.7})=0.95$

Calculation the value from standard normal z table, we have,

$\displaystyle\frac{x - 73.3}{9.7} = 1.645\\\\x = 89.26$

Thus, the numerical value of score to achieve grade B is

$76 \leq x \leq 89$

7 0

3 years ago

Consider a series circuit consisting of a resistor of R ohms, an inductor of L henries, and variable voltage source of V(t) volt

ser-zykov [4K]

Answer:

$I(t)=\frac{1}{3}(1-e^{-30t})$

Step-by-step explanation:

We are given that

$\frac{dI}{dt}+\frac{R}{L}I=\frac{V(t)}{L}$

R=150 ohm

L=5 H

V(t)=10 V

$P=\frac{R}{L}=\frac{150}{5}=30$

$I.F=e^{\int Pdt}=e^{\int 30 dt}=e^{30 t}$

$I(t)\times I.F=\int e^{30 t}\times 10 dt+C$

$I(t)\times e^{30 t}=\frac{10}{30}e^{30 t}+C$

$I(t)=\frac{1}{3}+Ce^{-30 t}$

I(0)=0

Substitute t=0

$0=\frac{1}{3}+C$

$C=-\frac{1}{3}$

Substitute the values

$I(t)=\frac{1}{3}-\frac{1}{3}e^{-30 t}$

$I(t)=\frac{1}{3}(1-e^{-30t})$

5 0

3 years ago

Can you help me please.

Kamila [148]

9514 1404 393

Answer:

a. (-1/2, 5)

Step-by-step explanation:

The coefficients of x have opposite signs, so the x-terms can be eliminated by adding the two equations.

(2y -12x) +(3y +12x) = (16) +(9)

5y = 25 . . . . . . simplify

y = 5 . . . . . . . . .divide by 5

This matches only answer choice A.

<em>Check</em>

The other part of answer choice A is x = -1/2. We can test this to make sure it works:

2(5) -12(-1/2) = 16

10 + 6 = 16 . . . . . . true, the selected answer works OK

5 0

3 years ago

I need help please.

Mumz [18]

Answer:

the result is : 14x + 27

4 0

3 years ago

Read 2 more answers

There is a proportional relationship between x and y complete the table with the missing values

Masja [62]

Answer: The first blank space is 45 and the one at the bottom is 10

8 0

3 years ago