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4 years ago

A clock radio phone cost 83.95. Tax is $4.49 for each radio phone how much do 3 clock radio phones cost

Mathematics

1 answer:

zimovet [89]4 years ago

5 0

Step-by-step explanation:

If each clock radio costs $83.95 + taxes ($4.49), each clock radio has the real cost of $88.44

Therefore if you multiply the real cost of the 3 pieces, you'll get the result:

$88.44 x 3 = $265.32

Answer:

$265.32

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I need help with Law of Sines; I posted pictures but I need help ASAPP

r-ruslan [8.4K]

Answer:

B = 26.407 degrees

C = 73.593 degrees

c = 30.196 units

Step-by-step explanation:

Use the law of sines

sin(B) / 14 = sin(80) / 31

B = sin-1( sin(80) / 31 x 14)

B = 26.407

C = 180 - 80 - 26.407

C = 73.593

c / sin(73.593) = 31 / sin(80)

c = 31 / sin(80) x sin(73.593)

c = 30.196

4 0

3 years ago

36) The speed of a boat in still water is 5km/hr. If the speed of the boat against the stream is 3

Radda [10]

Answer:

2km/hr

Step-by-step explanation:

Given the following

Speed of boat in water = 5km/hr

Speed of boat against stream = 3km/hr

Since the speed of boat is going against that of the stream, hence, the speed of the stream will be the difference between the speed of boat and that when the boat is against the stream

Speed of stream = 5km/hr - 3km/hr

Speed of stream = 2km/hr

hence the required speed of stream is 2km/hr

5 0

3 years ago

I have calculus problems that I need help with.

aleksklad [387]

a. Note that $f(x)=x^ne^{-2x}$ is continuous for all $x$ . If $f(x)$ attains a maximum at $x=3$ , then $f'(3) = 0$ . Compute the derivative of $f$ .

$f'(x) = nx^{n-1} e^{-2x} - 2x^n e^{-2x}$

Evaluate this at $x=3$ and solve for $n$ .

$n\cdot3^{n-1} e^{-6} - 2\cdot3^n e^{-6} = 0$

$n\cdot3^{n-1} = 2\cdot3^n$

$\dfrac n2 = \dfrac{3^n}{3^{n-1}}$

$\dfrac n2 = 3 \implies \boxed{n=6}$

To ensure that a maximum is reached for this value of $n$ , we need to check the sign of the second derivative at this critical point.

$f(x) = x^6 e^{-2x} \\\\ \implies f'(x) = 6x^5 e^{-2x} - 2x^6 e^{-2x} \\\\ \implies f''(x) = 30x^4 e^{-2x} - 24x^5 e^{-2x} + 4x^6 e^{-2x} \\\\ \implies f''(3) = -\dfrac{486}{e^6} < 0$

The second derivative at $x=3$ is negative, which indicate the function is concave downward, which in turn means that $f(3)$ is indeed a (local) maximum.

b. When $n=4$ , we have derivatives

$f(x) = x^4 e^{-2x} \\\\ \implies f'(x) = 4x^3 e^{-2x} - 2x^4 e^{-2x} \\\\ \implies f''(x) = 12x^2 e^{-2x} - 16x^3e^{-2x} + 4x^4e^{-2x}$

Inflection points can occur where the second derivative vanishes.

$12x^2 e^{-2x} - 16x^3 e^{-2x} + 4x^4 e^{-2x} = 0$

$12x^2 - 16x^3 + 4x^4 = 0$

$4x^2 (3 - 4x + x^2) = 0$

$4x^2 (x - 3) (x - 1) = 0$

Then we have three possible inflection points when $x=0$ , $x=1$ , or $x=3$ .

To decide which are actually inflection points, check the sign of $f''$ in each of the intervals $(-\infty,0)$ , $(0, 1)$ , $(1, 3)$ , and $(3,\infty)$ . It's enough to check the sign of any test value of $x$ from each interval.

$x\in(-\infty,0) \implies x = -1 \implies f''(-1) = 32e^2 > 0$

$x\in(0,1) \implies x = \dfrac12 \implies f''\left(\dfrac12\right) = \dfrac5{43} > 0$

$x\in(1,3) \implies x = 2 \implies f''(2) = -\dfrac{16}{e^4} < 0$

$x\in(3,\infty) \implies x = 4 \implies f''(4) = \dfrac{192}{e^8} > 0$

The sign of $f''$ changes to either side of $x=1$ and $x=3$ , but not $x=0$ . This means only $\boxed{x=1}$ and $\boxed{x=3}$ are inflection points.

4 0

1 year ago

Read 2 more answers

Can anyone explain how to get to the answer ? I’ll mark brainliest! Please

Svetllana [295]

Answer:

Step-by-step explanation:

It is linear because it is a pattern of adding 1 and not quadratic for there is no squaring. It is no exponential because the ratio stays the same.

7 0

4 years ago

In order to slove the following system of equations by addition, which of the following could you do before adding the equations

Vladimir [108]

Answer:

B. use multipliers -3 and 2

Step-by-step explanation:

If you're solving the system by "addition," multiplying the equations by the chosen numbers needs to result in opposite coefficients for one of the variables.

<h3>Effect of answer choices</h3>

A. The system becomes ...

12x -6y = 21
12x -12y = 60

No variable has opposite coefficients.

B. The system becomes ...

-12x +6y = -21
6x -6y = 30

The y-variable has opposite coefficients. This is a good choice.

C. The system becomes ...

12x -6y = 21
6x -6y = 30

No variable has opposite coefficients.

D. The system becomes ...

4/3x -2/3y = 7/3
3x -3y = 15

No variable has opposite coefficients.

<em>Additional comment</em>

The simplest "no brainer" solution is to use the coefficients of one of the variables, negating one of them. Multiply each equation by the opposite equation's coefficient. Here, the y-coefficients are -2 and -3, and the multipliers of answer choice B are -3 and 2, consistent with this advice. (The sign of 2 was changed.)

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2 years ago

Read 2 more answers