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2 years ago

13

A rectangular room is 6 meters longer than it is wide, and its perimeter is 32 meters. Find the dimension of the room.

1 answer:

nikdorinn [45]2 years ago

6 0

Step-by-step explanation:

I am not sure but i think is this

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"suppose that it is decided to reject the null hypothesis if the sample mean time lag is less than 18.5 days. what is the probab

sweet-ann [11.9K]

Answer : 17
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4 years ago

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An elevator is supported by a cable and is moving downward through the elevator shaft at constant speed.

Nana76 [90]

Answer: D The upward tension force in the cable is equal to the downward force of gravity.

Step-by-step explanation:

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3 years ago

Of a journey of 216km, 56km were on untarred road. for this part of the journey, a motorist could only average 28km/h. if his av

kenny6666 [7]

Answer:

64 km/hr

Step-by-step explanation:

216 - 56 = 160 km on good road

216km / 48 km/hr = 4.5 hours total time

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160 km / 2.5 hr = 64 km/hr on the good road portion

3 0

2 years ago

Can you guys help me fill this out?

Nataly_w [17]

Answer: x=6

Step-by-step explanation:

2x+5= 17

-5=-5

(subtract 5 on both side)

2x=12

(divid 2 on each side)

x=6

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3 years ago

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U-substitutions only work for specific kinds of expressions. Below, you are asked to choose a value of n for which u-substitutio

Dahasolnce [82]

Answer:

Step-by-step explanation:

(a) $\int x^n e^{5x^4+1} \ dx$

Suppose $5x^4 + 1 = f$

by differentiation;

$\implies \ 20 x^3 dx = df --- (1)$

Suppose n = 3

Then, the integral

$I = \int x^ 3 e^{5x^4 + 1} \ dx$

$= \int e^f \ \dfrac{df}{20}$

$= \dfrac{1}{20} \int e^f \ dt$

$= \dfrac{1}{20} e^f + C$

recall that $f = 5x^4 + 1$

Then;

$\mathbf{ I = \dfrac{1}{20}e^{5x^4+1}+C}$

(b) $\int \dfrac{cos (\dfrac{1}{x^3})}{x^n } \ dx$

suppose; $\dfrac{1}{x^3} = f$

$x^3 = f$

$\implies -3x^{-4} \ dx = df$

$\implies \dfrac{1}{x^4} \ dx =-\dfrac{1}{3} df$

If n = u, then the integration is:

$I = \int \dfrac{1}{x^4} \ cos (\dfrac{1}{x^4}) \ dx$

$= \int -\dfrac{1}{3} \ cos \ f \ df$

$= -\dfrac{1}{3} \int \ cos \ f \ df$

$= -\dfrac{1}{3} \ sin \ f + C$

Since; $x^3 = f$

Then;

$\mathbf {I = -\dfrac{1}{3} \ sin \ \Big( \dfrac{1}{x^3}\Big) + C}$

(c) $\int \dfrac{x+n}{x^2 + 8x -4} \ dx$

Suppose $x^2 + 8x - 4 = f$

Then, by differentiation of both sides

$(2x + 8) \ dx = df$

$(x + 4) \ dx = \dfrac{1}{2} \ df$

Suppose n = 4 in integration, then:

$I = \int \dfrac{(x + 4) }{x^2 +8x -4} \ dx$

By substitution;

$I = \int \dfrac{1}{2}\dfrac{1}{f} \ df$

$= \dfrac{1}{2} \ \ { In |f|} + C$

$\mathbf{= \dfrac{1}{2} \ \ { In |x^2+8x -4|} + C}$

7 0

3 years ago