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Nutka1998 [239]

2 years ago

11

Find the length of bc Round answer to the nearest hundredth

1 answer:

Degger [83]2 years ago

6 0

Answer use sine rule
$\frac{16}{sin39} = \frac{ab}{sin21}$
answer is 9.11

You might be interested in

A rock is thrown from the top of a tall building.The distance in feet between the rock and the ground t seconds after thrown is

marusya05 [52]

Answer:

0.448 seconds

Step-by-step explanation:

d = -16t² -4t + 412

find t when d = 407

substituting d = 407 into the equation:

407 = -16t² -4t + 412 (subtract 407 from both sides)

-16t² -4t + 412 - 407 = 0

-16t² -4t + 5 = 0 (multiply both sides by -1)

16t² + 4t - 5 = 0

solving using your method of choice (i.e completing the square or using the quadratic equation), you will end up with

t = (-1/8) (1 + √21)= -0.70 seconds (not possible because time cannot be negative)

or

t = (-1/8) (1 -√21) = 0.448 seconds (answer)

4 0

3 years ago

Staples sell for $.60 a box. How many boxes can be bought for $3.45?

True [87]

Answer:

You can buy 5 boxes of staples

Step-by-step explanation:

3.45 divided by $.60 is 5.75 and u cannot get .75 boxes of staples so it rounds to 5.

6 0

3 years ago

Read 2 more answers

Evaluate using integration by parts

PolarNik [594]

Rather than carrying out IBP several times, let's establish a more general result. Let

$I(n)=\displaystyle\int x^ne^x\,\mathrm dx$

One round of IBP, setting

$u=x^n\implies\mathrm du=nx^{n-1}\,\mathrm dx$

$\mathrm dv=e^x\,\mathrm dx\implies v=e^x$

gives

$\displaystyle I(n)=x^ne^x-n\int x^{n-1}e^x\,\mathrm dx$

$I(n)=x^ne^x-nI(n-1)$

This is called a power-reduction formula. We could try solving for $I(n)$ explicitly, but no need. $n=5$ is small enough to just expand $I(5)$ as much as we need to.

$I(5)=x^5e^x-5I(4)$

$I(5)=x^5e^x-5(x^4e^x-4I(3))=(x-5)x^4e^x+20I(3)$

$I(5)=(x-5)x^4e^x+20(x^3e^x-3I(2))=(x^2-5x+20)x^3e^x-60I(2)$

$I(5)=(x^2-5x+20)x^3e^x-60(x^2e^x-2I(1))=(x^3-5x^2+20x-60)x^2e^x+120I(1)$

$I(5)=(x^3-5x^2+20x-60)x^2e^x+120(xe^x-I(0))$

Finally,

$I(0)=\displaystyle\int e^x\,\mathrm dx=e^x+C$

so we end up with

$I(5)=(x^4-5x^3+20x^2-60x+120)xe^x-120e^x+C$

$I(5)=(x^5-5x^4+20x^3-60x^2+120x-120)e^x+C$

and the antiderivative is

$\displaystyle\int2x^5e^x\,\mathrm dx=(2x^5-10x^4+40x^3-120x^2+240x-240)e^x+C$

8 0

2 years ago

The coordinates of the vertices of △JKL are J(3, 4), K(3, 1) , and L(1, 1) . The coordinates of the vertices of △J′K′L′ are J′(−

timurjin [86]

The <em><u>correct answer</u></em> is:

A 180° rotation followed by a translation 1 unit down.

Explanation:

The points are mapped as follows:

J(3, 4)→J'(-3, -5)

K(3, 1)→K'(-3, -2)

L(1, 1)→L'(-1, -2)

A 180° rotation maps a point (x, y) to (-x, -y). This would map

J(3, 4)→(-3, -4)

K(3, 1)→(-3, -1)

L(1, 1)→(-1, -1)

The difference between these points and the image points are that each y-coordinate of the image is 1 lower than these. This means a translation 1 unit down would result in the image points.

7 0

2 years ago

Read 2 more answers

2. JI. has coordinates J(-6, 1) and Z(-4, 3). Find the coordinates of the midpoint.

AleksandrR [38]

Answer:

(- 5, 2 )

Step-by-step explanation:

Given endpoints (x₁, y₁ ) and (x₂, y₂ ) then the midpoint is

( $\frac{x_{1}+x_{2} }{2}$ , $\frac{y_{1}+y_{2} }{2}$ )

Here (x₁, y₁ ) = (- 6, 1) and (x₂, y₂ ) = (- 4, 3) thus

midpoint = $\frac{-6-4}{2}$ , $\frac{1+3}{2}$ ) = ( $\frac{-10}{2}$ , $\frac{4}{2}$ ) = (- 5, 2 )

7 0

2 years ago