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1 year ago

5-16 find the general indefinite integral.

1 answer:

6 0

$\displaystyle\int \left( u^6 - 2u^5 - u^3 + \cfrac{2}{7} \right)du\implies \cfrac{u^7}{7}-2\cdot \cfrac{u^6}{6}-\cfrac{u^4}{4}+\cfrac{2}{7}u \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \cfrac{u^7}{7}- \cfrac{u^6}{3}-\cfrac{u^4}{4}+\cfrac{2u}{7}+C~\hfill$

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Georgia [21]

I just gave you the answers

7 0

3 years ago

Work out (1.2 x 104 ) 2.5 x 103<br>Give<br>your answer in standard form.

ExtremeBDS [4]

Answer:

32136

Step-by-step explanation:

it is basically (1.2*104 )*2.5*103

4 0

2 years ago

Read 2 more answers

The width of a rectangle is 2 inches less than its length. If the perimeter of the rectangle is 96 inches, what is its width?

Hatshy [7]

Answer:

Step-by-step explanation:

a) Perimeter of a rectangle = 2 *(L +w)

b) w = L - 2

Answer: B

c) Perimeter = 96 inches

2* (l +w) = 96

2*( L + L - 2 ) = 96 {Combine like terms}

2 * (2L - 2) = 96 {use distributive law}

2*2L - 2*2 = 96

4L - 4 = 96 {Add 4 to both sides}

4L = 96 +4

4L = 100 {divide both sides by 4}

4L/4 = 100/4

L = 25 inches

d) w = L - 2

= 25 - 2

w = 23 inches

4 0

2 years ago

I don't understand this question??

Len [333]

Answer:

Step-by-step explanation:

3 0

2 years ago

Solve for x and y please!

galina1969 [7]

Answer:

$x = 8$

$y=2\sqrt{3}$

Step-by-step explanation:

The figure is composed of 3 Right triangles. To find the values of the variables x and y we use the Pythagorean theorem to propose one equation.

$x^2 =4^2 + (4\sqrt{3})^2$

Now we solve for x

$x=\sqrt{4^2 + (4\sqrt{3})^2}$

$x=\sqrt{16 +16*3}\\\\x=\sqrt{64}\\\\x=8$

Let's call z at the angle opposite to y

Then we have that:

$sin(z) =\frac{opposite}{hypotenuse}$

Where

$hypotenuse = 8$

$opposite=4$

$sin(z) =\frac{4}{8}$

$z=sin^{-1}(\frac{1}{2})$

$z=30$

Now we use this angle to find the length y

$sin(z) =\frac{opposite}{hypotenuse}$

Where in this case

$hypotenuse = 4\sqrt{3}$

$opposite=y$

$z=\°30$

$sin(30\°) =\frac{y}{4\sqrt{3}}$

$y=sin(30\°)*4\sqrt{3}$

$y=2\sqrt{3}$

6 0

3 years ago