Can i have some help please ?!?!

Geometry i need help unit 10 assignment

Vladimir79 [104]

Answer:

20 senameters 2

Step-by-step explanation:

7 0

4 years ago

URGENT!<br> How do you solve a word problem using the direct variation formula?

Marat540 [252]

Answer:

U DIVIDE IT!!

Step-by-step explanation:

3 0

3 years ago

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Find the length of a rectangle if its

Alina [70]

Answer: L = 29cm

Step-by-step explanation:

Let's go!!

To find any rectangle area, you have to apply this formula: A = L . w , where L is the length and w is the width.

So, we have w = L - 13 (the width is 13 less than the Length) and then, perimeter is 90 cm.

Well, I hope you remember that the perimeter of rectangle is 2L + 2W. In this case, 2L + 2w = 90

Then, you might to solve this system of equation:

2L + 2w = 90

w = L - 13

Simplifying the first equation, you'll have L + w = 45 (you can divide everything of 2).

Our new system:

L + w = 45

w = L - 13

Using the substituition method:

L + L - 13 = 45

2L = 58

L = 29 cm

the width is 29 - 13 = 16 cm

4 0

3 years ago

Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab

Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

$SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx$

If $f(x)=x^2,$ then

$f'(x)=2x$

and

$b=0\\ \\a=2$

Therefore,

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx$

Apply substitution

$x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du$

Then

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du$

Now

$\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))$

Hence,

$SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}$

3 0

3 years ago

How do i find the area of this composite figure?

diamong [38]

To find the area of a composite figure, separate the figure into simpler shapes whose area can be found. Then add the areas together. Be sure than none of the simpler figures have overlapping areas

6 0

3 years ago

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