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2 years ago

How do you write 6,007,200 in words

Mathematics

2 answers:

valentinak56 [21]2 years ago

8 0

Six million and seven thousands and two hundreds

kirill [66]2 years ago

3 0

Six million, seven thousand two hundred

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5.2.AP-3

Ludmilka [50]

.6p + 4.5 =22.5
.6p=18
P=30

4 0

3 years ago

My sister needs help<br> "Find the perimeter of the figure to the nearest hundredth."

Flauer [41]

Answer: 27.42 ft

Step-by-step explanation:

To find the perimeter first we must find the circumference of the circles.

You can easily find the diameter by subtracting and you get 6.

Using the circle circumference formula c=2piR you get 9.42.

9.42 is our circumference of one circle.

You don't need to divide this by 2 because you already have 2 halves of a circle.

Next add all the sides which is 18.

Add this to the circumference we calculated earlier which gives you 27.42 ft.

5 0

3 years ago

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-7.832198653 . . . is: rational irrational

olga2289 [7]

Definitely irrational because the decimal never repeat

3 0

2 years ago

Read 2 more answers

How many hundreds in 148305

Wittaler [7]

$148305:100=\dfrac{148305}{100}=1483.05\\\\Answer:\ In\ 148305\ is\ 1483\ hundreds$

6 0

2 years ago

Read 2 more answers

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

2 years ago