All categories

2 years ago

How to estimate to the nearest tens 671×32

Mathematics

1 answer:

aleksandrvk [35]2 years ago

8 0

20, 100

Step-by-step explanation:

You look at the tens column, in this case the numbers would be "7" in the "671" and the "3" in the "32". If you're rounding the tens column you look at the ones column and if the number is bellow 5 you round down, if its 5 nd over you round up. Since the number is 1 in the ones column that means the 7 will round down so it would be "670", for "32" the number is "2" (which is bellow 5) so we will round down again to "30". This will leave you with the equation 670×30 which is 20, 100.

You might be interested in

Freee brainliest for iconic

jenyasd209 [6]

Answer:

hi lol I need to type 20 characters

8 0

3 years ago

Read 2 more answers

How to solve this. I am very confused.

sergiy2304 [10]

To solve, we will need to plug in -5 for x in both instances.

|-5 + 2| / -5 + 2

|-3| / -3

Now, these absolute value bars may be a bit puzzling. What we have to do is take the absolute value of the number inside of the brackets. Meaning, that if the number is negative, we make it positive, and if the number is positive, then it stays positive.

3 / -3

-1

Hope this helps!! :)

6 0

2 years ago

Read 2 more answers

I NEED YOUR HELP FAST!!!!

nekit [7.7K]

Answer:

$c. \: {w}^{64}$

Sorry if i get it wrong

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

Which expression is equivalent to (q^4)^-3/q^-15 for all values of q where the expression is defined?

vodka [1.7K]

Answer: (h)q^3

Step-by-step explanation:

4 0

2 years ago