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ryzh [129]
3 years ago
12

What is the place value of the 9 in 219, 775?

Mathematics
2 answers:
Lady_Fox [76]3 years ago
7 0
B) thousands

5 is ones
7 is tens
7 is hundreds
9 is thousands
1 is ten thousands
2 is hundred thousands.
mash [69]3 years ago
6 0

Answer:

<em>B) Thousands</em>

Step-by-step explanation:

Hth  Tth  Th   H   T   O

2       1      9,     7    7   5

Hth = Hundred thousands

Tth = Ten thousands

Th = Thousands

H = Hundreds

T = Tens

O = Ones/units

Use the above format to input your number, then it will be easier to identify the place of value of each number.

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True or false: in order to fill the following box, you would need to fill in 2 unit cubes in length, 2 unit cubes in width, and
tensa zangetsu [6.8K]

Answer:

Step-by-step explanation:

True

7 0
3 years ago
If your smart please help me answer this question .
diamong [38]

Answer:

it should be 30 or something I'm pretty sure that's the answer

8 0
3 years ago
What substitution should be used to rewrite 16(x3 + 1)2 – 22(x3 + 1) – 3 = 0 as a quadratic equation?
Vesna [10]

Answer:

  z = x^3 +1

Step-by-step explanation:

Noting the squared term, it makes sense to substitute for that term:

  z = x^3 +1

gives ...

  16z^2 -22z -3 = 0 . . . . the quadratic you want

_____

<em>Solutions derived from that substitution</em>

Factoring gives ...

  16z^2 -24z +2z -3 = 0

  8z(2z -3) +1(2z -3) = 0

  (8z +1)(2z -3) = 0

  z = -1/8  or  3/2

Then we can find x:

  x^3 +1 = -1/8

  x^3 = -9/8 . . . . . subtract 1

  x = (-1/2)∛9 . . . . . one of the real solutions

__

  x^3 +1 = 3/2

  x^3 = 1/2 = 4/8 . . . . . . subtract 1

  x = (1/2)∛4 . . . . . . the other real solution

The complex solutions will be the two complex cube roots of -9/8 and the two complex cube roots of 1/2.

7 0
3 years ago
Read 2 more answers
ПОЖАЛУЙСТА ПОМОГИТЕ РЕШИТЬ ДАЮ 23 ОЧКА!!!
posledela

(Простите, пожалуйста, мой английский. Русский не мой родной язык. Надеюсь, у вас есть способ перевести это решение. Если нет, возможно, прилагаемое изображение объяснит достаточно.)

Use the shell method. Each shell has a height of 3 - 3/4 <em>y</em> ², radius <em>y</em>, and thickness ∆<em>y</em>, thus contributing an area of 2<em>π</em> <em>y</em> (3 - 3/4 <em>y</em> ²). The total volume of the solid is going to be the sum of infinitely many such shells with 0 ≤ <em>y</em> ≤ 2, thus given by the integral

\displaystyle 2\pi \int_0^2 y \left(3-\frac34 y^2\right) \,\mathrm dy = 2\pi \left(\frac32 y^2 - \frac3{16} y^4\right)\bigg|_0^2 = 6\pi

Or use the disk method. (In the attachment, assume the height is very small.) Each disk has a radius of √(4/3 <em>x</em>), thus contributing an area of <em>π</em> (√(4/3 <em>x</em>))² = 4<em>π</em>/3 <em>x</em>. The total volume of the solid is the sum of infinitely many such disks with 0 ≤ <em>x</em> ≤ 3, or by the integral

\displaystyle \pi \int_0^3 \left(\sqrt{\frac43x}\right)^2 \,\mathrm dx = \frac{2\pi}3 x^2\bigg|_0^3 = 6\pi

Using either method, the volume is 6<em>π</em> ≈ 18,85. I do not know why your textbook gives a solution of 90,43. Perhaps I've misunderstood what it is you're supposed to calculate? On the other hand, textbooks are known to have typographical errors from time to time...

8 0
3 years ago
10 more than 405. 4 hhundred 0 tens 5 ones
vova2212 [387]

(4 hundred + 0 tens + 5 ones) + (1 ten) = 4 hundred + 1 ten + 5 ones

... = 415

4 0
3 years ago
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