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Karolina [17]
3 years ago
6

12. Between what two numbers would 9 be located on a number line?

Mathematics
2 answers:
lozanna [386]3 years ago
5 0

Answer:

B

Step-by-step explanation:

1,2,3,4,5,6,7,than 8,9,10

stepan [7]3 years ago
4 0
9 would be located between b. 8 and 10
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Reduce -56/-70 to lowest terms.<br> 8/10 <br> -4/5 <br> -2/3 <br> 4/5
VLD [36.1K]
This one is just common sense - you don't even have to do any work. Since there's a negative on both the top and bottom you cancel them, which eliminates both the second and third answer choices. Also, 8/10 and 4/5 are the same, but 4/5 is simplified more so that is your answer.
7 0
4 years ago
Put 1.09, 1.0, 1.9, 1.19, 1.1 on a number line in order?
Luden [163]

Answer:

1.0, 1.09, 1.1, 1.19, 1.9

Step-by-step explanation:

Basic ordering of decimals

7 0
3 years ago
Whatis 4b to the -6 power in its simplest form
Olenka [21]
4b^-6

When u have a negative exponent, u simply take the reciprocal. So the answer is

4/b^6
6 0
4 years ago
Read 2 more answers
Fill the blank to make these statement true
BigorU [14]

Answer:

Step-by-step explanation:

1a. -3 * -4 = 4 * 3

b.7(4+11) = 7(4) + 7(11)

c.6(7-4) = 6(3) - (0) *4

d. -2(6*(-5)) = -2*6*(-5)

2a.-60+4 = -56 = 50

b.-20 / 3 = 6.6 = 6.0 = 6

c. 2.4 = 2.0 = 2

d. 47.19 =40.00 = 40

e. 35.624+ (-4.17) = 31.454 =30.000 = 30

3 0
3 years ago
Use calculus to find the volume of the following solid S:
Sveta_85 [38]

I assume the base of S is the disk centered at the origin, or x² + y² ≤ r². Solving for y gives two solutions corresponding to the upper and lower halves of the circular boundary,

x^2+y^2 = r^2 \implies y = \pm \sqrt{r^2-x^2}

and the vertical distance between them - i.e. the side length of each cross section - is

\sqrt{r^2-x^2} - \left(-\sqrt{r^2-x^2}\right) = 2\sqrt{r^2-x^2}

Then the volume of each square cross section 4 (r² - x²) ∆x, and the volume of S is

\displaystyle \int_{-r}^r 4(r^2-x^2) \, dx

By symmetry, this is the same as

\displaystyle 8 \int_0^r (r^2-x^2) \, dx

which gives a volume of

\displaystyle 8 \int_0^r (r^2-x^2) \, dx = 8 \left(r^3 - \frac{r^3}3\right) = \boxed{\frac{16r^3}3}

4 0
3 years ago
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