Ask question

Ask question

All categories

3 years ago

12

What is 3 47/99 in decimal form?

2 answers:

ddd [48]3 years ago

5 0

The answer is 3.<span>474747474747475</span>

LiRa [457]3 years ago

3 0

The answer is 3.47 repeating

You might be interested in

5x+6y=7 need to find the slope intercept and standard form

Oksi-84 [34.3K]

The problem is already in standard form but to get it into slope intercept form you rewrite the equation so it looks like
6y=-5x+7 divide all by 6
y=-5/6x+7/6
slope being -5/6 and b being 7/6

7 0

3 years ago

What is the speed (approximately) of a 2.5-kilogram mass after it has fallen freely from rest through a distance of 12 meters?

Norma-Jean [14]

Answer:

15.3 m/s

Step-by-step explanation:1) Find the gravitational potential energy using the equation :

$PE_g=mgh$

$PE_g=2.5(12)(9.8)\\PE_g=294$

2) Then use the equation for kinetic energy to solve for the velocity:

$KE=\frac{1}{2} mv^{2}$

$294=\frac{1}{2}(2.5)(v^2)\\235.2=v^2\\\sqrt{235.2} =v\\15.3 m/s=v$

6 0

3 years ago

WILL MARK BRAINLIEST

____ [38]

Answer:

512

Step-by-step explanation:

∛n = 8

n = 512

double check

∛512 = ∛(8)^3 = 8

3 0

3 years ago

Read 2 more answers

Arrange the cones in order from lease volume to greatest volume

bearhunter [10]

Answer:

Volume of the cone in ascending order.

$V_{2}=270\pi\ units^{3}$

cone with DIAMETER of 18 & height of 10

cone with RADIUS of 10 & height of 9

cone with RADIUS of 11 & height of 9

cone with DIAMETER of 20 & height of 12

Step-by-step explanation:

Let $V_{2}. V_{3}. and\ V_{4}.$ be the volume of the cone.

Let d, r and h be the diameter, radius and height of the cone.

Given:

$d_{1} = 20\ and\ h_{1}=12$

$d_{2} = 18\ and\ h_{2}=10$

$r_{3} = 10\ and\ h_{3}=9$

$r_{4} = 11\ and\ h_{14}=9$

Arrange the cones in order from lease volume to greatest volume.

Solution:

The volume of the cone is given below.

$V=\pi r^{2} \frac{h}{3}$ ----------------(1)

where: r is radius of the base of cone.

and h is height of the cone.

The volume of the cone for $d_{1} = 20\ and\ h_{1}=12$

$r_{1} = \frac{d_{1}}{2}$

$r_{1} = \frac{20}{2}=10\ units$

$V_{1}=\pi (r_{1})^{2} \frac{h_{1}}{3}$

$V_{1}=\pi (10)^{2} \frac{12}{3}$

$V_{1}=\pi\times 100\times 4$

$V_{1}=400\pi\ units^{3}$

Similarly, for volume of the cone for $d_{2} = 18\ and\ h_{2}=10$

$r_{2} = \frac{d_{2}}{2}$

$r_{2} = \frac{18}{2}=9\ units$

$V_{2}=\pi (r_{2})^{2} \frac{h_{2}}{3}$

$V_{2}=\pi (9)^{2} \frac{10}{3}$

$V_{2}=\pi\times 81\times \frac{10}{3}$

$V_{2}=\pi\times 27\times 10$

$V_{2}=270\pi\ units^{3}$

Similarly, for volume of the cone for $r_{3} = 10\ and\ h_{3}=9$

$V_{3}=\pi (r_{3})^{2} \frac{h_{3}}{3}$

$V_{3}=\pi (10)^{2} \frac{9}{3}$

$V_{3}=\pi\times 100\times 3$

$V_{3}=\pi\times 300$

$V_{3}=300\pi\ units^{3}$

Similarly, for volume of the cone for $r_{4} = 11\ and\ h_{4}=9$

$V_{4}=\pi (r_{4})^{2} \frac{h_{4}}{3}$

$V_{4}=\pi (11)^{2} \frac{9}{3}$

$V_{4}=\pi\times 121\times 3$

$V_{4}=\pi\times 363$

$V_{4}=363\pi\ units^{3}$

So, the volume of the cone in ascending order.

$V_{2}=270\pi\ units^{3}$

7 0

3 years ago

What does James need to justify this statement in a proof.

ahrayia [7]

Definitions of Right Angles

6 0

4 years ago