All categories

3 years ago

382.993 to the nearest hundredth

Mathematics

2 answers:

tia_tia [17]3 years ago

8 0

Answer:

That number rounds to the nearest hundredth to this number: 382.99

s2008m [1.1K]3 years ago

6 0

Answer:382.990

Step-by-step explanation:

if its below 5 then round down somtimes it stays the same but 6 or higher you round up

You might be interested in

If there are 205 students in the school, which is the best estimate of the number of students that are in the 5th grade?

Mila [183]

Answer:

34 people

Step-by-step explanation:

8 0

3 years ago

Prove that ΔABC and ΔEDC are similar. triangles ABC and DEC where angles A and E are right angles, AC equals 4, AB equals 3, BC

nikklg [1K]

Answer:B

Step-by-step explanation:

4E and LA are right angles; therefore, these angles are congruent since all right

angles are congruent. Is - 15 over 5 and 12 over 4shows the corresponding sides are proportional;

therefore, ABC ~ AEDC by the SSS Similarity Postulate.

3 0

2 years ago

Plss help me do this o-o

Vinvika [58]

Answer:

$x=\sqrt6\\y=\sqrt{12}$

Step-by-step explanation:

We know that since angles in a triangle add up to 180º, the remaining angle must be 45º.

So the side with length $x$ must be equal to the side with length $\sqrt6$ . That is:

$x=\sqrt6$

Now, by Pythagoras:

$y=\sqrt{(\sqrt6)^2+(\sqrt6)^2}\\=\sqrt{6+6}\\=\sqrt{12}$

4 0

2 years ago

Read 2 more answers

On Mars the acceleration due to gravity is 12 ft/sec^2. (On Earth, gravity is much stronger at 32 ft/sec^2.) In the movie, John

insens350 [35]

Solution :

Given initial velocity, v= 48 ft/s

Acceleration due to gravity, g = $12\ ft/s^2$

a). Therefore the maximum height he can jump on Mars is

$H_{max}=\frac{v^2}{2g}$

$H_{max} = \frac{(48)^2}{2 \times 12}$

= 96 ft

b). Time he can stay in the air before hitting the ground is

$T=\frac{2v}{g}$

$T=\frac{2 \times 48}{12}$

= 8 seconds

c). Considering upward motion as positive direction.

v = u + at

We find the time taken to reach the maximum height by taking v = 0.

v = u + at

0 = 16 + (12) t

$t=\frac{16}{12}$

$=\frac{4}{3} \ s$

We know that, $S=ut + \frac{1}{2}at^2$

Taking t = $=\frac{4}{3} \ s$ , we get

$S=16 \times\frac{4}{3} + \frac{1}{2}\times(-12) \times \left(\frac{4}{3}\right)^2$

$S=\frac{32}{3}$ feet

Thus he can't reach to 100 ft as it is shown in the movie.

d). For any jump whose final landing position will be same of the take off level, the final velocity will be the initial velocity.

Therefore final velocity is = -16 ft/s

3 0

2 years ago

Which of the following is a statistical question

zmey [24]

Answer:

its d because who askes that question

4 0

2 years ago

382.993 to the nearest hundredth​

382.993 to the nearest hundredth