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inn [45]
2 years ago
14

1. ONE AND TWO HUNDREDTHS AS A DÉCIMAl AND FRACTION

Mathematics
1 answer:
astraxan [27]2 years ago
5 0

Answer:

1.02 as a decimal

1 and 1/50 as a fraction

Step-by-step explanation:

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DE =8 CM<br> DE = 4 CM<br> DE = 10 CM<br> DE = 6CM
Svetradugi [14.3K]

Answer:

just add

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
Solve for h.<br><br> –0.62 + 8.35 + 3.9h = –6.55 + 1.1h
stepan [7]

Answer:

- 0.62 + 8.35 + 3.9h =  - 6.55 + 1.1h \\  \\ 3.9h - 1.1h = - 6.55 + 0.62 - 8.35 \\  \\ 2.8h =  - 14.28 \\  \\ h =  \frac{ - 14.28}{2.8}  \\  \\ h =   \frac{ - 142.8}{28}  \\  \\ h =  - 5.1

I hope I helped you^_^

8 0
3 years ago
How far above sea level is the ranger station?
Schach [20]
Since he descended 12 meters, we subtract this from the overall height of Mount Ka'ala, so then we are only calculating how high ABOVE the sea level it is.

1232 - 12 = 1220

The height of Mount Ka'ala is therefore 1,220 meters.
To calculate how much a fifth of Mount Ka'ala is (since the ranger station is 2/5's up), we would divide this number by 5

1220 ÷ 5 = 244

Since ONE fifth of the height is 244 meters, TWO fifths would be double that amount.

244 x 2 = 488

488 meters.
The ranger station is 488m above sea level.
3 0
2 years ago
A mass of 5 kg stretches a spring 10 cm. The mass is acted on by an external force of 10sin( t ) N(newtons) and moves in a mediu
yKpoI14uk [10]

Answer:

A) C1 = 0.00187 m = 0.187 cm,  C2 = 0.0062 m = 0.62 cm

B)  A sample of how the graph looks like is attached below ( periodic sine wave )

C) w = \sqrt[4]{3} is when the amplitude of the forced response is maximum

Step-by-step explanation:

Given data :

mass = 5kg

length of spring = 10 cm = 0.1 m

f(t) = 10sin(t) N

viscous force = 2 N

speed of mass = 4 cm/s = 0.04 m/s

initial velocity = 3 cm/s = 0.03 m/s

Formulating initial value problem

y = viscous force / speed = 2 N / 0.04 m/s = 50 N sec/m

spring constant = mg/ Length of spring = (5 * 9.8) / 0.1 = 490 N/m

f(t) = 10sin(t/2) N

using the initial conditions of u(0) = 0 m and u"(0) = 0.03 m/s to express the equation of motion

the equation of motion = 5u" + 50u' + 490u = 10sin(t/2)

A) finding the solution of the initial value

attached below is the solution and

B) attached is a periodic sine wave replica of how the grapgh of the steady state solution looks like

C attached below

3 0
3 years ago
A car travelling at a constant speed travels 90km in 50 min .How far travelling at the same constant speed will the car travel i
vovikov84 [41]

Answer:

261 km

Step-by-step explanation:

given,s=constant

d1/T1=d2/T2

d1=90km,t1=50 min

d2=? ,t2=(120+25)min

=145min

therefore,d2=(90×145)/50

=261 km

6 0
3 years ago
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