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lorasvet [3.4K]
3 years ago
11

What is the value of x?

Mathematics
2 answers:
EastWind [94]3 years ago
8 0

6+16=5x

5x=22

x=22/5

x=4.4

kotykmax [81]3 years ago
7 0

Answer:

The correct answer is 3

Hope this helped

Step-by-step explanation:


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2:45pm to 5:30 pm how long did it take?
Marta_Voda [28]

Answer:

2 hours and 45 minutes. or 165 minutes

Step-by-step explanation:

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3 years ago
A vessel has 4 liters and 500 ml of curd. In how many glasses, each of 25 ml capacity, can it be filled?
dalvyx [7]

Step-by-step explanation:

easy to look up : mili stands for 1/1000

so,

500 ml are 500/1000 liters.

25 ml are 25/1000 liters (a tiny glass).

and 1 Iiter is of course, 1000/1000 liters (1000 ml).

4 liters 500 ml are then

4×1000/1000 + 500/1000 = 4500/1000 l = 4500 ml.

how many glasses of 25 ml can be filled with that ?

this is the same question as into how many parts of 25 can 4500 be split ? or, how often does 25 fit into 4500 ?

all this is done by division :

4500 / 25 = 180

so, 180 of such glasses can be filled.

7 0
2 years ago
Helpppppppppppppo anyone
Alenkasestr [34]

Answer: 50653

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= 64000 - 14400 + 1080 - 27

= 50653

Step-by-step explanation:

3 0
3 years ago
JDIDJRJDJF PLS HELP IM STUCK
Dovator [93]

Step-by-step explanation:

we have

B = J + 2

as the number of laps Bill ran.

to get the number of laps Jill ran we need to transform this equation, so that we get

J = ...

or

... = J

J has to be on one side, and the rest on the other (just as it is with B = ...).

so,

B = J + 2

B - 2 = J

or

J = B - 2

hey ! we are finished already !

6 0
2 years ago
An object is launched from a launching pad 144 ft. above the ground at a velocity of 128ft/sec. what is the maximum height reach
ollegr [7]

Answer:

18) a. h(x) = -16x² + vx + h(0) ⇒ h(x) = -16x² + 128x + 144

b. The maximum height = 400 feet

c. Attached graph

19) The rocket will reach the maximum height after 4 seconds

20) The rocket hits the ground after 9 seconds

Step-by-step explanation:

* Lets study the rule of motion for an object with constant acceleration

# The distance S = ut ± 1/2 at², where u is the initial velocity, t is the time

  and a is the acceleration of gravity

# The vertical distances h in x second is h(x) - h(0), where h(0)

   is the initial height of the object above the ground

∵ h(x) = vx + 1/2 ax², where h is the vrtical distance, v is the initial

  velocity, a is the acceleration of gravity (32 feet/second²) and x

  is the time

18)a.

∵ The value of a = -32 ft/sec² ⇒ negative because the direction

   of the motion

  is upward

∴ h(x) - h(0) = vx - (1/2)(32)(x²) ⇒ (1/2)(32) = 16

∴ h(x) = vx - 16x² + h(0)

∴ h(x) = -16x² + vx + h(0) ⇒ proved

* Find the height of the object after x seconds from the ground

∵ h(0) = 144 and v = 128 ft/sec

∴ h(x) = -16x² + 128x + 144

b.

* At the maximum height h'(x) = 0

∵ h'(x) = -32x + 128

∴ -32x + 128 = 0 ⇒ subtract 128 from both sides

∴ -32x = -128 ⇒ ÷ -32

∴ x = 4 seconds

- The time for the maximum height = 4 seconds

- Substitute this value of x in the equation of h(x)

∴ The maximum height = -16(4)² + 128(4) + 144 = 400 feet

c. Attached graph

19)

- The object will reach the maximum height after 4 seconds

20)

- When the rocket hits the ground h(x) = 0

∵ h(x) = -16x² + 128x + 144

∴ 0 = -16x² + 128x + 144 ⇒ divide the two sides by -16

∴ x² - 8x - 9 = 0 ⇒ use the factorization to find the value of x

∵ x² - 8x - 9 = 0

∴ (x - 9)( x + 1) = 0

∴ x - 9 = 0 OR x + 1 = 0

∴ x = 9 OR x = -1

- We will rejected -1 because there is no -ve value for the time

* The time for the object to hit the ground is 9 seconds

8 0
3 years ago
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