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ASHA 777 [7]
3 years ago
10

910/14=? please solve.

Mathematics
2 answers:
12345 [234]3 years ago
8 0

If you wanted to know what this fraction is simplified, it would be 65.

Nookie1986 [14]3 years ago
3 0

910 \div 14 = 65
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A cube has the dimensions 5/8 inches, 5/8 inches, 5/8 inches. what is the volume of the cube? how many smaller cubes with an edg
REY [17]
<span>- Volume of a cube = length³
                                 = \frac{5}{8} ³
                                 = 0.625³
~ Answer ~              = </span><span>0.244140625</span> ≈ 0.24
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3 years ago
??????????????????????????????????
yanalaym [24]

Answer:

It is b^6

Step-by-step explanation:

6 0
3 years ago
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The equation of a line is y = 2x + 3. What is the equation of the line that is parallel to the first line and passes through (2,
Masja [62]

Answer:

y = 2x - 5

Step-by-step explanation:

The slope for a parallel line is the same as the slope in the given equation.  Therefore the slope of the new line will be 2    m= 2

use that slope and (2, -1) to find b

-1 = 2(2) + b

-1 = 4 + b

-5 = b

Plug m and b into y = mx + b

y = 2x - 5

8 0
2 years ago
The sum of two numbers is 54.The smaller number is 22 less than the larger number.what are the numbers?
lukranit [14]

Answer:

x=16 y=38

Step-by-step explanation:

x+y=54.......equation 1

x=y-22........ equation 2

substitute

y-22+y=54

2y-22=54

2y=54+22

2y=76

y=76/2

y=38''

x=y-22

x=38-22

x=16''

8 0
2 years ago
A) Use the definition of Laplace transform to find L{f }. (Do the integrals.) For what values of s is L{f } defined?f(t) = (2t+1
kiruha [24]

For the given function f(t) = (2t + 1) using definition of Laplace transform the required solution is L(f(t))s = [ ( 2/s²) + ( 1/s) ].

As given in the question,

Given function is equal to :

f(t) = 2t + 1

Simplify the given function using definition of Laplace transform we have,

L(f(t))s = \int\limits^\infty_0 {f(t)e^{-st} } \, dt

          =  \int\limits^\infty_0[2t +1] e^{-st} dt

          = 2\int\limits^\infty_0 te^{-st} + \int\limits^\infty_0e^{-st} dt

         = 2 L(t) + L(1)

L(1) = \int\limits^\infty_0e^{-st} dt

     = (-1/s) ( 0 -1 )

     = 1/s , ( s >  0)

2L ( t ) = 2\int\limits^\infty_0 te^{-st}

        =  2[t\int\limits^\infty_0 e^{-st} - \int\limits^\infty_0 ({(d/dt)(t) \int\limits^\infty_0e^{-st} \, dt )dt]

        =  2/ s²

Now ,

L(f(t))s = 2 L(t) + L(1)

          = 2/ s² + 1/s

Therefore, the solution of the given function using Laplace transform the required solution is L(f(t))s = [ ( 2/s²) + ( 1/s) ].

Learn more about Laplace transform here

brainly.com/question/14487937

#SPJ4

8 0
1 year ago
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