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andrey2020 [161]
3 years ago
12

How do you Solve the proportion 4/n = 4/5

Mathematics
2 answers:
Allisa [31]3 years ago
8 0

Answer:

4/n=4/5

4n = 4×5

4n = 20

n = 20/4

n=5

Sedaia [141]3 years ago
5 0

Answer:n=5

Step-by-step explanation:

4/n=4/5 then cross multiplication

4×n and 4×5

Then you have 4×n=4n and 4×5=20

Then divide 4 everywhere so you could find n

4n/4 and 20/4

N=5

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Multiplying mixed numbers and whole numbers 1 1/2 x 2/1 =
Nezavi [6.7K]

Answer: 3

Step-by-step explanation:

1. Convert the mixed number to fraction:

- Multiply the denominator of the fraction by the whole number.

- Add the product obtained and the numerator of the fraction.

- Write the sum obtained as the numerator and rewrite the original denominator of the fraction.

Then:

1\ 1/2=\frac{(1)(2)+1}{2}=\frac{3}{2}

2. Multiply the numerators.

3. Multiply the denominator.

4. Reduce the fraction.

Then:

(\frac{3}{2})(\frac{2}{1})=\frac{6}{2}=3

5 0
3 years ago
Why will percent of change always be represented by a positive number?
Murljashka [212]
Percent of change will always be positive because the change is an absolute value. As we know, absolute value is always a positive value because you don't pay attention to the direction, only the distance is important. 
3 0
3 years ago
1. The position of a particle moving along a coordinate axis is given by: s(t) = t^2 - 5t + 1. a) Find the speed of the particle
zimovet [89]

Answer: \left |  2t-5\right |,\ 2,\ 2t-5

Step-by-step explanation:

Given

Position of the particle moving along the coordinate axis is given by

s(t)=t^2-5t+1

Speed of the particle is given by

\Rightarrow v=\dfrac{ds}{dt}\\\\\Rightarrow v=\dfrac{d(t^2-5t+1)}{dt}\\\\\Rightarrow v=\left |  2t-5\right |

Acceleration of the particle is

\Rightarrow a=\dfrac{dv}{dt}\\\\\Rightarrow a=2

velocity can be negative, but speed cannot

\Rightarrow v=\dfrac{ds}{dt}\\\\\Rightarrow v=\dfrac{d(t^2-5t+1)}{dt}\\\\\Rightarrow v=2t-5

3 0
3 years ago
What is the least common multiple (LCM) that could be
tensa zangetsu [6.8K]

Answer:

the greatest common factor will be 3

8 0
3 years ago
If f(x) = 2x^3 - 14x^2 + 38x -26 and x-1 is a factor of f(x) find all of the zeros of f(x) algebraically.
Anestetic [448]

Answer:

Step-by-step explanation:

First confirm that x = 1 is one of the zeros.

f(1) = 2(1)^3 - 14(1)^2 + 38(1) - 26

f(1) = 2 - 14 + 38 - 26

f(1) = -12 + 38 = + 26

f(1) = 26 - 26

f(1) = 0

=========================

next perform a long division

x -1  || 2x^3 - 14x^2 + 38x - 26 || 2x^2 - 12x + 26

          2x^3 - 2x^2

          ===========

                    -12x^2 + 28x

                     -12x^2 +12x

                     ==========

                                  26x -26

                                  26x - 26

                                 ========

                                      0

Now you can factor 2x^2 - 12x + 26

                                 2(x^2 - 6x + 13)

The discriminate of the quadratic is negative. (36 - 4*1*13) = - 16

So you are going to get a complex result.

x = -(-6) +/- sqrt(-16)

     =============

                 2

x  = 3 +/- 2i

f(x) = 2*(x - 1)*(x - 3 + 2i)*(x - 3 - 2i)

The zeros are

1

3 +/- 2i

8 0
3 years ago
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