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geniusboy [140]
3 years ago
15

Vf²=vi²+ 2ad A) solve for vi B) solve for d Using the same equation

Mathematics
1 answer:
diamong [38]3 years ago
7 0
A)
               Vf² = vi² + 2ad   Subtract 2ad from both sides
     Vf² - 2ad = vi²              Find the square roots of both sides
√(Vf² - 2ad) = √(vi²)         Cancel out the squares with the square roots
  Vf - √(2ad) = vi               Switch the sides to make it easier to read
                 vi = Vf - √(2ad)

B)                 Vf² = vi² + 2ad   Subtract vi² from both sides
             Vf² - vi² = 2ad           Divide both sides by 2a
    (Vf² - vi²) / 2a = d               Switch the sdies to make it easier to read
                         d = (Vf² - vi²) / 2a
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1/6x =12
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1x/6=12
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x/6=12
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3 0
2 years ago
Which expression is equivalent to One-fifth (150 x minus 80 y + 50 minus 50 x minus 25 y + 20)?
inna [77]

Answer:

The answer to your question is  20x + 21y + 14        

Step-by-step explanation:

Data

Expression             1/5 (150x - 80y + 50 - 50x - 25y + 20)

Process

1.- Multiply 1/5 by each term

                                     150/5x - 80/5y + 50/5 - 50/5x - 25/5y + 20/5

2.- Simplify

                                      30x - 16y + 10 - 10x - 5y + 4

3.- Simplify like terms

                                      (30x - 10x) + (-16y - 5y) + (10 + 4)

4.- Result

                                        20x + 21y + 14                                        

3 0
3 years ago
Read 2 more answers
HELPme
aalyn [17]

Answer:

1. Rectangle

2. 376.8 cm

3. 379.94 square feet

4. –4 and 4

Step-by-step explanation:

A square is always a rectangle.

If a wheel has a radius of 15 cm, it would travel approximately 376.8 cm per 4 revolutions.

If the diameter of a circular garden is 22 feet, the approximate area of the garden is 379.94 square feet.

The solutions to the equation y2 – 1 = 15 is –4 and 4.

3 0
3 years ago
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Describe the cross section, please help
levacccp [35]
A cross section is the shape we get when cutting straight through an object. The cross section of this object is a triangle. It is like a view into the inside of something made by cutting through it.
6 0
3 years ago
Write a polynomial function, p(x) with degree 3 that has p(7)=0
MArishka [77]

Answer:

p (x) = x^{3} - 21x^{2}+ 147x - 343

is the required polynomial with degree 3 and p ( 7 ) = 0

Step-by-step explanation:

Given:

p ( 7 ) = 0

To Find:

p ( x ) = ?

Solution:

Given p ( 7 ) = 0 that means substituting 7 in the polynomial function will get the value of the polynomial as 0.

Therefore zero's of the polynomial is seven i.e 7

Degree : Highest raise to power in the polynomial is the degree of the polynomial

We have the identity,

(a -b)^{3} = a^{3}-3a^{2}b +3ab^{2} - b^{3}

Take a = x

        b = 7

Substitute in the identity we get

(x -7)^{3} = x^{3}-3x^{2}(7) +3x(7)^{2} - 7^{3}\\(x -7)^{3} = x^{3}-21x^{2} +147x - 343

Which is the required Polynomial function in degree 3 and if we substitute 7 in the polynomial function will get the value of the polynomial function zero.

p ( 7 ) = 7³ - 21×7² + 147×7 - 7³

p ( 7 ) = 0

p (x) = x^{3} - 21x^{2}+ 147x - 343

4 0
3 years ago
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