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OLEGan [10]
3 years ago
14

What expressions are equivalent to -3*4/-5

Mathematics
1 answer:
lara31 [8.8K]3 years ago
6 0

Answer: Equivalent expressions have the same value for every possible value of

the variables they include.  5*(x-4) and 5*x - 20 are equivalent

expressions because no matter what the value of x, they are equal in  

value.

Equivalent equations are ones such that the truth of one implies and

is implied by the truth of the other.  (y-2)^2 = 0 and y = 2 are

equivalent equations, even though the expressions y and (y-2)^2 are

not equivalent expressions.

Step-by-step explanation:

You might be interested in
f(X) = 9x^3 + 2x^2 - 5x^3 - 7x + 4 and g(X) = 5x^3 - 7x + 4. what is f(X) - g(X)? Show all of your steps and write your final an
dolphi86 [110]
First we can simplify f(x) to make it a little easier. We can write it as
4x^3+2x^2-7x+4 by combining like terms

The we just subtract g(x) from f(x) and simplify
4x^3+2x^2-7x+4-(5x^3-7x+4) and remember to distribute the negative to g(x)
4x^3+2x^2-7x+4-5x^3+7x-4 (combine like terms)
-x^3+2x^2
In factored form, this becomes
-x^2(x-2)

Hope this helps
5 0
3 years ago
Read 2 more answers
What is the distance between the coordinates (8,-5) and (8,13).
Darina [25.2K]

Answer:

18

Step-by-step explanation:

use the distance forumla to detwemine the distance between the points.

7 0
3 years ago
What is the expression 6π -3(y+2)
romanna [79]
Y=4


Don’t judge if I’m wrong

Simplifies it would be 18.34-6y-6
7 0
3 years ago
The height H of an ball that is thrown straight upward from an initial position 3 feet off the ground with initial velocity of 9
mariarad [96]

Answer:

The ball will be 84 feet above the ground 1.125 seconds and 4.5 seconds after launch.

Step-by-step explanation:

Statement is incorrect. Correct form is presented below:

<em>The height </em>h(t)<em> of an ball that is thrown straight upward from an initial position 3 feet off the ground with initial velocity of 90 feet per second is given by equation </em>h(t) = 3 +90\cdot t -16\cdot t^{2}<em>, where </em>t<em> is time in seconds. After how many seconds will the ball be 84 feet above the ground. </em>

We equalize the kinematic formula to 84 feet and solve the resulting second-order polynomial by Quadratic Formula to determine the instants associated with such height:

3+90\cdot t -16\cdot t^{2} = 84

16\cdot t^{2}-90\cdot t +81 = 0 (1)

By Quadratic Formula:

t_{1,2} = \frac{90\pm \sqrt{(-90)^{2}-4\cdot (16)\cdot (81)}}{2\cdot (16)}

t_{1} = 4.5\,s, t_{2} = 1.125\,s

The ball will be 84 feet above the ground 1.125 seconds and 4.5 seconds after launch.

3 0
3 years ago
Help!!!!!?!!!!!!!?!!?!!!!
Andreas93 [3]

Answer:

yellow

Step-by-step explanation:

7 0
4 years ago
Read 2 more answers
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