All categories

3 years ago

Weight is simply the force of gravity acting on an object. TRUE or FALSE

2 answers:

6 0

The correct answer for the given statement above would be TRUE. It is true that weight is simply the force of gravity, acting on an object. S<span>ince weight is a force, the equation formulated for it is F = mg, where f is weight in Newtons, m is for mass, and g is for gravitational acceleration. Hope this is the answer that you are looking for.</span>

Digiron [165]3 years ago

3 0

The answer is true. .........

You might be interested in

Seven more than the cube of a number is what

zloy xaker [14]

7+x^3. It cannot be simplified further without knowing the value of x

7 0

3 years ago

Factorise using the difference of two squares:<br> I)4x squared-(2x+3) squared

alex41 [277]

4x² - (2x + 3)²
4x² - (2x + 3)(2x + 3)
4x² - (2x(2x + 3) + 3(2x + 3))
4x² - (2x(2x) + 2x(3) + 3(2x) + 3(3))
4x² - (4x² + 6x + 6x + 9)
4x² - (4x² + 12x + 9)
4x² - 4x² - 12x - 9
-12x - 9
-3(4x) - 3(3)
-3(4x + 3)

8 0

3 years ago

Rewrite the expression as a single exponent

kifflom [539]

Answer:

9^7

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Which statement best reflects the solution(s) of the equation?

Inessa [10]

x=2 is only solution while x=1 is extraneous solution

Option C is correct.

Step-by-step explanation:

We need to solve the equation $\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}$ and find values of x.

Solving:

Find the LCM of denominators x-1,x and x-1. The LCM is x(x-1)

Multiply the entire equation with x(x-1)

$\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}\\\frac{1}{x-1}*x(x-1)+\frac{2}{x}*x(x-1)=\frac{x}{x-1}*x(x-1)\\Cancelling\,\,out\,\,the\,\,same\,\,terms:\\x+2(x-1)=x^2\\x+2x-2=x^2\\3x-2=x^2\\x^2-3x+2=0$

Now, factoring the term:

$x^2-2x-x+2=0\\x(x-2)-1(x-2)=0\\(x-1)(x-2)=0\\x-1=0\,\,and\,\, x-2=0\\x=1\,\,and\,\, x=2$

The values of x are x=1 and x=2

Checking for extraneous roots:

Extraneous roots: The root that is the solution of the equation but when we put it in the equation the answer turns out not to be right.

If we put x=1 in the equation, $\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}$ the denominator becomes zero i.e

$\frac{1}{1-1}+\frac{2}{1}=\frac{1}{1-1}\\\frac{1}{0}+2=\frac{1}{0}$

which is not correct as in fraction anything divided by zero is undefined. So, x=1 is an extraneous solution.

If we put x=2 in the equation,

$\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}$

$\frac{1}{2-1}+\frac{2}{2}=\frac{2}{2-1}\\\frac{1}{1}+1=\frac{2}{1}\\1+1=2\\2=2$

So, x=2 is only solution while x=1 is extraneous solution

Option C is correct.

Keywords: Solving Equations and checking extraneous solution

Learn more about Solving Equations and checking extraneous solution at:

#learnwithBrainly

5 0

3 years ago

2. Into what three zones is India divided and how are these zones different from each other?

LenaWriter [7]

Answer:

Into what three zones is India divided and how are these zones different from each other? Deccan Plateau: triangular shaped, juts into the Indian Ocean, is dry, unproductive and sparsely populated.

Step-by-step explanation:

6 0

3 years ago