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3 years ago

Write as a single fraction in it's simplest form:

Mathematics

1 answer:

e-lub [12.9K]3 years ago

5 0

Answer:

-8/3x

Step-by-step explanation:

-8 / 3x / 1

-8 / 3x

.....

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Prove algebraically that (m + 2)2 – m 2 – 12 is always a multiple of 4

Julli [10]

Answer:

$(m + 2)^{2} - m^{2} - 12 = 4(m - 2)$

Step-by-step explanation:

Step 1:

Write the expression

$(m+2)^{2} - m^{2} - 12$

Step 2: Expand $(m + 2)^{2}$

$(m+2)^{2} - m^{2} - 12\\(m+2)(m+2) - m^{2} - 12\\m^{2} + 2m + 2m + 4 - m^{2} - 12$

Step 3: Collect similar terms

$m^{2} - m^{2} + 4m + 4 - 12\\4m - 8$

Step 4: Factor 4 out of the expression to prove that the expression is a multiple of 4.

$Therefore\\4m - 8 = 4(m - 2)\\Hence,\\(m+2)^{2} - m^{2} - 12 is a multiply of 4 because the expression is equal to 4(m-2)$

3 0

3 years ago

Cary claimed that the expression − 5 + m − 5 + m is negative. Determine whether Cary's claim is always true, sometimes true, or

atroni [7]

Answer:

The given Expression is sometimes true.

Step-by-step explanation:

from the question,

The given expression by Cary is $-5 + m -5+ m$ .

So, solving the expression we get

⇒ $-10+2m$

Let $m$ is an integer.

Whenever,

(a) $m < 5$ { $m = 4,3,2,1,0,-1,-2.......................$ }

The given Expression always gives negative value.

(b) $m\geq 5$ { $m = 5,6,7,8,9.......................$ }

The given Expression always gives positive value.

Here, we get that the given Expression is not providing the Negative value always, that's why it is not always true and can't be never true.

Hence,

According to Cary's claimantion the given Expression is sometimes true.

5 0

3 years ago

5 + 3x – 2x = 31 – 6 – 5<br> What is the answer?

san4es73 [151]

Answer:

x=15

Step-by-step explanation:

i just know its right

3 0

4 years ago

Find the inverse of f informally. verify that f(f^-1(x))=x and f^-1(f(x))=x<br><br> f(x)= (x-1)/5

Lera25 [3.4K]

Find inverse function.
$f(x)= \frac{x-1}{5} 
\\y= \frac{x-1}{5} 
\\x= \frac{y-1}{5} 
\\y-1=5x
\\y=5x+1
\\f^{-1}(x)=5x+1
\\
\\f(x)= \frac{x-1}{5} 
\\f^{-1}(x)=5x+1 \\ \\f(f^{-1}(x))=f(5x+1)= \frac{(5x+1)-1}{5} = \frac{5x}{5}= x \\ \\ f^{-1}(f(x))=f^{-1}(\frac{x-1}{5} )=5\frac{x-1}{5}+1=x-1+1=x$

8 0

3 years ago

How do you solve a system of equations approximately using tables, without using graphs or equations ? Please I need to figure o

Phantasy [73]

Answer:

Ok, a system of equations means that we have a given number of equations with the same solutions.

If we only have tables, this means that we need to have one table for each equation:

For example, if we are working only with two variables, x and y, in those tables we can see the pints (x, y) that belong to each equation.

Now, a point (x, y) will be a solution of the system of equations only if it belongs to the data table for each equation

This would mean that if we graph those data sets, the graphs will intersect at the point (x, y) that belongs to all the tables of data.

Other way may be using the data in the tables to construct the equations, but you said that you only want to use the tables, so this method can be discarded.

4 0

3 years ago