All categories

3 years ago

Simplify each expression 3(m+4)-5m

Mathematics

2 answers:

mash [69]3 years ago

3 0

Answer:

You can start by distributing the 3 to the m+4

(3m + 12)

Now you can add on the -5m and simplify

(3m+12-5m)

-2m+12

Shalnov [3]3 years ago

3 0

3m+12=5m
-3m. -3m
——————-
12. = 2m.

*if you need to divide it’s gonna me 6

You might be interested in

What is the range of the function f(x) = 4x + 9, given the domain D = {-4, -2, 0, 2}?

anyanavicka [17]

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

$f( - 4) = 4( - 4) + 9 = - 7$

_________________________________

$f( - 2) = 4( - 2) + 9 = 1$

_________________________________

$f(0) = 4(0) + 9 = 9$

_________________________________

$f(2) = 4(2) + 9 = 17$

_________________________________

Thus ;

R = { - 7 , 1 , 9 , 17 }

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

The correct answer is (( D )) .

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

3 0

2 years ago

Least common multiple of 2,7, and 13

san4es73 [151]

To work out the LCM, we need their prime factors. So:

2 is simple 2 x 1.

7 is also 7 x 1.

13 is also 13 and 1.

They all have in common "1" - so we keep that aside as 1 figure, rather than 3.

We still have the 2, 7 and 13 leftover.

We then multiply the 1 with the other figures.

So 1 x 2 x 7 x 13 is our answer.

Which is = 182.

Hope I helped:)

~FabSafxo

6 0

3 years ago

Read 2 more answers

Compare the motion of the boy and the man. Who was going faster? Was their acceleration about the same? Who had more frictional

aivan3 [116]

Answer:

can you make the question clearer old?

5 0

2 years ago

Challenge You have 2 different savings accounts. For Account A, the simple interest earned after months is $. For Account B,

Debora [2.8K]

Answer and Step-by-step explanation:

The computation is shown below;

As we know that

The simple interest is

Simple interest = Principal × rate × time

For Account A, the principal would be

2 = Principal × 3.2% × 3

So, the principal is $20.83

For account B, the principal would be

38.50 = Principal × 2.2% × 30

So, the principal is $58.33

Now the interest for both the accounts are as follows

For account 1

= $20.83 × 3.2%

= $0.67

And, for the account 2

= $58.33 × 2.2%

= $1.28

As it can be seen that the account 2 has the highest interest

The same is to be considered

7 0

2 years ago

Chose parameters h and k such that the system has a) a unique solution, b) many solutions, and c) no solution. X1 + 3x2 = 4 2x1

AysviL [449]

Answer:

a) The system has a unique solution for $k\neq 6$ and any value of $h$ , and we say the system is consisted

b) The system has infinite solutions for $k=6$ and $h=8$

c) The system has no solution for $k=6$ and $h\neq 8$

Step-by-step explanation:

Since we need to base the solutions of the system on one of the independent terms ( $h$ ), the determinant method is not suitable and therefore we use the Gauss elimination method.

The first step is to write our system in the augmented matrix form:

$\left[\begin{array}{cc|c}1&3&4\\2&k&h\end{array}\right]$

The we can use the transformation $r_0\rightarrow r_0 -2r_1$ , obtaining:

$\left[\begin{array}{cc|c}1&3&4\\0&k-6&h-8\end{array}\right]$ .

Now we can start the analysis:

If $k\neq 6$ then, the system has a unique solution for any value of $k$ , meaning that the last row will transform back to the equation as:

$(k-6)x_2=h-8\\x_2=h-8/(k-6)$

from where we can see that only in the case of $k=6$ the value of $x_2$ can not be determined.

if $k=6$ and $h=8$ the system has infinite solutions: this is very simple to see by substituting these values in the equation resulting from the last row:

$(k-6)x_2=h-8\\0=0$ which means that the second equation is a linear combination of the first one. Therefore, we can solve the first equation to get $x_1$ as a function of $x_2$ o viceversa. Thus, $x_2$ ( $x_1$ ) is called a parameter since there are no constraints on what values they can take on.

if $k=6$ and $h\neq 8$ the system has no solution. Again by substituting in the equation resulting from the last row:

$(k-6)x_2=h-8\\0=h-8$ which is false for all values of $h\neq 8$ and since we have something that is not possible $(0\neq h-8,\ \forall \ h\neq 8)$ the system has no solution

6 0

3 years ago