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2 years ago

Why can you multiply unlike terms but not add them?

1 answer:

4 0

When you’re adding unlike terms think of the term as a last name. You could never add two last names because that wouldn’t make sense. But with multiplying you don’t need to worry about the “last name”.

You might be interested in

Sandeep used 8 litres of fuel to travel 76 miles.

OleMash [197]

Answer:

I got, when calculated, 209 miles. (It may be wrong)

Step-by-step explanation:

First, I calculated how many miles can Sandeep travel with 1L, therefore I divided 76 miles by 8L and got 9.5 miles on 1L.

Then I multiplied 9.2 miles by 22L and got 209 miles.

This may be wrong but if anyone else is able to work it out much appreciated if it's correct.. Your most welcome!

4 0

2 years ago

A particle travels20m in 4 seconds.what is its speed?

padilas [110]

5m/s

Divide it by the unit. It’s basic simply if

5 0

2 years ago

Tell whether the graph of the equation is a horizontal or a vertical line. Explain your choice.

Orlov [11]

Answer:

horizontal line

Step-by-step explanation:

The equation of a horizontal line parallel to the x- axis is

y = c

where c is the value of the y- coordinates the line passes through.

The equation of a vertical line parallel to the y- axis is

x = c

where c is the value of the x- coordinates the line passes through.

Hence

y = - 1 is a horizontal line passing through all points with a y- coordinate - 1

4 0

2 years ago

I NEED MORE HELP CAN"T FIGURE OUT

Fittoniya [83]

Answer:

y≤5

Step-by-step explanation:

3 ≥ y - 2

Add 2 to both sides

3 (+2) ≥ y - 2 (+2)

5 ≥ y

y ≤ 5

Hope this helps!

7 0

3 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