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julia-pushkina [17]

3 years ago

5

Will 4 - 6 be a negative or positive answer? Negative Positive

1 answer:

Rufina [12.5K]3 years ago

7 0

Answer:

it will be negative

Step-by-step explanation:

it will be -2 because 6 cannot be taken out of 4, when looking on a number line if you move your point six points to the left (because it is minus six) of 4 you will end up at negative two. Any equation that is subtracting a number bigger than itself will be negative.

Hope this helps :)

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What is x² + 7x - 30

Aleonysh [2.5K]

Are you looking for the value of x?

We can complete the square/factor:

x^2+7x-30=0

(x+10)(x-3)=0

By the zero product property rule, we have two equations, x+10=0 and x-3=0.

So the zeroes of x are -10 and 3

Hope this helped. Comment below if this didn't make sense.

6 0

4 years ago

What is the equation of the line that passes through the point (-2, 1) and has a<br> slope of – ?

KengaRu [80]

Accurate Explanation: Aloha! Basically, as I can see here, you're missing a slope. But you need at least 1 point and a slope in order to write an equation. Hope it helps.

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3 years ago

Glven: 2x+3y= 6.

ruslelena [56]

Answer:

The answer is B 0,2

Step-by-step explanation:

I hope that correct

7 0

3 years ago

cora arrived at school for 1 hr for doctores appointment she was dismissed from school at 3 15 pm how long was cora at school

Ostrovityanka [42]

Answer:

7 hours

Step-by-step explanation:

5 0

3 years ago

Perform the indicated row operations, then write the new matrix.

Studentka2010 [4]

The matrix is not properly formatted.

However, I'm able to rearrange the question as:

$\left[\begin{array}{ccc}1&1&1|-1\\-2&3&5|3\\3&2&4|1\end{array}\right]$

Operations:

$2R_1 + R_2 ->R_2$

$-3R_1 +R_3 ->R_3$

Please note that the above may not reflect the original question. However, you should be able to implement my steps in your question.

Answer:

$\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]$

Step-by-step explanation:

The first operation:

$2R_1 + R_2 ->R_2$

This means that the new second row (R2) is derived by:

Multiplying the first row (R1) by 2; add this to the second row

The row 1 elements are:

$\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]$

Multiply by 2

$2 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}2&2&2|-2\end{array}\right]$

Add to row 2 elements are: $\left[\begin{array}{ccc}-2&3&5|3\end{array}\right]$

$\left[\begin{array}{ccc}2&2&2|-2\end{array}\right] + \left[\begin{array}{ccc}-2&3&5|3\end{array}\right]$

$\left[\begin{array}{ccc}0&5&7|1\end{array}\right]$

The second operation:

$-3R_1 +R_3 ->R_3$

This means that the new third row (R3) is derived by:

Multiplying the first row (R1) by -3; add this to the third row

The row 1 elements are:

$\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]$

Multiply by -3

$-3 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right]$

Add to row 2 elements are: $\left[\begin{array}{ccc}3&2&4|1\end{array}\right]$

$\left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right] + \left[\begin{array}{ccc}3&2&4|1\end{array}\right]$

$\left[\begin{array}{ccc}0&-1&1|4\end{array}\right]$

Hence, the new matrix is:

$\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]$

3 0

3 years ago

Read 2 more answers