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Blizzard [7]
3 years ago
7

Which equation demonstrates the additive identity property?

Mathematics
2 answers:
xxMikexx [17]3 years ago
7 0

Answer:

See Explanation

Step-by-step explanation:

<em>The options are not given; however, you can take a clue from my explanation to answer your question</em>

Let x be a real number;

Additive identity property implies that; adding x to 0 or 0 to x gives x;

In other words;

x + 0 = x

0 + x = x

Note that x can be replaced with any real number; Take for instance

1 + 0 = 1

0 + 2.5 = 2.5

3 + 0 = 3

There are uncountable number of examples;

<em>However, take note that adding 0 to a given digit results in the exact digit and that's the implication of addition identity property</em>

nalin [4]3 years ago
5 0

Answer:

(7+4i)+0=7+4i

Step-by-step explanation:

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Find the value of −12 + 8 −(−9). Question 20 options: 5 −13 −5 13
larisa86 [58]

Answer:

5

-12 + 8 = -4

-4 - (-9) = -4 + 9 = 5

Step-by-step explanation:

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3 years ago
What are the roots of the polynomial equation
rosijanka [135]

Answer:

-4 or -3

Step-by-step explanation:

Calculate the determinant:

D = b^2 - 4ac = 49 - 48 = 1

Apply the formula:

x = (-b +- sqrt(D))/2a = (-7 +- 1)/2 = -4 or -3

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2 years ago
On a number line, the coordinates of j, k, l, m, n are −10.5, −4, −1, 1.5, 4 respectively. find the lengths of the two segments.
nordsb [41]

Answer:

  • JK = 6.5
  • KM = 5.5
  • not congruent

Step-by-step explanation:

JK = K-J = (-4) -(-10.5) = 6.5

KM = M -K = 1.5 -(-4) = 5.5

6.5 ≠ 5.5, so the segments are not congruent.

6 0
3 years ago
Please help with 1 , and 2 I really don’t understand this!
elena-s [515]
1. B
2. A

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6 0
3 years ago
Read 2 more answers
Use matrices and elementary row to solve the following system:
LiRa [457]

I assume the first equation is supposed to be

5x-3y+2z=13

and not

5x-3x+2x=4x=13

As an augmented matrix, this system is given by

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\4&-2&4&12\end{array}\right]

Multiply through row 3 by 1/2:

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\2&-1&2&6\end{array}\right]

Add -1(row 2) to row 3:

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&5&5\end{array}\right]

Multiply through row 3 by 1/5:

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&1&1\end{array}\right]

Add -2(row 3) to row 1, and add 3(row 3) to row 2:

\left[\begin{array}{ccc|c}5&-3&0&11\\2&-1&0&4\\0&0&1&1\end{array}\right]

Add -3(row 2) to row 1:

\left[\begin{array}{ccc|c}-1&0&0&-1\\2&-1&0&4\\0&0&1&1\end{array}\right]

Multiply through row 1 by -1:

\left[\begin{array}{ccc|c}1&0&0&1\\2&-1&0&4\\0&0&1&1\end{array}\right]

Add -2(row 1) to row 2:

\left[\begin{array}{ccc|c}1&0&0&1\\0&-1&0&2\\0&0&1&1\end{array}\right]

Multipy through row 2 by -1:

\left[\begin{array}{ccc|c}1&0&0&1\\0&1&0&-2\\0&0&1&1\end{array}\right]

The solution to the system is then

\boxed{x=1,y=-2,z=1}

5 0
3 years ago
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