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katrin2010 [14]
2 years ago
15

Please help it’s easy, State the property that is used. 10=10•1

Mathematics
1 answer:
Molodets [167]2 years ago
5 0

The property identity of multiplication is applied in the given statement.

<h3>Properties of Multiplication</h3>

The properties of multiplication are:

  • Distributive:  a(b±c)=  ab±ac
  • Comutative:   a . b = b. a
  • Associative:    a(b+c)=  c(a+b)
  • Identity: b.1=b

From the property identity, you know that the product between any number by the number 1 equals that number. Example: 3 • 1=3.

The question shows 10=10•1. Like, it was shown previously, this occurs due to the identity property of multiplication.

Read more about the identity property here:

brainly.com/question/23977324

#SPJ1

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14. If the domain of f(x)= x2 +1 is limited to {0,1,2,3), what is the maximum value of the range?
fiasKO [112]

the domain of f(x)= x2 +1 is limited to {0,1,2,3)

the maximum value will be at the maximum value of x which is at x = 3

So, the maximum value of the range will be = 3^2 + 1 = 9 + 1 = 10

7 0
1 year ago
What is bigger 24 fl. Oz. Or 8 cups?
Rudik [331]

Answer:

8 Cups.

Step-by-step explanation:

8 Cups is 64 Fl Oz, while 24 Fl Oz is only 3 cups.

3 0
3 years ago
Read 2 more answers
Can somebody please help me
guajiro [1.7K]

Here is your answer

B. (12,30)

REASON:

Given,

y= 2x+6 .......(i)

3x-y= 6 .......(ii)

Putting the value of y from eq.i in eq.ii we get

3x-(2x+6)=6

3x-2x-6=6

=6+6

x=12

Putting x=12 in eq.i we get

y= 2x+6

y= 2×12+6

y= 24+6

y= 30

So,

\bold{x=12 and y= 30}

Hence answer is (12,30) in which x=12 and y=30

HOPE IT IS USEFUL

5 0
3 years ago
G(-4)=3(-4)+4<br><br> Find g(-4)
soldier1979 [14.2K]
G(x) = 3x + 4
g(-4) = 3(-4) + 4
g(-4) = -12 + 4
g(-4) = -8
6 0
4 years ago
Hard math question!
luda_lava [24]
First, let me show you some notation.

To show a matrix is an inverse of another matrix, we write A^{-1}

-1 is not an exponent. It just shows that a matrix is an inverse of another matrix.

For a 2x2 matrix, we can get the inverse by first making b and c negatives and swap the positions of a and d.

Then multiply each entry in the matrix by 1 divided by the determinant.

\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]^{-1} = &#10;  \frac{1}{ad - bc}\left[\begin{array}{ccc}d&{-b}\\{-c}&a\end{array}\right] =  \\  \\ \\ \left[\begin{array}{ccc}d(\frac{1}{ad-bc})&{-b}(\frac{1}{ad-bc}) \\ {-c}(\frac{1}{ad-bc}) &a(\frac{1}{ad-bc}) \end{array}\right]

I hope this helped!
8 0
3 years ago
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