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3 years ago

Which of the following number sentences is an example of the associative property of multiplication?

Mathematics

2 answers:

koban [17]3 years ago

4 0

20 × (10 + 6) = (20 × 10) + (20 × 6)

(12 × 3) × 8 = 12 (3 × 8)

sertanlavr [38]3 years ago

3 0

Answer:

Option C and D are examples of associative property of multiplication.

Option A is examples of commutative property of multiplication.

Option B is normal multiplication.

Hope this helps!

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PLSS HELP write the precent as a fraction or mixed number in simplest form (reduce) 36%

Ivanshal [37]

Answer:

9/25

Step-by-step explanation:

36% is equal to 36/100

You can simplify it

36/2 = 18

100/2=50

18/2=9

50/2=25

That is as much as you can simplify it so 36% is equal to 9/25.

5 0

3 years ago

Read 2 more answers

Simplify (1234567)^0

Daniel [21]

Answer:

$every \: number \: by \: power \: 0 \: is \: equal1 \\ {x}^{0} = 1$

4 0

3 years ago

Find the slope and the y-intercept of the graph of the linear equation. y=3−12x

pychu [463]

Slope is -12 and y-intercept is 3

3 0

3 years ago

L :V --> W is a linear transformation. Prove each of the following (a) ker L is a subspace of V. (b) range L is a subspace of

iragen [17]

Answer:

a) Assume that $x,y\in\ker L$ , and $\alpha$ is a scalar (a real or complex number).

First. Let us prove that $\ker L$ is not empty. This is easy because $L(0_V)=0_W$ , by linearity. Here, $0_V$ stands for the zero vector of V, and $0_W$ stands for the zero vector of W.

Second. Let us prove that $\alpha x\in\ker L$ . By linearity

$L(\alpha x) = \alpha L(x)=\alpha 0_W=0_W$ .

Then, $\alpha x\in\ker L$ .

Third. Let us prove that $y+ x\in\ker L$ . Again, by linearity

$L(x+y)=L(x)+L(y) = 0_W + 0_W=0_W$ .

And the statement readily follows.

b) Assume that $u$ and $v$ are in range of $L$ . Then, there exist $x,y\in V$ such that $L(x)=u$ and $L(y)=v$ .

First. Let us prove that range of $L$ is not empty. This is easy because $L(0_V)=0_W$ , by linearity.

Second. Let us prove that $\alpha u$ is on the range of $L$ .

$\alpha u = \alpha L(x) = L(\alpha x) = L(z)$ .

Then, there exist an element $z\in V$ such that $L(z)=\alpha u$ . Thus $\alpha u$ is in the range of $L$ .

Third. Let us prove that $u+v$ is in the range of $L$ .

$u+v = L(x)+L(y) = L(x+y)=L(z)$ .

Then, there exist an element $z\in V$ such that $L(z)= u +v$ . Thus $u +v$ is in the range of $L$ .

Notice that in this second part of the problem we used the linearity in the reverse order, compared with the first part of the exercise.

Step-by-step explanation:

6 0

3 years ago

-16 = m + 71 Please help

Andre45 [30]

-16 = m + 71

Move +71 to the other side. Sign changes from +71 to -71. To get m by itself.

-16-71=m+71-71

-16-71=m

-87=m or m=-87

Answer: -87=m or m=-87

6 0

3 years ago