How do you simplify 9*499 using the distributive property?

2 answers:

4 0

1. 9(499) = 4491 = (200 + 299)9

For the basic properties examples are:
Addition, you can have 6 + 107 = 113 
Commutative Property by moving: 107 + 6 = 113 
Associative Property by grouping: (3 + 3) + (100 + 7 ) = 113 
Distributive Property by allotting: 2 (3) + 107 = 113 

Multiplication, you can have 6 x 107 = 642 
Commutative Property by moving: 107 x 6 = 642 
Associative Property by grouping: (3 + 3) x (100 + 7 ) = 642 
Distributive Property by allotting: 2(3) x 107 = 642

Snowcat [4.5K]3 years ago

3 0

Answer:

642 is your sum

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x2 - 10x + _ "find value that completes the square and rewrite as perfect square" what does it mean by "rewrite as perfect squar

abruzzese [7]

A perfect square is something times that same thing. Example (x+2)(x+2)

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8 0

3 years ago

Read 2 more answers

Find the dot product of the position vectors whose terminal points are (14, 9) and (3, 6).

erma4kov [3.2K]

Answer:

Step-by-step explanation:

The formula for the dot product of vectors is

u·v = |u||v|cosθ

where |u| and |v| are the magnitudes (lengths) of the vectors. The formula for that is the same as Pythagorean's Theorem.

$|u|=\sqrt{14^2+9^2}$ which is $\sqrt{277}$

$|v|=\sqrt{3^2+6^2}$ which is $\sqrt{45}$

I am assuming by looking at the above that you can determine where the numbers under the square root signs came from. It's pretty apparent.

We also need the angle, which of course has its own formula.

$cos\theta=\frac{uv}{|u||v|}$ where uv has ITS own formula:

uv = (14 * 3) + (9 * 6) which is taking the numbers in the i positions in the first set of parenthesis and adding their product to the product of the numbers in the j positions.

uv = 96.

To get the denominator, multiply the lengths of the vectors together. Then take the inverse cosine of the whole mess:

$cos^{-1}\theta=\frac{96}{111.64676}$ which returns an angle measure of 30.7. Plugging that all into the dot product formula: