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Yuri [45]
3 years ago
9

In the multiplication sentence below, which numbers are the factors? Check

Mathematics
1 answer:
IRISSAK [1]3 years ago
3 0

Answer:

The factors are 7 and 3

Step-by-step explanation:

The factors of a multiplication sentence are the numbers that are being multiplied for the product (or answer).

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Express 80 as the product of its prime factors.<br> Write the prime factors in ascending order.
stepan [7]

Answer:

the answer of that is 16 x 5

Step-by-step explanation:

7 0
1 year ago
Use complete sentences to describe why √-1 ≠ -√1
tekilochka [14]

Well let's say that to compare these two numbers, we have to start with the definition first.

<u>D</u><u>e</u><u>f</u><u>i</u><u>n</u><u>i</u><u>t</u><u>i</u><u>o</u><u>n</u>

\displaystyle \large{ {y}^{2}  = x} \\  \displaystyle \large{ y =  \pm  \sqrt{x} }

Looks like we can use any x-values right? Nope.

The value of x only applies to any positive real numbers for one reason.

As we know, any numbers time itself will result in positive. No matter the negative or positive.

<u>D</u><u>e</u><u>f</u><u>i</u><u>n</u><u>i</u><u>t</u><u>i</u><u>o</u><u>n</u><u> </u><u>I</u><u>I</u>

\displaystyle \large{  {a}^{2}  = a \times a =  |b| }

Where b is the result from a×a. Let's see an example.

<u>E</u><u>x</u><u>a</u><u>m</u><u>p</u><u>l</u><u>e</u><u>s</u>

\displaystyle \large{  {2}^{2}  = 2 \times 2 = 4} \\  \displaystyle \large{  {( - 2)}^{2}  = ( - 2) \times ( - 2) =  | - 4|  = 4}

So basically, their counterpart or opposite still gives same value.

Then you may have a question, where does √-1 come from?

It comes from this equation:

\displaystyle \large{   {y}^{2}  =  - 1}

When we solve the quadratic equation in this like form, we square both sides to get rid of the square.

\displaystyle \large{   \sqrt{ {y}^{2} } =   \sqrt{ - 1}  }

Then where does plus-minus come from? It comes from one of Absolute Value propety.

<u>A</u><u>b</u><u>s</u><u>o</u><u>l</u><u>u</u><u>t</u><u>e</u><u> </u><u>V</u><u>a</u><u>l</u><u>u</u><u>e</u><u> </u><u>P</u><u>r</u><u>o</u><u>p</u><u>e</u><u>r</u><u>t</u><u>y</u><u> </u><u>I</u>

\displaystyle \large{  \sqrt{ {x}^{2}  } =  |x|  }

Solving absolute value always gives the plus-minus. Therefore...

\displaystyle \large{  y =   \pm \sqrt{ - 1}  }

Then we have the square root of -1 in negative and positive. But something is not right.

As I said, any numbers time itself of numbers squared will only result in positive. So how does the equation of y^2 = -1 make sense? Simple, it doesn't.

Because why would any numbers squared result in negative? Therefore, √-1 does not exist in a real number system.

Then we have another number which is -√1. This one is simple.

It is one of the solution from the equation y^2 = 1.

\displaystyle \large{   {y}^{2}  = 1} \\  \displaystyle \large{    \sqrt{ {y}^{2} }  =  \sqrt{1} } \\  \displaystyle \large{  y  =  \pm  \sqrt{1} }

We ignore the +√1 but focus on -√1 instead. Of course, we know that numbers squared itself will result in positive. Since 1 is positive then we can say that these solutions exist in real number.

<u>C</u><u>o</u><u>n</u><u>c</u><u>l</u><u>u</u><u>s</u><u>i</u><u>o</u><u>n</u>

So what is the different? The different between two numbers is that √-1 does not exist in a real number system since any squared numbers only result in positive while -√1 is one of the solution from y^2 = 1 and exists in a real number system.

5 0
2 years ago
Read 2 more answers
?????????????????????
valentinak56 [21]

Step-by-step explanation:

\huge \: 2 \sqrt{x} .4  {x}^{ -  \frac{5}{2} } \\  \\   \huge \: = 2.4 {x}^{ \frac{1}{2} }  . {x}^{ -  \frac{5}{2} } \\  \\ \huge \:   = 8 {x}^{ \frac{1}{2}  -  \frac{5}{2} }  \\  \\   \huge \: = 8 {x}^{ \frac{1 - 5}{2} }  \\  \\ \huge \:   = 8 {x}^{ -  \frac{4}{2} }  \\  \\ \huge \:   = 8. {x}^{ - 2}  \\

4 0
3 years ago
Find an equation for the line tangent to the circle x^2 +y^2=25 at the point (3, -4)
Lostsunrise [7]
Hello : 
the center of this cercle is : O( 0, 0)
calculate the slope  m of the line : OA    .. A (3,-4)
m= (yA - yO) / (xA- xO) 
m = (-4-0)/(3-0)
m= -4/3
if : k the slope of the tangent : k×m = -1   (the tangent perpendiclar of : OA)
k× (-4/3) = -1
k = 3/4
an equation of this tangent is : y - (-4) = (3/4)(x-3)

7 0
3 years ago
Solve using the substitution method 8x-2y=-14 5x+y=33 show work
SOVA2 [1]
13x - y = 19
-y = - 13x + 19
y = 13x + 19
8 0
2 years ago
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