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UNO [17]
3 years ago
14

If a student gets an A on a final exam, then the student will pass the course.

Mathematics
1 answer:
Mrac [35]3 years ago
3 0

Answer: PERIOD

Step-by-step explanation:

Get good grades on your assignments and the student will pass the course with an A.

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Find all the zeros of the equation.<br> -x^3-x^2=11x+11
anyanavicka [17]

Answer:

x= -1 , i(sqrt11) , -i(sqrt11)

Step-by-step explanation:

subtract 11x from both sides

-x^3 - x^2 - 11x = 11

subtract 11 from both sides

-x^3 - x^2 - 11x - 11 = 0

Factor the left side of the equation

- (x - 1)(x^2 + 11) = 0

Set both factors to zero and solve for x

x + 1 = 0 x^2 + 11 = 0

-1 -11

x^2 = -11

x = -1 x = +/- i(sqrt11)

7 0
3 years ago
Simplify (write without the absolute value sign): 120+x, if x &lt;- 120
Tju [1.3M]

Answer:

Step-by-step explanation:

hello :

120+x, if x <- 120

given : x <- 120    add 120      x+120<- 120+120

x+120< 0

8 0
3 years ago
Please help me find the y-intercept! :(
zvonat [6]
5 inches

Explanation: y = Mx + b, m is your slope (1/2), b is your y-intercept (5 inches)

Equation would be y= (1/2)x + 5
4 0
3 years 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
3 years ago
Read 2 more answers
Bly makes fleece scarves and sells them for $11 each. The supplies cost her $5 per scarf. Last month, Bly had a profit of $139.
seraphim [82]

Bly sold 29 scarves last month

<em><u>Solution:</u></em>

Let "x" be the number of scarves sold

Bly made 7 more scarves than she sold

Then, number of scarves made = x + 7

Bly makes fleece scarves and sells them for $11 each

Thus for selling "x" scraves , he will get 11x dollars

Total sales revenue = 11x

number of scarves made = x + 7

The supplies cost her $5 per scarf

Therefore, cost for scarves made = 5(x + 7)

Therefore,

profit = Total sales revenue - cost for scarves made

profit = 11x - 5(x + 7)

profit = 11x - 5x - 35

profit = 6x - 35

Last month, Bly had a profit of $139

139 = 6x - 35

6x = 174

x = 29

Thus Bly sold 29 scarves last month

5 0
3 years ago
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