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3 years ago

If x = -3, then x2-7x + 10 equals

Mathematics

2 answers:

belka [17]3 years ago

6 0

Answer:

4-7x is your answer unless you have not included parentheses

Step-by-step explanation:

Nastasia [14]3 years ago

5 0

The problem would equal 25

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Helppppppp please first is brainliest

Ad libitum [116K]

The answer is C my boi

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3 years ago

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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.

Definition

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

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.

Definition II

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

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

Examples

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

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.

Absolute Value Property I

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

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.

Conclusion

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

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Which is the equation of a circle with its center at (-4,2) and a radius of 4?

Snowcat [4.5K]

Answer:

(x +4)^2 + (y-2)^2 = 16

Step-by-step explanation:

We can write the equation for a circle with the formula

(x-h)^2 + (y-k)^2 = r^2

Where (h,k) is the center and r is the radius

We know the center is at (-4,2) and the radius is 4

(x- -4)^2 + (y-2)^2 = 4^2

(x +4)^2 + (y-2)^2 = 16

4 0

3 years ago

2 (5p - 4) = 3p -7 need this very soon

MakcuM [25]

Answer:

p=7

Step-by-step explanation:

2 (5p-4)=3p-7

(10p-8)=3p-7

10p=3p+1

7p=1

p=7

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3 years ago

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For each ordered pair, determine whether it is a solution to 2x - 5y=11.

Sidana [21]

Answer:

(3, -1)

Step-by-step explanation:

5 0

3 years ago