Find the equation in slope-intercept form of a line with slope -2 and y-intercept 4.

A rancher has a roll of fencing to enclose a rectangular area. The table shows how the area that the rancher can enclose with th

Free_Kalibri [48]

Checking for each option

Option A: $A(w) = -w^2+200w$

Substitute w = 10 and check if we'd get 900 as the answer
$A(10) = -(10)^2+200(10) = -100+2000 = 1900$

Option B: $A(w) = -w^2+100w$
Substitute w = 10 and check if we'd get 900 as the answer
$A(10)=-(10)^2+100(10)=-100+1000=900$

Option C: $A(w)=w^2+40w$
Substitute w = 10 and check if we'd get 900 as the answer
$A(10) = 10^2+40(10)=100+400=500$

Option D: $A(w) = w^2+90w$
Substitue w = 10 and check if we'd get 900 as the answer
$A(10)=10^2+90(10) = 100+900=1000$

Answer: Option B

8 0

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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En la fábrica de Don Toño hoy fabricaron 320 pantalones, si los guardan en cajas donde caben 50 ¿cuántas cajas llenaron?

Scrat [10]

Answer:

6.4 Cajas

Step-by-step explanation:

320/50 = 6.4 Cajas

8 0

3 years ago

lesya692 [45]

Answer:

x = 1.2

Step-by-step explanation:

AD/AB = DE/BC (Similarity Theorem)

AD = 6 + 4 = 10

AB = 6

DE = 2

BC = x

Plug in the values

10/6 = 2/x

Cross multiply

10*x = 2*6

10x = 12

Divide both sides by 10

x = 12/10

x = 1.2

8 0

3 years ago

What is the value of b ?

HACTEHA [7]

Two polynomials are equal if the coefficients are equal.

$x^3+(3+b)x^2+(3b-2)x-6\\\downarrow\qquad\ \ \downarrow\qquad\qquad\ \ \ \ \downarrow\\x^3\ +\ \ 6x^2\ \ \ \ +\ \ \ \ \ 7x\ \ -\ 6\\\\x^3\to1=1\\x^2\to3+b=6\\x\to3b-2=7\\constans\to-6=-6\\\\3+b=6\ \wedge\ 3b-2=7\\\\3+b=6\ \ \ |-3\\b=3\\\\3b-2=\ \ \ \ |+2\\3b=9\ \ \ \ |:3\\b=3\\\\Answer:\ \boxed{b=3}\to\boxed{B.}$

6 0

3 years ago