All categories

3 years ago

What is the width of a rectangle with a length of 14 cm and an area of 161 cm squared

1 answer:

8 0

Area of a rectangle=length×width. If the area is 161 and the length is 14 then that means: 161=14×width. You would then divide 161÷14 and you will find that the width is 11.5.

You might be interested in

What is equivalent to -1/4 - 5/3

Stolb23 [73]

-3/12 - 20/12 is equivalent to -1/4 - 5/3

7 0

3 years ago

Identify the horizontal asymptote of f(x) = quantity 2 x minus 1 over quantity x squared minus 7 x plus 3. y = 0 y = 1 over 2 y

Ratling [72]

The horizontal asymptote is the value at the y-axis where the graph approaches the line but not necessarily touching it. Hence, the asymptotic characteristic of the graph. The standard form of a function in fraction form is y = (ax^n +...)/(bx^m+...). There are rules to follow to determine the horizontal asymptote of a function.
1) if n = m , then the horizontal equation is y = a/b
2) if n>m, then there is no horizontal equation
3) if n<m, then the horizontal equation is the x axis ; y = 0.

The function given falls on the third rule hence the horizontal asymptote of the function is at y = 0.

8 0

3 years ago

Read 2 more answers

Marya plans to participate in a 20 km walk-a-thon. To prepare, she walked 4 km on Monday. She plans to increase her walk by 2 km

Afina-wow [57]

Answer: 8 weeks

Step-by-step explanation:

20 - 4 = 16

16 / 2 = 8

7 0

2 years ago

Read 2 more answers

PLEASE ANSWER THIS QUESTION !! 20 POINTS AND BRAINLIEST !!

Anastaziya [24]

Answer:

x^2+7x+12

Step-by-step explanation:

(x+3) (x+4)

FOIL

first: x*x = x^2

outer: 4x

inner: 3x

last: 3*4 =12

Add them together

x^2 +4x+3x+12

x^2+7x+12

8 0

3 years ago

Read 2 more answers

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

2 years ago

Read 2 more answers