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FrozenT [24]
3 years ago
5

A rectangle's length to width ratio is 9:4. If its length is 14 cm more than its width, what is its area?​

Mathematics
1 answer:
sergiy2304 [10]3 years ago
6 0
The area of the rectangle should be 87.1cm^2
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Rashid has one dollar.he wants to buy a ball for 50 cents. he also wants to buy two other toys and still have money left over. e
Alex777 [14]

Answer:

He needs to makes sure both equal to 0.50 or less


Step-by-step explanation:


8 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bx%5E%7B2%7D-5x%2B6%7D%7B2x%5E%7B2%7D-7x%2B6%20%7D" id="TexFormula1" title="\frac{x^{
il63 [147K]

Answer:

\frac{x-3}{2x-3}. hole or removable discontinuity at x=2

Step-by-step explanation:

Well generally if you want the simplest form, you factor each the denominator and numerator and then see if you can cancel any of the factors out (because they're in the denominator and numerator)

So let's start by factoring the first equation:

x^2-5x+6

Now let's find what ac is (it's just c since a=1...)

AC= 6

List factors of -6

\pm1, \pm2, \pm3, \pm6.

Now we have to look for two numbers that add up to -5. It's a bit obvious here since there isn't many factors, but it's -2 and -3, and they're both negative since 6 is positive, and -5 is negative...

So using these two factors we get

(x-2)(x-3)

Ok now let's factor the second equation:

2x^2-7x+6

Multiply a and c

AC = 12

List factors of 12:

\pm1, \pm2, \pm3, \pm4, \pm6, \pm12.

Factors that add up to -7 and multiply to 12:

-3\ and\ -4

Rewrite equation:

2x^2-4x-3x+6

Group terms:

(2x^2-4x)+(-3x+6)

Factor out GCF:

2x(x-2)-3(x-2)

Rewrite:

(2x-3)(x-2)

Now let's write out the equation using these factors:

\frac{(x-2)(x-3)}{(2x-3)(x-2)}.

Here we can factor out the x-2 and the simplified form is:

\frac{x-3}{2x-3}

So we can "technically" define f(2) using the most simplified form, but it's removable discontinuity, so it has a hole as x=2. since it makes (x-2) equal to 0 (2-2) = 0.

8 0
2 years ago
Consider a single spin of the spinner.
alex41 [277]

Answer:

"landing on a shaded portion and landing on a 3"

"landing on an unshaded portion and landing on a number less than 2 "

Step-by-step explanation:

Mutually exclusive means the events will have no intersection.

Let's look at your first choice:

"landing on a shaded portion and landing on an even number"

Landing on a shaded portion would be 1 or 4.

Landing on an even number would be 2 or 4.

There is an intersection (they contain a common element), the 4.

These events are not mutually exclusive.

Let's look at your second choice:

"landing on a shaded portion and landing on a number greater than 3"

Landing on a shaded portion would be 1 or 4.

Landing on a number greater than 3 would be just 4.

There is an intersection; they both contain 4.

These events are not mutually exclusive.

Let's look at your third choice:

"landing on a shaded portion and landing on a 3"

Landing on a shaded portion would be 1 or 4.

Landing on 3 would just be 3.

There is no common elements in the lists listed.  These events have no intersection.

These events are mutually exclusive.

Let's look at your fourth choice:

"landing on an unshaded portion and landing on an odd number"

Landing on a unshaded portion would be 2 or 3.

Landing on an odd number would be 1 or 3.

There is an intersection; they both have 3 in common.

These events are not mutually exclusive.

Let's look at your fifth choice:

"landing on an unshaded portion and landing on a number less than 2 "

Landing on an unshaded portion would 2 or 3.

Landing on a number less than 2 would be 1.

There is no intersection.

These events are mutually exclusive.

6 0
3 years ago
Read 2 more answers
In a race in which eleven automobiles are entered and there are no ties, in how many ways can the first three finishers come in?
viva [34]

Answer:

990 ways

Step-by-step explanation:

The total number of automobiles we have is 11.

Now, what this means is that for the first position , we shall be selecting 1 out of 11 automobiles, this can be done in 11 ways( 11C1 = 11!/(11-1)!1! = 11!/10!1! = 11 ways)

For the second position, since we have the first position already, the number of ways we can select the second position is selecting 1 out of available 10 and that can be done in 10 ways(10C1 ways = 10!9!1! = 10 ways)

For the third position, we have 9 automobiles and we want to select 1, this can be done in 9 ways(9C1 ways = 9!/8!1! = 9 ways)

Thus, the total number of ways the first three finishers come in = 11 * 10 * 9 = 990 ways

8 0
4 years ago
Study the statements carefully.
Elan Coil [88]

Answer

x = unidentified

Explanation

If 12÷3=4 and 12 = 4×3, that means 6÷0=x and x×0=6.

This isn't possible because any number divided by 0 is unidentified.

7 0
3 years ago
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