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3 years ago

The express the area of the entire rectangle

Mathematics

1 answer:

horrorfan [7]3 years ago

5 0

Answer:

x^2 +11x +30

Step-by-step explanation:

(x+6) * ( x+5)

The blue area is x^2

pink is 6x

green is 5x

orange is 5*6 = 30

Add them together

x^2 +6x+5x+30

x^2 +11x +30

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18 times what is 504

schepotkina [342]

Answer:

Step-by-step explanation:

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3 years ago

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Geometry segment proofs

Lemur [1.5K]

solving a linear equation we conclude that the value of x must be 6.

<h3>How to find the value of x?</h3>

On the statement we have the relations:

AB = BC

BC = CD

For the transitive property, then we have that:

AB = CD

Replacing the equations in the image, we get:

2x + 1 = 4x - 11

Solving this linear equation for x we get:

1 + 11 = 4x - 2x

12 = 2x

12/2 = x

6 = x

So yes, solving that linear equation we conclude that the value of x must be 6.

Learn more about linear equations:

brainly.com/question/1884491

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8 0

1 year ago

There are 3 red chips and 2 blue chips. If they form a certain color pattern when arranged in a row, for example RBRRB, how many

svp [43]

Answer:

Option A - 10

Step-by-step explanation:

Given : There are 3 red chips and 2 blue chips. If they form a certain color pattern when arranged in a row, for example RBRRB.

To find : How many color patterns are possible?

Solution :

Total number of chips = 5

So, 5 chips can be arranged in 5! ways.

There are 3 red chips and 2 blue chips.

So, choosing 3 red chips in 3! ways

and choosing 2 blue chips in 2! ways.

As changing the places of similar chip will not create new pattern.

The total pattern is given by,

$T=\frac{5!}{3!\times 2!}$

$T=\frac{5\times 4\times 3!}{3!\times 2}$

$T=10$

Therefore, color patterns are possible are 10.

Option A is correct.

4 0

3 years ago

What is the nth term?

kumpel [21]

Let $a_k$ denote the kth term of the sequence. Then

$a_k=a_1+d(k-1)$

where d is the common difference between consecutive terms in the sequence and a₁ is the first term.

The sum of the first n terms is

$S_n=\displaystyle\sum_{k=1}^na_k=a_1+a_2+\cdots+a_{n-1}+a_n$

From the formula for $a_k$ , we get

$S_n=\displaystyle\sum_{k=1}^n(a_1+d(k-1))=a_1\sum_{k=1}^n1+d\sum_{k=1}^n(k-1)$

$S_n=\displaystyle na_1+d\sum_{k=0}^{n-1}k$

$S_n=na_1+\dfrac{d(n-1)n}2$

$S_n=\dfrac n2(2a_1-d+dn)$

So we have $d=-5$ , and $2a_1-d=16$ so that $a_1=\frac{11}2$ .

Then the nth term in the sequence is

$a_n=\dfrac{11}2-5(n-1)=\boxed{\dfrac{21-10n}2}$

7 0

3 years ago

PLEASE HURRY! ^^

GarryVolchara [31]

Answer: 3 and -3

Your welcome

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3 years ago

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