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neonofarm [45]
3 years ago
14

The area of a rectangular garden is given by the trinomial x^2+6x-27. what are the possible dimensions of the rectangle? use fac

toring
Mathematics
2 answers:
snow_tiger [21]3 years ago
6 0
A = x^2 + 6x - 27 A = x^2 + 9x - 3x - 27 A = x(x+9)-3(x+9) A = (x+9)(x-3) A = lw l = (x+9) w =(x-3)
nikitadnepr [17]3 years ago
4 0

Answer:

Step-by-step explanation:

Given that a garden is in rectangular form

Area of the garden = x^2+6x-27

We know that area of a rectangle = length x width

Hence length x width = x^2+6x-27

There can be infinite number of answers for this equation.

Let us assume that both length and width are rational.

Then the factors would be answers.

Factorise

x^2+6x-27 as

(x+9)(x-3)

Since normally length would be longer, we can say

length = x+9 and width = x-3, for x >3.

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Answer:

1/6

Step-by-step explanation:

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2 years ago
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327.8 in scientific notation
Allushta [10]

Answer:

3.278 * 10 ^ 2

Step-by-step explanation:

To change a number to scientific notation you:

Move the decimal point to the left to make the number have 1 number before the decimal point.

From there you multiply by 10 to the power 2 the number of decimal points you moved.

So the answer is 3.278 * 10^2

HOPE IT HELPED

5 0
2 years ago
Help me please I really need!!! :)
WITCHER [35]

Answer:

\huge\boxed{\sf x = 1.8}

Step-by-step explanation:

\sf 2x-1.5 = 2.1\\\\Add \ 1.5 \ to \ both \ sides\\\\2x = 2.1 + 1.5\\\\2x = 3.6\\\\Divide \ 2 \ to \ both \ sides\\\\x = 3.6 / 2\\\\x = 1.8\\\\\rule[225]{225}{2}

Hope this helped!

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5 0
2 years ago
What is the sum of (1-5q)+2(2.5q+8)
bogdanovich [222]

Answer:

17

Step-by-step explanation:

Eliminating parentheses, we have ...

... 1 - 5q +2·2.5q +2·8

... = 1 -5q +5q +16 . . . . . carry out the multiplication

... = q(-5+5) +(1 +16) . . . . group like terms

... = q·0 +17 . . . . . . . . . . . combine like terms

... = 17

8 0
2 years ago
Select all of the following true statements if R = real numbers, I = integers, and W = {0, 1, 2, ...}.
Vinvika [58]

Answer and Step-by-step explanation:

We will begin to solve this problem by defining first what the sets' elements really are.

R consists of real numbers. This means that this set contains all the numbers, rational or not.

Z is composed of whole numbers. Integers include all negative and positive numbers as well as zero (it's basically a set of whole numbers and their negated values).

W, on the other hand, has 0,1,2, and its elements are onward. Those numbers are referred to as whole numbers.

W ⊂ Z is TRUE. Z contains all the numbers as stated earlier, and W is a subset of it.

R ⊂ W is FALSE. Not all numbers are complete numbers. Complete numbers must be rational and represented fractionless. These requirements are not met by those real numbers.

0 ∈ Z is TRUE.  Zero is just an integer so it is a component of Z.

∅ ⊂ R is TRUE. A set i.e null be R subset, and each and every set is a general set. Moreover, there were not single elements in a null set, so it spontaneous became a non empty set subset through description as there is no element of R.

{0,1,2,...} ⊆ W is TRUE. The set on the left is precisely what is specified in the statement for problem for W. (The bar below the subset symbol simply implies that the subset is not rigid, because the set on the left may be equal to the set on the right. Without it, the argument would be incorrect, because a strict subset needs that the two sets not be identical).

-2 ∈ W is FALSE. W's only made up of whole numbers and not their negated equivalents.

4 0
3 years ago
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