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Firdavs [7]
3 years ago
6

The rectangle below has an area of x^2-16 square meters and a width of x+4 meters.

Mathematics
2 answers:
Basile [38]3 years ago
5 0
X^2-16 is a difference of squares of the form:

(a^2-b^2) which always factors to:

(a+b)(a-b), in this case we have:

(x+4)(x-4), we are told that width is (x+4) so the length is:

(x-4)
Flura [38]3 years ago
3 0
Hello Mate!

Area=LW
x^2-16=L(x+4)
Remember difference of 2 perfect squares to factor x^2-16
(x-4)(x+4)=L(x+4)
Divide both sides by (x+4)
x-4=L

Therefore, length=x-4

I Hope my answer has come to your Help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead! :)

(Ps. Mark As Brainliest IF Helped!)

-TheOneAboveAll :D
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7 0
2 years ago
Read 2 more answers
6x - 2y = 5
laiz [17]

Answer:

The given system has NO SOLUTION.

Step-by-step explanation:

Here, the given system of equation is:

6 x -  2 y = 5     .......... (1)

3 x  - y = 10          .... (2)

Multiply equation 2 with (-2), we get:

3 x  - y = 10     ( x -2)

⇒  - 6 x + 2 y = - 20

Now, ADD this to equation (1) , we get:

6 x - 2 y  - 6 x + 2 y  = 5 - 20

or, 0 = - 15

WHICH IS NOT POSSIBLE as 0 ≠ -15

Hence, the given system has NO SOLUTION.

7 0
3 years ago
Let vector F = (6 x^2 y + 2 y^3 + 4 e^x) i + (7 e^{y^2} + 54 x) j . Consider the line integral of vector F around the circle of
balu736 [363]

Denote the circle of radius a by C. C is simple and closed, so by Green's theorem the line integral reduces to a double integral over the interior of C (call it D):

\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\int_C(6x^2y+2y^3+4e^x)\,\mathrm dx+(7e^{y^2}+54x)\,\mathrm dy

=\displaystyle\iint_D\left(\frac{\partial(7e^{y^2}+54x)}{\partial x}-\frac{\partial(6x^2y+2y^3+4e^x)}{\partial y}\right)\,\mathrm dx\,\mathrm dy

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D is a circle of radius a, so we can write the double integral in polar coordinates as

\displaystyle\iint_D(54-6x^2-6y^2)\,\mathrm dx\,\mathrm dy=\int_0^{2\pi}\int_0^a(54-6r^2)r\,\mathrm dr\,\mathrm d\theta

a. For a=1, we have

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I(a) has critical points when

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8 0
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