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Genrish500 [490]

3 years ago

11

Zane needs 20 yards of fencing to go around his rectangular yard. What are possible dimensions for Zane's yard? Select all that

apply .

1 answer:

stiv31 [10]3 years ago

7 0

Answer:

1 yard by 9 yards

2 yards by 8 yards

3 yards by 7 yards

4 yards by 6 yards

Step-by-step explanation:

the perimeter of the rectangular yard needs to be 20 yards

Perimeter of a rectangle = 2 x ( length + width)

20 = 2 x ( length + width)

( length + width) = 20 /2

( length + width) = 10

The sum of the length and width should equal 10

Possible dimensions that would equal 10 are :

1 yard by 9 yards

2 yards by 8 yards

3 yards by 7 yards

4 yards by 6 yards

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Kryger [21]

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4 years ago

Please help its urgent

hram777 [196]

Answer:

<h2>Function is y = x.</h2><h2>Domain: $-3 \leq x \leq 2$ .</h2><h2>Range: $-5 \leq y \leq 4$ .</h2>

Step-by-step explanation:

In the given image, the line passes through (-1, -1) and (1, 1).

Let the equation of the line is $y = mx + c$ , where m is the tangent of the line and c is a constant.

Putting the co-ordinates of the points in the equation, we get $-1 = -m + c$ and $1 = m + c$

From the two equations we get, c = 0 and m = 1.

Hence, the function is y = x.

Domain is $-3 \leq x \leq 2$ .

Range is $-5 \leq y \leq 4$

5 0

3 years ago

50 points ! marking brainly to whoever gets all 5 correct !

Svetradugi [14.3K]

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3 0

4 years ago

Differentiate with respect to X <br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B%20%5Cfrac%7Bcos2x%7D%7B1%20%2Bsin2x%20%7D%20

Mice21 [21]

Power and chain rule (where the power rule kicks in because $\sqrt x=x^{1/2}$ ):

$\left(\sqrt{\dfrac{\cos(2x)}{1+\sin(2x)}}\right)'=\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'$

Simplify the leading term as

$\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}$

Quotient rule:

$\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'=\dfrac{(1+\sin(2x))(\cos(2x))'-\cos(2x)(1+\sin(2x))'}{(1+\sin(2x))^2}$

Chain rule:

$(\cos(2x))'=-\sin(2x)(2x)'=-2\sin(2x)$

$(1+\sin(2x))'=\cos(2x)(2x)'=2\cos(2x)$

Put everything together and simplify:

$\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{(1+\sin(2x))(-2\sin(2x))-\cos(2x)(2\cos(2x))}{(1+\sin(2x))^2}$

$=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{-2\sin(2x)-2\sin^2(2x)-2\cos^2(2x)}{(1+\sin(2x))^2}$

$=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{-2\sin(2x)-2}{(1+\sin(2x))^2}$

$=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac{\sin(2x)+1}{(1+\sin(2x))^2}$

$=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac1{1+\sin(2x)}$

$=-\dfrac1{\sqrt{\cos(2x)}}\dfrac1{\sqrt{1+\sin(2x)}}$

$=\boxed{-\dfrac1{\sqrt{\cos(2x)(1+\sin(2x))}}}$

5 0

3 years ago

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Grace [21]

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7 0

3 years ago