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2 years ago

True or false? Based only on the given information, it is guaranteed that

Mathematics

1 answer:

Black_prince [1.1K]2 years ago

8 0

Answer:

True

Step-by-step explanation:

The triangles are both right triangles, meaning they can be used with pythagorean theorem. If the triangles CAD and CBD share the same "hypotenuse" (CD), and they have a pair of congruent sides, they other pair of sides must be able to fit into the pythagorean theorem/

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A rectangular table is four times as long as it is wide. If the area is 36 ft2, find the length and the width of the table.

FrozenT [24]

Answer:

The width is 3 ft

The length is 12 ft

Step-by-step explanation:

Let w = width

l = 4w

A = l*w

Replace length with 4w

A = 4w*w

36 = 4 w^2

Divide each side by 4

36/4 = 4w^2/4

9 = w^2

Take the square root of each side

sqrt(9) = sqrt(w)

3 =w

The width is 3 ft

The length is 4*3 = 12 ft

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3 years ago

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erma4kov [3.2K]

7+5 is 12. 114 divided by 12 is 9.5 so one part is 9.5. Dipaks share is 5 so 9.5 multiplied by 5 is 47.5 so the answer is $47.50

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2 years ago

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3 years ago

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Someone please help, i’m stuck

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Answer:

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2x-2

0 is answer

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2 years ago

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Bogdan [553]

$f(x)=\dfrac{x^2}{x^2+k^2}$

By definition of the derivative,

$f'(x)=\displaystyle\lim_{h\to0}\frac{\frac{(x+h)^2}{(x+h)^2+k^2}-\frac{x^2}{x^2+k^2}}h$

$f'(x)=\displaystyle\lim_{h\to0}\frac{(x+h)^2(x^2+k^2)-x^2((x+h)^2+k^2)}{h(x^2+k^2)((x+h)^2+k^2)}$

$f'(x)=\dfrac{k^2}{x^2+k^2}\displaystyle\lim_{h\to0}\frac{(x+h)^2-x^2}{h((x+h)^2+k^2)}$

$f'(x)=\dfrac{k^2}{x^2+k^2}\displaystyle\lim_{h\to0}\frac{2xh+h^2}{h((x+h)^2+k^2)}$

$f'(x)=\dfrac{k^2}{x^2+k^2}\displaystyle\lim_{h\to0}\frac{2x+h}{(x+h)^2+k^2}$

$f'(x)=\dfrac{2xk^2}{(x^2+k^2)^2}$

$\dfrac{k^2}{(x^2+k^2)^2}$ is positive for all values of $x$ and $k$ . As pointed out, $x\ge0$ , so $f'(x)\ge0$ for all $x\ge0$ . This means the proportion of occupied binding sites is an increasing function of the concentration of oxygen, meaning the presence of more oxygen is consistent with greater availability of binding sites. (The question says as much in the second sentence.)

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2 years ago