Write the number

Mathematics

1 answer:

iragen [17]3 years ago

6 0

Answer:

700,21,1009.104 or 721,904.

Step-by-step explanation:

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A ladder is resting against a wall. the ladder and the ground make an angle of 39° and the ladder is 14 ft long. how far away fr

zhannawk [14.2K]

Let:

the ladder length = l
the distance from the wall and the ladder = x

cos (39) = x / l

x = l * cos (39) = 14 * cos (39) = 10.88 ft

Hope that helps

8 0

4 years ago

Lucy’s chocolate bar is 54% cocoa. If the weight of the chocolate bar is 54 grams, how many grams of cocoa does it contain? Roun

Masteriza [31]

Answer:

The chocolate bar contains 29.2 grams of cocoa.

Step-by-step explanation:

Given that Lucy’s chocolate bar is 54% cocoa and the weight of the chocolate bar is 54 grams.

Hence, the amount of cocoa in the chocolate bar is given by

54% of 54 grams

54% percentage means 54/100 = 0.54

Hence, the amount of cocoa in the chocolate bar is

$0.54\times54\\=29.16$

When rounded to the nearest tenth, 29.16 should be equal to 29.2

Hence, the chocolate bar contains 29.2 grams of cocoa.

4 0

3 years ago

Determine the horizontal vertical and slant asymptote y=x^2+2x-3/x-7

lilavasa [31]

Answer:

<h2>A.Vertical:x=7</h2><h2>Slant:y=x+9</h2>

Step-by-step explanation:

$f(x)=\dfrac{x^2+2x-3}{x-7}\\\\vertical\ asymptote:\\\\x-7=0\qquad\text{add 7 to both sides}\\\\\boxed{x=7}\\\\horizontal\ asymptote:\\\\\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3}{x-7}=\lim\limits_{x\to\pm\infty}\dfrac{x^2\left(1+\frac{2}{x}-\frac{3}{x^2}\right)}{x\left(1-\frac{7}{x}\right)}=\lim\limits_{x\to\pm\infty}\dfrac{x\left(1+\frac{2}{x}-\frac{3}{x^2}\right)}{1-\frac{7}{x}}=\pm\infty\\\\\boxed{not\ exist}$

$slant\ asymptote:\\\\y=ax+b\\\\a=\lim\limits_{x\to\pm\infty}\dfrac{f(x)}{x}\\\\b=\lim\limits_{x\to\pm\infty}(f(x)-ax)\\\\a=\lim\limits_{x\to\pm\infty}\dfrac{\frac{x^2+2x-3}{x-7}}{x}=\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3}{x(x-7)}=\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3}{x^2-7x}\\\\=\lim\limits_{x\to\pm\infty}\dfrac{x^2\left(1+\frac{2}{x}-\frac{3}{x^2}\right)}{x^2\left(1-\frac{7}{x}\right)}=\lim\limits_{x\to\pm\infty}\dfrac{1+\frac{2}{x}-\frac{3}{x^2}}{1-\frac{7}{x}}=\dfrac{1}{1}=1$

$b=\lim\limits_{x\to\pm\infty}\left(\dfrac{x^2+2x-3}{x-7}-1x\right)=\lim\limits_{x\to\pm\infty}\left(\dfrac{x^2+2x-3}{x-7}-\dfrac{x(x-7)}{x-7}\right)\\\\=\lim\limits_{x\to\pm\infty}\left(\dfrac{x^2+2x-3}{x-7}-\dfrac{x^2-7x}{x-7}\right)=\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3-(x^2-7x)}{x-7}\\\\=\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3-x^2+7x}{x-7}=\lim\limits_{x\to\pm\infty}\dfrac{9x-3}{x-7}=\lim\limits_{x\to\pm\infty}\dfrac{x\left(9-\frac{3}{x}\right)}{x\left(1-\frac{7}{x}\right)}$