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2 years ago

2/3 cup of sugar for 24 cookies equal how many cup in 1 cookie?

Mathematics

1 answer:

sweet [91]2 years ago

5 0

Answer:

0.02777777777

Step-by-step explanation: 2/3 divided by 24 = 0.02777777777

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A recipe has a ratio of 2 cups of sugar to 6 cups of flour. How many cups of flour are used for each cup of sugar in the recipe?

statuscvo [17]

The ratio 2 to 6 can be reduced to 1 to 3. That means for every 3 cups of flour there is one cup of sugar.

6 0

3 years ago

Read 2 more answers

Please be correct!!!!!!!

DerKrebs [107]

Answer:

D is the answer

4 0

2 years ago

Katy has 100 quarters and dimes. Their total value is $19. How many of each coin does Katy have?

jekas [21]

Answer:

she has D=40, and Q=60

Step-by-step explanation:

7 0

2 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:

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

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

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

$=\lim\limits_{x\to\pm\infty}\dfrac{9-\frac{3}{x}}{1-\frac{7}{x}}=\dfrac{9}{1}=9\\\\\boxed{y=1x+9}$

8 0

3 years ago

By what factor would your weight be multiplied if earth's diameter were 1/ 4 as big and earth's mass remained unchanged?.

GenaCL600 [577]

Answer:

Step-by-step explanation:

Your weight is (roughly) proportional to the inverse of the square of the distance you are from the center of the earth. If that distance is reduced to 1/4 its previous value, then your weight will be multiplied by the factor ...

1/(1/4)² = 16

You would weigh 16 times as much.

_____

<em>Additional comment</em>

The distribution of mass within the earth also affects your weight.

3 0

2 years ago