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Amiraneli [1.4K]
3 years ago
15

Mai finished her chemistry assignment in 1/2 hours. Then She completed her history assignment in 1/4 hours. How much more time d

id Mai spend on her chemistry assignment ?
Mathematics
2 answers:
Aleksandr [31]3 years ago
8 0
Since 1/2 an hour is 30 minutes and 1/4 is 15 minutes than the answer has to be 15 minutes 15+15=30
hjlf3 years ago
6 0
SHE SPENT 15 MINUTES MORE. 1 hour is 60 minutes. 1/2 hour is 30 minutes and 1/4 hour is 15 minutes. 30-15=15
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Li’s brother used the work shown to determine that he must grow at most 6 inches to be able to ride the roller coaster. Is this
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Lynee bought some produce. She bought a bag of grapefruit, 1 5/8 pounds of apples, and 2 3/16 pounds of bananas. The total weigh
iragen [17]

Given:

Lynee bought a bag of grapefruit, 1\dfrac{5}{8} pounds of apples, and 2\dfrac{3}{16} pounds of bananas.

The total weight of her purchase was 7\dfrac{1}{2} pounds.

To find:

The weight of the bag of grapefruit.

Solution:

We have,

Weight of apples = 1\dfrac{5}{8} pounds.

Weight of bananas = 2\dfrac{3}{16} pounds.

Total weight of her purchase = 7\dfrac{1}{2} pounds.

Let x be the weight of the bag of grapefruit.

x+1\dfrac{5}{8}+2\dfrac{3}{16}=7\dfrac{1}{2}

x+\dfrac{1(8)+5}{8}+\dfrac{2(16)+3}{16}=\dfrac{7(2)+1}{2}

x+\dfrac{13}{8}+\dfrac{35}{16}=\dfrac{15}{2}

\dfrac{16x+26+35}{16}=\dfrac{15}{2}

Multiply both sides by 16.

16x+61=120

16x=120-61

16x=59

x=\dfrac{59}{16}

x=3\dfrac{11}{16}

Therefore, the weight of a bag of grapefruit is 3\dfrac{11}{16} pounds.

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3 years ago
What is cos^{-1} (0.393) [/tex]
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3 years ago
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Stella [2.4K]

Answer:  -\frac{\sqrt{2a}}{8a}

=======================================================

Explanation:

The (x-a) in the denominator causes a problem if we tried to simply directly substitute in x = a. This is because we get a division by zero error.

The trick often used for problems like this is to rationalize the numerator as shown in the steps below.

\displaystyle \lim_{x\to a} \frac{\sqrt{3a-x}-\sqrt{x+a}}{4(x-a)}\\\\\\\lim_{x\to a} \frac{(\sqrt{3a-x}-\sqrt{x+a})(\sqrt{3a-x}+\sqrt{x+a})}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{(\sqrt{3a-x})^2-(\sqrt{x+a})^2}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{3a-x-(x+a)}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{3a-x-x-a}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\

\displaystyle \lim_{x\to a} \frac{2a-2x}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{-2(-a+x)}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{-2(x-a)}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{-2}{4(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\

At this point, the (x-a) in the denominator has been canceled out. We can now plug in x = a to see what happens

\displaystyle L = \lim_{x\to a} \frac{-2}{4(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\L = \frac{-2}{4(\sqrt{3a-a}+\sqrt{a+a})}\\\\\\L = \frac{-2}{4(\sqrt{2a}+\sqrt{2a})}\\\\\\L = \frac{-2}{4(2\sqrt{2a})}\\\\\\L = \frac{-2}{8\sqrt{2a}}\\\\\\L = \frac{-1}{4\sqrt{2a}}\\\\\\L = \frac{-1*\sqrt{2a}}{4\sqrt{2a}*\sqrt{2a}}\\\\\\L = \frac{-\sqrt{2a}}{4\sqrt{2a*2a}}\\\\\\L = \frac{-\sqrt{2a}}{4\sqrt{(2a)^2}}\\\\\\L = \frac{-\sqrt{2a}}{4*2a}\\\\\\L = -\frac{\sqrt{2a}}{8a}\\\\\\

There's not much else to say from here since we don't know the value of 'a'. So we can stop here.

Therefore,

\displaystyle \lim_{x\to a} \frac{\sqrt{3a-x}-\sqrt{x+a}}{4(x-a)} = -\frac{\sqrt{2a}}{8a}\\\\\\

3 0
3 years ago
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