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vova2212 [387]
4 years ago
12

Caleb and Sergio stacked boxes on a shelf. Caleb lifted 14 boxes and Sergio lifted 12 boxes. The boxes that Sergio lifted each w

eighed 10 lb more than the boxes Caleb lifted.
Which expression represents the total number of pounds Sergio lifted?
Let c represent the weight of each box Caleb lifted.
Mathematics
1 answer:
scoundrel [369]4 years ago
5 0

Answer:

14 ( c ) = Caleb's

12 ( c + 10 ) = Sergio's


Step-by-step explanation:

Caleb lifted 14 boxes so you multiply 14 by c which is the weight of each box and for Sergio he lifted 12 boxes although his weighed 10 pounds more so you add 10 to c which would equal c + 10 multiplied by 12

hope this did not confuse you

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John has a bag of marbles that contains 12 red marbles, 20 green marbles, and 17 blue marbles.  Problem 1) What is the probabili
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12 red marbles
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Total amount: 39 marbles

39 (total) - 12 (red) = 27.

John has a probability of 27/39 that he will not pick a red marble out of the bag.
If you want to simplify your fraction by dividing both numbers by 3 to make it even more accurate, an even better answer would be 9/13.

(so all you do is divide 27/3 and divide 39/3 aswell which gives you 9 and 13.)
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$4860= 1 year
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Jill buys a t-shirt for $12.79 and a pair of socks for $6.75 at the mall. How much did she spend?
Pie

Answer:

$12.79 + $6.75 = $19.54

Step-by-step explanation:

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Can you Solve this plzzz nerd ur help<br><br>6÷2(1+2)=​
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3 0
3 years ago
Find cot and cos <br> If sec = -3 and sin 0 &gt; 0
Natali5045456 [20]

Answer:

Second answer

Step-by-step explanation:

We are given \displaystyle \large{\sec \theta = -3} and \displaystyle \large{\sin \theta > 0}. What we have to find are \displaystyle \large{\cot \theta} and \displaystyle \large{\cos \theta}.

First, convert \displaystyle \large{\sec \theta} to \displaystyle \large{\frac{1}{\cos \theta}} via trigonometric identity. That gives us a new equation in form of \displaystyle \large{\cos \theta}:

\displaystyle \large{\frac{1}{\cos \theta} = -3}

Multiply \displaystyle \large{\cos \theta} both sides to get rid of the denominator.

\displaystyle \large{\frac{1}{\cos \theta} \cdot \cos \theta = -3 \cos \theta}\\\displaystyle \large{1=-3 \cos \theta}

Then divide both sides by -3 to get \displaystyle \large{\cos \theta}.

Hence, \displaystyle \large{\boxed{\cos \theta = - \frac{1}{3}}}

__________________________________________________________

Next, to find \displaystyle \large{\cot \theta}, convert it to \displaystyle \large{\frac{1}{\tan \theta}} via trigonometric identity. Then we have to convert \displaystyle \large{\tan \theta} to \displaystyle \large{\frac{\sin \theta}{\cos \theta}} via another trigonometric identity. That gives us:

\displaystyle \large{\frac{1}{\frac{\sin \theta}{\cos \theta}}}\\\displaystyle \large{\frac{\cos \theta}{\sin \theta}

It seems that we do not know what \displaystyle \large{\sin \theta} is but we can find it by using the identity \displaystyle \large{\sin \theta = \sqrt{1-\cos ^2 \theta}}  for \displaystyle \large{\sin \theta > 0}.

From \displaystyle \large{\cos \theta = -\frac{1}{3}} then \displaystyle \large{\cos ^2 \theta = \frac{1}{9}}.

Therefore:

\displaystyle \large{\sin \theta=\sqrt{1-\frac{1}{9}}}\\\displaystyle \large{\sin \theta = \sqrt{\frac{9}{9}-\frac{1}{9}}}\\\displaystyle \large{\sin \theta = \sqrt{\frac{8}{9}}}

Then use the surd property to evaluate the square root.

Hence, \displaystyle \large{\boxed{\sin \theta=\frac{2\sqrt{2}}{3}}}

Now that we know what \displaystyle \large{\sin \theta} is. We can evaluate \displaystyle \large{\frac{\cos \theta}{\sin \theta}} which is another form or identity of \displaystyle \large{\cot \theta}.

From the boxed values of \displaystyle \large{\cos \theta} and \displaystyle \large{\sin \theta}:-

\displaystyle \large{\cot \theta = \frac{\cos \theta}{\sin \theta}}\\\displaystyle \large{\cot \theta = \frac{-\frac{1}{3}}{\frac{2\sqrt{2}}{3}}}\\\displaystyle \large{\cot \theta=-\frac{1}{3} \cdot \frac{3}{2\sqrt{2}}}\\\displaystyle \large{\cot \theta=-\frac{1}{2\sqrt{2}}

Then rationalize the value by multiplying both numerator and denominator with the denominator.

\displaystyle \large{\cot \theta = -\frac{1 \cdot 2\sqrt{2}}{2\sqrt{2} \cdot 2\sqrt{2}}}\\\displaystyle \large{\cot \theta = -\frac{2\sqrt{2}}{8}}\\\displaystyle \large{\cot \theta = -\frac{\sqrt{2}}{4}}

Hence, \displaystyle \large{\boxed{\cot \theta = -\frac{\sqrt{2}}{4}}}

Therefore, the second choice is the answer.

__________________________________________________________

Summary

  • Trigonometric Identity

\displaystyle \large{\sec \theta = \frac{1}{\cos \theta}}\\ \displaystyle \large{\cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta}}\\ \displaystyle \large{\sin \theta = \sqrt{1-\cos ^2 \theta} \ \ \ (\sin \theta > 0)}\\ \displaystyle \large{\tan \theta = \frac{\sin \theta}{\cos \theta}}

  • Surd Property

\displaystyle \large{\sqrt{\frac{x}{y}} = \frac{\sqrt{x}}{\sqrt{y}}}

Let me know in the comment if you have any questions regarding this question or for clarification! Hope this helps as well.

5 0
2 years ago
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