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4 years ago

Which of the following phases are expressions

Mathematics

1 answer:

valentinak56 [21]4 years ago

4 0

Attach a picture of the question please.

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How do I I solve this equation? -18 = x/3 + 3

LuckyWell [14K]

Subtract 3 from both sides

-18 - 3 = x/3

Simplify -18 - 3 to -21

-21 = x/3

Multiply both sides by 3

-21 * 3 = x

Simplify -21 * 3 to -63

-63 = x

Switch sides

x = -63

7 0

3 years ago

Three bags of chips costs m cents. At this rate, what is the cost, in cents, of one bag of chips?

olya-2409 [2.1K]

Answer: m/3 cents

Step-by-step explanation:

Hi, to answer this question we have to divide the total cost of the chips (m) by the number of bags of chips (3).

Mathematically speaking:

Cost of one bag chip: m/3 cents

Since the total cost is represented by a variable, the individual cost is an expression.

If we replace the value of m by a number we will obtain the result as a number.

Feel free to ask for more if needed or if you did not understand something.

8 0

3 years ago

Find cot and cos If sec = -3 and sin 0 > 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

What is the area of the rhombus?

nikklg [1K]

Answer:

if you need the formula, it is pq/2

Step-by-step explanation:

the "p" is a diagonal of the rhombus

the "q" is the other diagonal of the rhombus

you multiply p and q

the you divide that number by 2

then, you have found the area

5 0

3 years ago

The 6th term is 22 and the common difference is 6. what is the 50th term

Tom [10]

The general formula for an arithmetic sequence is:
a(n) = a + d(n - 1) 

a : first term 
d : common difference (6) 
n : number of the term 

In your example, you are told the 6th term is 22. 
a(6) = a + 6(6 - 1) = 22 
a + 30 = 22 
a = 22 - 30 
a = -8 

So now you have the general formula: 
a(n) = -8 + 6(n - 1) 

If you like you can simplify it: 
a(n) = -8 + 6n - 6 
a(n) = 6n - 14 

Then you can plug in n = 50: 
a(50) = 6(50) - 14 
= 300 - 14 
= 286

4 0

3 years ago