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2 years ago

How many terms in this question?

Mathematics

1 answer:

trapecia [35]2 years ago

3 0

Answer:

Step-by-step explanation:

a and b are both terms

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87g of sugar is needed to make 10 cakes. How much sugar is needed for 7 cakes?

nataly862011 [7]

87g divided by 10 = 8.7
8.7 times 7 = 60.9
60.9g of sugar is needed to make 7 cakes
Hope this helps! Feel free to thanks me, rate me 5/5 and make me brainliest!!!

7 0

3 years ago

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Use the law of cosines to find the value of 2*3*4 cos theta.

daser333 [38]

The law of cosines :

A^2 = B^2 + C^2 - 2BC cos A

2^2 = 3^2 + 4^2 - 2 . 3. 4 cos theta

= 21

hope this helps

3 0

3 years ago

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Your high school is having a competition where students have to guess how many candy canes are in a jar. The tens digit is 4 mor

rodikova [14]

Answer:

Step-by-step explanation:

let ones be x

tens is 4 + 2x

4 + 2x + x = 4

3x = 0

x = 0

tens is 4 + 2*0

tens = 4

40 candy canes

5 0

2 years ago

First question, thanks. I believe there should be 3 answers

zysi [14]

Given: The following functions

$A)cos^2\theta=sin^2\theta-1$ $B)sin\theta=\frac{1}{csc\theta}$ $\begin{gathered} C)sec\theta=\frac{1}{cot\theta} \\ D)cot\theta=\frac{cos\theta}{sin\theta} \\ E)1+cot^2\theta=csc^2\theta \end{gathered}$

To Determine: The trigonometry identities given in the functions

Solution

Verify each of the given function

$\begin{gathered} cos^2\theta=sin^2\theta-1 \\ Note\text{ that} \\ sin^2\theta+cos^2\theta=1 \\ cos^2\theta=1-sin^2\theta \\ Therefore \\ cos^2\theta sin^2\theta-1,NOT\text{ }IDENTITIES \end{gathered}$

$\begin{gathered} sin\theta=\frac{1}{csc\theta} \\ Note\text{ that} \\ csc\theta=\frac{1}{sin\theta} \\ sin\theta\times csc\theta=1 \\ sin\theta=\frac{1}{csc\theta} \\ Therefore \\ sin\theta=\frac{1}{csc\theta},is\text{ an identities} \end{gathered}$

$\begin{gathered} sec\theta=\frac{1}{cot\theta} \\ note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ tan\theta cot\theta=1 \\ tan\theta=\frac{1}{cot\theta} \\ Therefore, \\ sec\theta\ne\frac{1}{cot\theta},NOT\text{ IDENTITY} \end{gathered}$

$\begin{gathered} cot\theta=\frac{cos\theta}{sin\theta} \\ Note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ cot\theta=1\div tan\theta \\ tan\theta=\frac{sin\theta}{cos\theta} \\ So, \\ cot\theta=1\div\frac{sin\theta}{cos\theta} \\ cot\theta=1\times\frac{cos\theta}{sin\theta} \\ cot\theta=\frac{cos\theta}{sin\theta} \\ Therefore \\ cot\theta=\frac{cos\theta}{sin\theta},is\text{ an Identity} \end{gathered}$

$\begin{gathered} 1+cot^2\theta=csc^2\theta \\ csc^2\theta-cot^2\theta=1 \\ csc^2\theta=\frac{1}{sin^2\theta} \\ cot^2\theta=\frac{cos^2\theta}{sin^2\theta} \\ So, \\ \frac{1}{sin^2\theta}-\frac{cos^2\theta}{sin^2\theta} \\ \frac{1-cos^2\theta}{sin^2\theta} \\ Note, \\ cos^2\theta+sin^2\theta=1 \\ sin^2\theta=1-cos^2\theta \\ So, \\ \frac{1-cos^2\theta}{sin^2\theta}=\frac{sin^2\theta}{sin^2\theta}=1 \\ Therefore \\ 1+cot^2\theta=csc^2\theta,\text{ is an Identity} \end{gathered}$

Hence, the following are identities

$\begin{gathered} B)sin\theta=\frac{1}{csc\theta} \\ D)cot\theta=\frac{cos\theta}{sin\theta} \\ E)1+cot^2\theta=csc^2\theta \end{gathered}$

The marked are the trigonometric identities

3 0

1 year ago

Ahhhh i need help again.

qwelly [4]

Answer:

Step-by-step explanation:

A. these simply to 21 + 7x and 21 + 3x

this one is incorrect because the 7x and 3x are different

B. these simplify to 10 + x and 10 + x

this is the correct one since they are exactly the same

C. these simplify to 21x and 7x + 21

these two are not the same because one expression has only a number with a variable but the other one has a number with a variable <em>and </em>another number

D. these simplify to $\frac{7}{3-x}$ and $\frac{7}{x-3}$

in subtraction, the order of the 3 and the x matters, so the equations are not the same
example: 6 - 4 = 2 but 4 - 6 = -2

3 0

2 years ago

How many terms in this question?​

How many terms in this question?