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2 years ago

If a =5 b =6 and c=a-2 . answer the question , 8a-2b+1c

Mathematics

1 answer:

azamat2 years ago

3 0

Answer:

here <em>given.</em><em>,</em>

a =5 b =6 and c=a-2(5-2=3)

putting the given value in the equation...we get..

=8*5-2*6+1*3

=40-12+3

=43-12

=31

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for her book club, Sharon reads for 45 minutes each day. write an expression for the number of hours she reads in d days

Dafna1 [17]

N=(3/4)*h*d

--------------------

n=Number of hours she reads

h=Hour(s)

d=Days

--------------------

You see, with this formula - when d=2,

n=(3/4)*h*2=(6/4)*h=(3/2)*h=1.5 hours

So, in the space of 2 days, she had spent 1.5 hours reading.

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2 years ago

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Determine the number of subsets of 3,4,5,6,7,8,9,10,11,12,13,14,15

lbvjy [14]

Answer:

8192

Step-by-step explanation:

The number of subsets of an n-element set is 2^n.

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This count includes the empty set and the full set.

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2 years ago

Solve the system of linear equations by substitution

ziro4ka [17]

$y - x = 0 ...equation 1$
$2x - 5y = 9 \: ...equation \: 2$
working with equation 1
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next subsitue equation 3 in equation 1
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$2x - 5x = 9$
$- 3x = 9$
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2 years ago

What is the value of the expression below?<br><br> 0.72/0.36*2

Tanya [424]

Answer:

Step-by-step explanation:

0.72/0.36 x 2

0.72/0.72

I'm always happy to help :)

8 0

3 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