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2 years ago

Please answer asap! Will mark brainliest.

Mathematics

2 answers:

choli [55]2 years ago

5 0

It is too blurry i will try too help if I tell me what too do

Tema [17]2 years ago

3 0

I do belive it's the first one

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. 4/18 times 2/92. 1/9 times 3/63. 12/14 times 1/84.1/7 times 5/65. 17/20 times 1/61. 4 1/4 times 3 1/22. 3 5/6 times 4 1/23. 2

AlladinOne [14]

Answer:

Step-by-step explanation:

4/18×2/92

=1/9×1/23

=1/207

1/9×3/63

=1/3×1/23

=1/69

12/14×1/84

=1/14×1/7

=1/98

1/7×5/65

=1/7×1/13

=1/91

17/20×1/61

=17/1220

4 1/4 ×3 1/22

=17/4×67/22

=1,139/88

=12 83/88

3 5/6×4 1/23

=23/6×93/23

=93/6

=15 3/6

=15 1/2

2 1/2×4 7/84

=5/2×343/84

=1715/168

=10 35/168

I have answered more than 2 from each set

6 0

3 years ago

PLEASE HELLPPPP What is 20 + 57 (45x56), (12÷50)

ValentinkaMS [17]

$\\ \sf\longmapsto 20+57(45\times 56)$

Solve the Bracket

$\\ \sf\longmapsto 20+57(2520)$

$\\ \sf\longmapsto 20+143640$

$\\ \sf\longmapsto 143660$

4 0

2 years ago

Read 2 more answers

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