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Blababa [14]
2 years ago
5

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

You might be interested in
. 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<br><br> What is 20 + 57 (45x56), (12÷50)
ValentinkaMS [17]

#1

\\ \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-1B)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}

B

\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}

C

\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}

D

\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}

E

\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
A car travels 9 miles in 7 minutes. How fast is it going in miles per hour? How fast is it going in meters per second?
Tems11 [23]

Answer:

he's going 77mph and then 2069 meters per hour.

8 0
1 year ago
How many times must we toss a coin to ensure that a 0.95-confidence interval for the probability of heads on a single toss has l
musickatia [10]

Answer:

(1) 97

(2) 385

(3) 9604

Step-by-step explanation:

The (1 - <em>α</em>) % confidence interval for population proportion is:

CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}

The margin of error in this interval is:

MOE= z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}

The formula to compute the sample size is:

\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}

(1)

Given:

\hat p = 0.50\\MOE=0.1\\z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

*Use the <em>z</em>-table for the critical value.

Compute the value of <em>n</em> as follows:

\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}\\=\frac{1.96^{2}\times0.50\times(1-0.50)}{0.1^{2}}\\=96.04\\\approx97

Thus, the minimum sample size required is 97.

(2)

Given:

\hat p = 0.50\\MOE=0.05\\z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

*Use the <em>z</em>-table for the critical value.

Compute the value of <em>n</em> as follows:

\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}\\=\frac{1.96^{2}\times0.50\times(1-0.50)}{0.05^{2}}\\=384.16\\\approx385

Thus, the minimum sample size required is 385.

(3)

Given:

\hat p = 0.50\\MOE=0.01\\z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

*Use the <em>z</em>-table for the critical value.

Compute the value of <em>n</em> as follows:

\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}\\=\frac{1.96^{2}\times0.50\times(1-0.50)}{0.01^{2}}\\=9604

Thus, the minimum sample size required is 9604.

8 0
2 years ago
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