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3 years ago

The sum of two numbers is 15 and their product is 54.What are the two numbers?

Mathematics

1 answer:

PSYCHO15rus [73]3 years ago

5 0

6 and 9. 6+9=15 and 6x9=54. hope this helps! :))

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3 years ago

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3 years ago

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Step-by-step explanation:

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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

2 years ago

An urn contains a very large number of marbles of four diﬀerent colors: red, orange, yellow, and green. A sample of 12 marbles d

Umnica [9.8K]

Answer:

We can reject the hypothesis at the 0.05 signiﬁcance level.

Step-by-step explanation:

given that an urn contains a very large number of marbles of four diﬀerent colors: red, orange, yellow, and green. A sample of 12 marbles drawn at random from the urn revealed 2 red, 5 orange, 4 yellow, and 1 green marble.

H0: All colours are equally likely

Ha: atleast two colours are not equally likely

(Two tailed chi square test at 5% significance level)

color Red Orange Yellow Green total

Observed O 2 5 4 1 12

Expected E 3 3 3 3 12

(O-E)^2/E 0.3333 5.3333 0.3333 5.3333 11.3333

chi square = 11.3333

df =3

p value = 0.010055

Since p < 0.05 we reject H0

We can reject the hypothesis at the 0.05 signiﬁcance level.

3 0

3 years ago