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3 years ago

6

Suppose the primes p and q used in the rsa algorithm are consecutive primes. how would you factor n

1 answer:

zmey [24]3 years ago

8 0

I would look for the first prime larger (or smaller) than √n.

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The angle θ 1 is located in Quadrant IV, and cos ⁡ ( θ 1 ) = 9/ 19 , theta, start subscript, 1, end subscript, right parenthesis

Firdavs [7]

Answer:

$sin\theta_1 = - \frac{2\sqrt{70}}{19}$

Step-by-step explanation:

We are given that $\theta_1$ is in <em>fourth</em> quadrant.

$cos\theta_1$ is always positive in 4th quadrant and

$sin\theta_1$ is always negative in 4th quadrant.

Also, we know the following identity about $sin\theta$ and $cos\theta$ :

$sin^2\theta + cos^2\theta = 1$

Using \theta_1 in place of \theta:

$sin^2\theta_1 + cos^2\theta_1 = 1$

We are given that $cos\theta_1 = \frac{9}{19}$

$\Rightarrow sin^2\theta_1 + \dfrac{9^2}{19^2} = 1\\\Rightarrow sin^2\theta_1 = 1 - \dfrac{81}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{361-81}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{280}{361}\\\Rightarrow sin\theta_1 = \sqrt{\dfrac{280}{361}}\\\Rightarrow sin\theta_1 = +\dfrac{2\sqrt{70}}{19}, -\dfrac{2\sqrt{70}}{19}$

$\theta_1$ is in <em>4th quadrant </em>so $sin\theta_1$ is negative.

So, value of $sin\theta_1 = - \frac{2\sqrt{70}}{19}$

6 0

2 years ago

What is the value of the square of i?

BlackZzzverrR [31]

I^2 = (sqrt-1)^2 then the square cancels out so the answer is just

-1

3 0

3 years ago

Aldon and Jamal raised the same amount of money for the school fundraiser. Aldon

Naddika [18.5K]

Answer: $5

Step-by-step explanation: This can be represented by the equation

40 + 12T = 25 + 15T

We just solve for T.

Subtract 12T from both sides

40 = 25 + 3T Subtract 25

15 = 3T Divide by 3

5 = T

7 0

3 years ago

Is y = -4 + 2x linear?

BlackZzzverrR [31]

Answer: it is linear

Step-by-step explanation: Find the degree of the equation to determine if linear.

5 0

2 years ago

Read 2 more answers

How do I convert standard form to point slope

Fittoniya [83]

The point-slope form:

$y-y_1=m(x-x_1)$

$(x_1,\ y_1)$ - given point

$m$ - given slope

The standard form:

$Ax+By=C$

$y-y_1=m(x-x_1)$ <em>use distributive property</em>

$y-y_1=mx-mx_1$ <em>add $y_1$ to both sides</em>

$y=mx-mx_1+y_1$ <em>subtract $mx$ from both sides</em>

$-mx+y=-mx_1+y_1$ <em>change the signs</em>

$mx-y=mx_1-y_1$

8 0

2 years ago