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4 years ago

For a-b, reduce to the least non-negative residue

Mathematics

1 answer:

Olenka [21]4 years ago

3 0

Answer:

a. 6

b. 9

Step-by-step explanation:

a. The product modulo 7 can be found from the product of the individual numbers modulo 7:

(88·95·36·702) mod 7 = (88 mod 7)·(95 mod 7)·(36 mod 7)·(703 mod 7) mod 7

= (4·4·1·3) mod 7 = 48 mod 7 = 6

b. Powers of 4 mod 11 repeat with period 5:

4 mod 11 = 4

4^2 mod 11 = 5

4^3 mod 11 = 9

4^4 mod 11 = 3

4^5 mod 11 = 1

So, 4^83 mod 11 = 4^3 mod 11 = 9

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8 is what percent of 400?

Nostrana [21]

8 = x% * 400

8 = x/100 * 400

8 = x * 4

4x = 8

Divide 4 on both sides.

x = 8/4

x = 2

2 percent<span> of 400 is 8.</span>

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Dima020 [189]

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3 years ago

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If sin theta = (4)/(7), theta in quadrant II, find the exact value of (a) cos theta (b) sin (theta + (pi) / (6) ) (c) cos (the

EleoNora [17]

Answer:

a) $\cos(\theta) = \frac{\sqrt[]{33}}{7}$

b) $\sin(\theta + \frac{\pi}{6})\frac{-3\sqrt[]{11}+4}{14}$

c) $\cos(\theta-\pi)=\frac{\sqrt[]{33}}{7}$

d) $\tan(\theta + \frac{\pi}{4}) = \frac{\frac{-4}{\sqrt[]{33}}+1}{1+\frac{4}{\sqrt[]{33}}}$

Step-by-step explanation:

We will use the following trigonometric identities

$\sin(\alpha+\beta) = \sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)$

$\cos(\alpha+\beta) = \cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)$ $\tan(\alpha+\beta) = \frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}$ .

Recall that given a right triangle, the sin(theta) is defined by opposite side/hypotenuse. Since we know that the angle is in quadrant 2, we know that x should be a negative number. We will use pythagoras theorem to find out the value of x. We have that

$x^2+4^2 = 7 ^2$

which implies that $x=-\sqrt[]{49-16} = -\sqrt[]{33}$ . Recall that cos(theta) is defined by adjacent side/hypotenuse. So, we know that the hypotenuse is 7, then

$\cos(\theta) = \frac{-\sqrt[]{33}}{7}$

b)Recall that $\sin(\frac{\pi}{6}) =\frac{1}{2} , \cos(\frac{\pi}{6}) = \frac{\sqrt[]{3}}{2}$ , then using the identity from above, we have that

$\sin(\theta + \frac{\pi}{6}) = \sin(\theta)\cos(\frac{\pi}{6})+\cos(\alpha)\sin(\frac{\pi}{6}) = \frac{4}{7}\frac{1}{2}-\frac{\sqrt[]{33}}{7}\frac{\sqrt[]{3}}{2} = \frac{-3\sqrt[]{11}+4}{14}$

c) Recall that $\sin(\pi)=0, \cos(\pi)=-1$ . Then,

$\cos(\theta-\pi)=\cos(\theta)\cos(\pi)+\sin(\theta)\sin(\pi) = \frac{-\sqrt[]{33}}{7}\cdot(-1) + 0 = \frac{\sqrt[]{33}}{7}$

d) Recall that $\tan(\frac{\pi}{4}) = 1$ and $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}=\frac{-4}{\sqrt[]{33}}$ . Then

$\tan(\theta+\frac{\pi}{4}) = \frac{\tan(\theta)+\tan(\frac{\pi}{4})}{1-\tan(\theta)\tan(\frac{\pi}{4})} = \frac{\frac{-4}{\sqrt[]{33}}+1}{1+\frac{4}{\sqrt[]{33}}}$

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3 years ago