Write the expression as a constant , a single trigonometric function, or a power of a trigonometric function. Sin x cos x/ tan x

Answers

Answer 1

Answer:

The expression a single Trigonometric function $(sin(x) * cos(x)) / tan(x) simplifies\ to \ 1.$

Let's express the expression $(sin(x) * cos(x)) / tan(x)$ in terms of a single trigonometric function or a power of a trigonometric function, we can simplify it as follows:

$(sin(x) * cos(x)) / tan(x)$

First, we can use the trigonometric identity: tan(x) = sin(x) / cos(x).

$(sin(x) * cos(x)) / (sin(x) / cos(x))$

Now, we can simplify by canceling out common terms in the numerator and denominator:

$(sin(x) * cos(x)) * (cos(x) / sin(x))$

Now, we can see that the sin(x) term in the numerator cancels out with the sin(x) term in the denominator, and the cos(x) term in the denominator cancels out with the cos(x) term in the numerator:

= 1

$the \ expression (sin(x) * cos(x)) / tan(x) simplifies\ to \ 1.$

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Answer 2

Answer:

Answer:

cos²x

Step-by-step explanation:

Using the trigonometric identity

• tanx = $(sinx)/(cosx)$ , then

$(sinxcosx)/(tanx)$

= $(sinxcosx)/((sinx)/(cosx) )$

= sinxcosx × $(cosx)/(sinx)$

cancel the sinx in the multiplier with the sin x on the denominator

= cos²x

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Find the remainder when 1 + 2 + 2^2 + 2^3 + ... + 2^100 is divided by 7.Thanks in advance!

Answers

Answer:

Step-by-step explanation:

This is a geometric series so the sum is:

a1 * r^n - 1 / (r - 1)

= 1 * (2^101 -1) / (2-1)

= 2^101 - 1.

Find the remainder when 2^101 is divided by 7:

Note that 101 = 14*7 + 3 so

2^101 = 2^(7*14 + 3) = 2^3 * (2^14)^7 = 8 * (2^14)^7.

By Fermat's Little Theorem (2^14) ^ 7 = 2^14 mod 7 = 4^7 mod 7.

So 2^101 mod 7 = (8 * 4^7) mod 7

= (8 * 4) mod 7

= 32 mod 7

= 4 = the remainder when 2^101 is divided by 7.

So the remainder when 2^101- 1 is divided by 7 is 4 - 1 = 3..

Which ordered pair in the form (a, b) is a solution of this equation?7a – 5b = 28

A.
(4, 0)

B.
(0, 4)

C.
(–2, –3)

D.
(–3, –2)

Answers

$A. \n (4,0) \na=4 \n b=0 \n \Downarrow \n 7 * 4-5 * 0 \stackrel{?}{=} 28 \n 28-0 \stackrel{?}{=} 28 \n 28\stackrel{?}{=} 28 \n 28=28 \ntrue$

$B. \n (0,4) \na=0 \n b=4 \n \Downarrow \n7 * 0 - 5 * 4 \stackrel{?}{=} 28 \n0-20 \stackrel{?}{=} 28 \n-20 \stackrel{?}{=} 28 \n-20 \not= 28 \nnot \ true$

$C. \n(-2,-3) \na=-2 \n b=-3 \n \Downarrow \n7 * (-2) - 5 * (-3) \stackrel{?}{=} 28 \n-14+15 \stackrel{?}{=} 28 \n1 \stackrel{?}{=} 28 \n1 \not= 28 \nnot \ true$

$D. \n(-3,-2) \na=-3 \n b=-2 \n \Downarrow \n7 * (-3) -5 * (-2) \stackrel{?}{=} 28 \n-21 +10 \stackrel{?}{=} 28 \n-11 \stackrel{?}{=} 28 \n-11 \not= 28 \n not \ true$

The answer is A.

Whats the distance between -30 and 91

Answers

121 I want some black

How do you solve number 13 and 14?

Answers

this is what I got hope u get it right gl

Determine whether the polynomial is a difference of squares and if it is, factor it.y2 − 25
A. Is not a difference of squares
B. Is a difference of squares: (y − 5)2
C. Is a difference of squares: (y + 5)(y − 5)
D. Is a difference of squares: (y + 5)2