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sergeinik [125]

1 year ago

8

Which equation shows that the Pythagorean identity is true for 0 = 180 degrees? select the equation that is in the form sin^2 0+

cos^2 0=1
A. 0^2 + (-1)^2 = 1
B. 1^2 + 0^2 = 1
C. 0^2 + 1^2 = 1
D. (-1)^2 + 0^2 = 1

1 answer:

elixir [45]1 year ago

8 0

The equation in the form of the given expression is (0)² + (1)² = 1

<h3>Trigonometry identity</h3>

According to some of the trigonometry identity

sin 0 = 0

cos 0 1

Given the expression below

sin^2 0+cos^2 0=1

This can also be expressed as:

(sin0)² + (cos0)² = 1

Substitute

(0)² + (1)² = 1

Hence the equation in the form of the given expression is (0)² + (1)² = 1

Learn more on trig identity here: brainly.com/question/20094605

#SPJ1

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Consider these functions:

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I think the answer might be B

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1 year ago

Tina creates a dot plot using the match scores 3, 2, 4, 5, 2, 3, 1, 2, 1, and 5. Which of the following dot plots represents the

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The answer is

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7 0

3 years ago

I know you want to answer this question.

Alik [6]

Answer:

D. x = 3

Step-by-step explanation:

$\frac{1}{2} ^{x-4}$ - 3 = $4^{x-3}$ - 2

First, convert $4^{x-3}$ to base 2:

$4^{x-3}$ = $(2^{2})^{x-3}$

$\frac{1}{2} ^{x-4}$ - 3 = $(2^{2})^{x-3}$ - 2

Next, convert $\frac{1}{2} ^{x-4}$ to base 2:

$\frac{1}{2} ^{x-4}$ = $(2^{-1})^{x-4}$

$(2^{-1})^{x-4}$ - 3 = $(2^{2})^{x-3}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$(2^{-1})^{x-4}$ = $2^{-1*(x-4)}$

$2^{-1*(x-4)}$ - 3 = $(2^{2})^{x-3}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$(2^{2})^{x-3}$ = $2^{2(x-3)}$

$2^{-1*(x-4)}$ - 3 = $2^{2(x-3)}$ - 2

Apply exponent rule: $a^{b+c}$ = $a^{b}$ $a^{c}$ :

$2^{-1(x-4)}$ = $2^{-1x}$ * $2^{4}$ , $2^{2(x-3)}$ = $2^{2x}$ * $2^{-6}$

$2^{-1 * x}$ * $2^{4}$ - 3 = $2^{2x}$ * $2^{-6}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$2^{-1x}$ = $(2^{x})^{-1}$ , $2^{2x}$ = $(2^{x})^{2}$

$(2^{x})^{-1}$ * $2^{4}$ - 3 = $(2^{x})^{2}$ * $2^{-6}$ - 2

Rewrite the equation with $2^{x} = u$ :

$(u)^{-1}$ * $2^{4}$ - 3 = $(u)^{2}$ * $2^{-6}$ - 2

Solve $u^{-1}$ * $2^{4}$ - 3 = $u^{2}$ * $2^{-6}$ - 2:

$u^{-1}$ * $2^{4}$ - 3 = $u^{2}$ * $2^{-6}$ - 2

Refine:

$\frac{16}{u}$ - 3 = $\frac{1}{64}$ $u^{2}$ - 2

Add 3 to both sides:

$\frac{16}{u}$ - 3 + 3 = $\frac{1}{64}$ $u^{2}$ - 2 + 3

Simplify:

$\frac{16}{u}$ = $\frac{1}{64}$ $u^{2}$ + 1

Multiply by the Least Common Multiplier (64u):

$\frac{16}{u}$ * 64u = $\frac{1}{64}$ $u^{2}$ + 1 * 64u

Simplify:

$\frac{16}{u}$ * 64u = $\frac{1}{64}$ $u^{2}$ + 1 * 64u

Simplify $\frac{16}{u}$ * 64u:

1024

Simplify $\frac{1}{64}$ $u^{2}$ * 64u:

$u^{3}$

Substitute:

1024 = $u^{3}$ + 64u

Solve for u:

u = 8

Substitute back u = $2^{x}$ :

8 = $2^{x}$

Solve for x:

x = 3

4 0

3 years ago

What are the ordered pairs of the

9966 [12]

Start by writing the system down, I will use $y$ to represent $f(x)$

$y=x^2-5x+3\wedge y=-3$

Substitute the fact that $y=-3$ into the first equation to get,

$-3=x^2-5x+3$

Simplify into a quadratic form ( $ax^2+bx+c=0$ ),

$x^2-5x+6=0$

Now you can use Vieta's rule which states that any quadratic equation can be written in the following form,

$x^2+x(m+n)+mn=0$

which then must factor into

$(x+m)(x+n)=0$

And the solutions will be $m,n$ .

Clearly for small coefficients like ours $5,6$ , this is very easy to figure out. To get 5 and 6 we simply say that $m=3, n=2$ .

This fits the definition as $5=3+2$ and $6=2\cdot3$ .

So as mentioned, solutions will equal to $m=3, n=2$ but these are just x-values in the solution pairs of a form $(x,y)$ .

To get y-values we must substitute 3 for x in the original equation and then also 2 for x in the original equation. Luckily we already know that substituting either of the two numbers yields a zero.

So the solution pairs are $(2,0)$ and $(3,0)$ .

Hope this helps :)

5 0

2 years ago