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2 years ago

Find missing angle 100° 100° X=

Mathematics

1 answer:

Ghella [55]2 years ago

4 0

Answer:

200 + X

Step-by-step explanation:

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Mice21 [21]

Answer:

The answer is C.

Step-by-step explanation:

When you look at the number line for C, you can see that it starts from positive 2 and shows the distance all the way to -5.

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3 years ago

55% of what number is 33%?

pishuonlain [190]

Answer:

33% is 55% of 0.6

Step-by-step explanation:

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3 years ago

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Ivenika [448]

Answer:

Step-by-step explanation:

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3 years ago

Determine whether each expression below is always, sometimes, or never equivalent to sin x when 0° < x < 90° ? Can someone

Lubov Fominskaja [6]

Answer:

$(a)\ \cos(180 - x)$ --- Never true

$(b)\ \cos(90 -x)$ --- Always true

$(c)\ \cos(x)$ ---- Sometimes true

$(d)\ \cos(2x)$ ---- Sometimes true

Step-by-step explanation:

Given

$\sin(x )$

Required

Determine if the following expression is always, sometimes of never true

$(a)\ \cos(180 - x)$

Expand using cosine rule

$\cos(180 - x) = \cos(180)\cos(x) + \sin(180)\sin(x)$

$\cos(180) = -1\ \ \sin(180) =0$

So, we have:

$\cos(180 - x) = -1*\cos(x) + 0*\sin(x)$

$\cos(180 - x) = -\cos(x) + 0$

$\cos(180 - x) = -\cos(x)$

$-\cos(x) \ne \sin(x)$

Hence: (a) is never true

$(b)\ \cos(90 -x)$

Expand using cosine rule

$\cos(90 -x) = \cos(90)\cos(x) + \sin(90)\sin(x)$

$\cos(90) = 0\ \ \sin(90) =1$

So, we have:

$\cos(90 -x) = 0*\cos(x) + 1*\sin(x)$

$\cos(90 -x) = 0+ \sin(x)$

$\cos(90 -x) = \sin(x)$

Hence: (b) is always true

$(c)\ \cos(x)$

$\sin(x) = \cos(x)$

Then:

$x + x = 90$

$2x = 90$

Divide both sides by 2

$x = 45$

(c) is only true for $x = 45$

Hence: (c) is sometimes true

$(d)\ \cos(2x)$

$\sin(x) = \cos(2x)$

Then:

$x + 2x = 90$

$3x = 90$

Divide both sides by 2

$x = 30$

(d) is only true for $x = 30$

Hence: (d) is sometimes true

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3 years ago

<img src="https://tex.z-dn.net/?f=%20%7Bx%7D%5E%7B2%7D%7Bsin%20x%7D%5E%7B2%7D%20" id="TexFormula1" title=" {x}^{2}{sin x}^{2} "

suter [353]

Answer:

2x(sinx^2 + x^2cosx^2).

Step-by-step explanation:

We use the product rule:

If u and v are functions of x then:-

d(uv)/dx = u dv/dx + v du/dx

So:

d(x^2 sinx^2)dx

= x^2 * d(sinx^2)/dx + sinx^2 * d(x^2)/dx

= x^2 * 2xcosx^2 + sinx^2 * 2x

= 2x^3cosx^2 + 2xsinx^2

= 2x(sinx^2 + x^2cosx^2).

5 0

3 years ago