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4 years ago

Another way to write 1/2

Mathematics

2 answers:

Pie4 years ago

5 0

Another way to write 1/2 is .5

earnstyle [38]4 years ago

3 0

½=0.5.................

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Which package has the lowest cost per ounce of rice ( 12, 18, 7)

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The package with the most low cost is 7.

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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

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3 years ago

A student has taken three math tests so far this semester. His scores for the first three tests were 71, 75, and 79. Part A: Sup

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Answer:

The grade of this sixth and (final) test <u>would be 91</u>.

Step-by-step explanation:

This is an arithmetic series since each time the data is increasing by the common difference of 4, and we already know a₁ which is the first term in the sequence is 71, therefore we can say:

a₁ = 71

r = 4

a₆ = the sixth term that needs to be found out

n = 6

Formula of an arithemetic series:

aₙ = a₁ + r(n - 1)

a₆ = 71 + 4(6 - 1)

a₆ = 71 + 4(5)

a₆ = 71 + 20

a₆ = 91

Hence, the grade of this sixth and (final) test would be 91.

Hope this helps!

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