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Leviafan [203]
3 years ago
13

Please answer the question in the image below ASAP

Mathematics
2 answers:
attashe74 [19]3 years ago
6 0
Option A would run parallel because they have the same slope.
dolphi86 [110]3 years ago
5 0
I would say option a because they slope is running above the x axis
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A 4-ounce serving of yogurt has 8 grams of protein. How many grams of protein
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Answer: which one

Step-by-step explanation:

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A student makes and sells designer T-shirts to earn extra cash. The material needed to make each T-shirt costs $4.25. Currently,
natita [175]

Step-by-step explanation:

Let x be the number of designer T shirts.

To break even,

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3 years ago
Write the first six terms of the geometric sequence with the first term 6 and common ratio 1/3
Alja [10]

Answer:

6, 2, 2/3, 2/9, 2/27, 2/81

Step-by-step explanation:

The nth term of a geometric progression is expressed as;

Tn  = ar^n-1

a is the first term

n is the number of terms

r is the common ratio

Given

a = 6

r = 1/3

when n = 1

T1 = 6(1/3)^1-1

T1 = 6(1/3)^0

T1 = 6

when n = 2

T2= 6(1/3)^2-1

T2= 6(1/3)^1

T2 = 2

when n = 3

T3 = 6(1/3)^3-1

T3= 6(1/3)^2

T3= 6 * 1/9

T3 = 2/3

when n = 4

T4 = 6(1/3)^4-1

T4= 6(1/3)^3

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T4 = 2/9

when n = 5

T5 = 6(1/3)^5-1

T5= 6(1/3)^4

T5= 6 * 1/81

T5 = 2/27

when n = 6

T6 = 6(1/3)^6-1

T6= 6(1/3)^5

T6= 6 * 1/243

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Hence the first six terms are 6, 2, 2/3, 2/9, 2/27, 2/81

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3 years ago
Sin theta+costheta/sintheta -costheta+sintheta-costheta/sintheta+costheta=2sec2/tan2 theta -1
sleet_krkn [62]

\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}=\dfrac{2sec^2\theta}{tan^2\theta-1}

From Left side:

\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}\bigg(\dfrac{sin\theta+cos\theta}{sin\theta+cos\theta}\bigg)+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}\bigg(\dfrac{sin\theta-cos\theta}{sin\theta-cos\theta}\bigg)

\dfrac{sin^2\theta+2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}+\dfrac{sin^2\theta-2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}

NOTE: sin²θ + cos²θ = 1

\dfrac{1 + 2cos\theta sin\theta}{sin^2\theta-cos^2\theta}+\dfrac{1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}

\dfrac{1 + 2cos\theta sin\theta+1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}

\dfrac{2}{sin^2\theta-cos^2\theta}

\dfrac{2}{\bigg(sin^2\theta-cos^2\theta\bigg)\bigg(\dfrac{cos^2\theta}{cos^2\theta}\bigg)}

\dfrac{2sec^2\theta}{\dfrac{sin^2\theta}{cos^2\theta}-\dfrac{cos^2\theta}{cos^2\theta}}

\dfrac{2sec^2\theta}{tan^2\theta-1}

Left side = Right side <em>so proof is complete</em>

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