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yarga [219]
3 years ago
15

Match the following items. Plane Point Line Ray Line segment

Mathematics
1 answer:
Sophie [7]3 years ago
8 0

Answer:

The first one is a line because a line extends infinitely in both directions which the arrows indicate.

The second one is a plane since it is a 2-dimensional shape.

The last one depicts 2 rays because rays have one endpoint, but extend infinitely in the other direction.

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You want to know how long it'll take for the bacteria to grow
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The number of double-times it'll take is 

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The double-time for these particular critters is 12 hours,
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6. If 55% of the selling price is markup, the remainder (45%) is the cost. Gaitan can pay up to 45% of 489.50 = $220.28.

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... 0.48 × selling price = cost

... selling price = cost/0.48 = $38.87/0.48 = $80.98

8. If 40% of the selling price is markup, the remainder (60%) is the cost, $12.95.

... 0.60 × selling price = $12.95

... selling price = $12.95/0.60 = $21.58

9. If 35% of the selling price is markup, the remainder (65%) is the cost.

... 0.65 × $24.00/dz = $15.60/dz

10. If 60% of the selling price is markup, the remainder (40%) is the cost. Veronica can pay up to 40% of $600.00 = $240 for the jacket.

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2 years ago
Simplify the expression.<br><br> 7y + 4 + 9 - 2y PLEASE HELP ME MY LIFE DEPENDS ON IT
balu736 [363]

Answer:5y+13

Step-by-step explanation:

6 0
3 years ago
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Find all solutions of each equation on the interval 0 ≤ x &lt; 2π.
Korvikt [17]

Answer:

x = 0 or x = \pi.

Step-by-step explanation:

How are tangents and secants related to sines and cosines?

\displaystyle \tan{x} = \frac{\sin{x}}{\cos{x}}.

\displaystyle \sec{x} = \frac{1}{\cos{x}}.

Sticking to either cosine or sine might help simplify the calculation. By the Pythagorean Theorem, \sin^{2}{x} = 1 - \cos^{2}{x}. Therefore, for the square of tangents,

\displaystyle \tan^{2}{x} = \frac{\sin^{2}{x}}{\cos^{2}{x}} = \frac{1 - \cos^{2}{x}}{\cos^{2}{x}}.

This equation will thus become:

\displaystyle \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} \cdot \frac{1}{\cos^{2}{x}} + \frac{2}{\cos^{2}{x}} - \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} = 2.

To simplify the calculations, replace all \cos^{2}{x} with another variable. For example, let u = \cos^{2}{x}. Keep in mind that 0 \le \cos^{2}{x} \le 1 \implies 0 \le u \le 1.

\displaystyle \frac{1 - u}{u^{2}} + \frac{2}{u} - \frac{1 - u}{u} = 2.

\displaystyle \frac{(1 - u) + u - u \cdot (1- u)}{u^{2}} = 2.

Solve this equation for u:

\displaystyle \frac{u^{2} + 1}{u^{2}} = 2.

u^{2} + 1 = 2 u^{2}.

u^{2} = 1.

Given that 0 \le u \le 1, u = 1 is the only possible solution.

\cos^{2}{x} = 1,

x = k \pi, where k\in \mathbb{Z} (i.e., k is an integer.)

Given that 0 \le x < 2\pi,

0 \le k.

k = 0 or k = 1. Accordingly,

x = 0 or x = \pi.

8 0
2 years ago
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