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azamat
2 years ago
4

What is the length of cd in this diagram ABC ~ EDC.

Mathematics
2 answers:
Mashutka [201]2 years ago
8 0
Notice the angles at vertex C, are congruent, and therefore, both triangles are similar by AA, thus

\bf \cfrac{large}{small}\qquad \cfrac{12-x}{x}=\cfrac{18}{6}\implies 72-6x=18x
\\\\\\
72=24x\implies \cfrac{72}{24}=x\implies 3=x
Lostsunrise [7]2 years ago
6 0

Answer:

B) 3

Step-by-step explanation:

Using the similarity statement, we know that the corresponding sides are AB and ED; AC and EC; BC and DC.  

The ratio of sides AB and ED is unknown, since we do not know either side.

The ratio of sides AC and EC is 18/6.

The ratio of sides BC and CD is (12-x)/x.

Since the triangles are similar, the ratio of corresponding sides is equal; this means we can set up a proportion:

\frac{18}{6}=\frac{12-x}{x}

Cross-multiplying, we have

18(x) = 6(12-x)

Using the dsitributive property, we have

18x = 6(12)-6(x)

18x = 72 - 6x

Adding 6x to each side,

18x+6x = 72-6x+6x

24x = 72

Divide both sides by 24:

24x/24 = 72/24

x = 3

Since the length of CD is x, this means the length of CD is 3.

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The population can be modeled by P(t) = 82.5 − 67.5cos⎡ ⎣(π/6)t ⎤ ⎦, where t is time in months (t = 0 represents January 1) and
Fed [463]

Answer:

The intervals in which the population is less than 20,000 include

(0 ≤ t < 0.74) and (11.26 < t ≤ 12)

Step-by-step explanation:

P(t) = 82.5 - 67.5 cos [(π/6)t]

where

P = population in thousands.

t = time in months.

During a year, in what intervals is the population less than 20,000?

That is, during (0 ≤ t ≤ 12), when is (P < 20)

82.5 - 67.5 cos [(π/6)t] < 20

- 67.5 cos [(π/6)t] < 20 - 82.5

-67.5 cos [(π/6)t] < -62.5

Dividing both sides by (-67.5) changes the inequality sign

cos [(π/6)t] > (62.5/67.5)

Cos [(π/6)t] > 0.9259

Note: cos 22.2° = 0.9259 = cos (0.1233π) or cos 337.8° = cos (1.8767π) = 0.9259

If cos (0.1233π) = 0.9259

Cos [(π/6)t] > cos (0.1233π)

Since (cos θ) is a decreasing function, as θ increases in the first quadrant

(π/6)t < 0.1233π

(t/6) < 0.1233

t < 6×0.1233

t < 0.74 months

If cos (1.8767π) = 0.9259

Cos [(π/6)t] > cos (1.8767π)

cos θ is an increasing function, as θ increases in the 4th quadrant,

[(π/6)t] > 1.8767π (as long as (π/6)t < 2π, that is t ≤ 12)

(t/6) > 1.8767

t > 6 × 1.8767

t > 11.26

Second interval is 11.26 < t ≤ 12.

Hope this Helps!!!

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In ΔABC, and m∠ABC = 90°. D and E are the midpoints of and , respectively. If the length of is 9 units, the length of is units a
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This is an isosceles right triangle (AB = BC & ∠ B=90° - Given)
Then the angles at the base are equal and ∠ CAB = ∠ ACB = 45°

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