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sdas [7]
3 years ago
8

Urgent help ! similar figures use the similarity relationship to find the indicated value

Mathematics
1 answer:
masya89 [10]3 years ago
5 0

Answer:

FE = 28 and YZ = 28

Step-by-step explanation:

In similar triangles the ratios of corresponding sides are equal.

(4)

\frac{BC}{FG} = \frac{BD}{FE} , substitute values

\frac{39}{4x+2} = \frac{42}{5x-2} ( cross- multiply )

39(5x - 2) = 42(4x + 2) ← distribute parenthesis on both sides

195x - 78 = 168x + 84 ( subtract 168x from both sides )

27x - 78 = 84 ( add 78 to both sides )

27x = 162 ( divide both sides by 27 )

x = 6

Thus

FE = 5x - 2 = 5(6) - 2 = 30 - 2 = 28

(5)

\frac{ST}{SZ} = \frac{RT}{YZ} , substitute values

\frac{40}{35} = \frac{3x-7}{2x+2} ( cross- multiply )

35(3x - 7) = 40(2x + 2) ← distribute parenthesis on both sides

105x - 245 = 80x + 80 ( subtract 80x from both sides )

25x - 245 = 80 ( add 245 to both sides )

25x = 325 ( divide both sides by 25 )

x = 13

Thus

YZ = 2x + 2 = 2(13) + 2 = 26 + 2 = 28

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<h3>Answer: 1</h3>

where x is nonzero

=======================================================

Explanation:

We'll use two rules here

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The portion [ x^(a-b) ]^(a+b) would turn into x^[ (a-b)(a+b) ] after using the first rule shown above. That turns into x^(a^2 - b^2) after using the difference of squares rule.

Similarly, the second portion turns into x^(b^2-c^2) and the third part becomes x^(c^2-a^2)

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After applying rule 1 to each of the three pieces, we will have 3 bases of x with the exponents of (a^2-b^2),  (b^2-c^2) and (c^2-a^2)

Add up those exponents (using rule 2 above) and we get

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2 years ago
Suppose the speeds of cars along a stretch of I-40 is normally distributed with a mean of 70 mph and standard deviation of 5 mph
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Answer:

a) 68%.

b) 84%.

c) Approximately 2.5% of cars are traveling at a speed greater than or equal to 80 mph.

d) Approximately 2.5% of cars are traveling at a speed greater than or equal to 80 mph.

e) Between 60 mph and 80 mph.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

Approximately 68% of the measures are within 1 standard deviation of the mean.

Approximately 95% of the measures are within 2 standard deviations of the mean.

Approximately 99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean of 70 mph, standard deviation of 5 mph.

(a) Approximately what percent of cars are travelling between 65 and 75 mph?

70 - 5 = 65

70 + 5 = 75

Within 1 standard deviation, so approximately 68%.

(b) If the speed limit on this stretch of highway is 65 mph, approximately what percent of cars are traveling faster than the speed limit?

The normal distribution is symmetric, which means that 50% of the measures are below the mean, and 50% are above.

65 is one standard deviation below the mean, so of the cars below the mean, 68% are above 65 mph.

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84%.

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2 standard deviations above the mean.

Approximately 95% of the measures are within 2 standard deviations of the mean. Due to the symmetry of the normal distribution, of the other 5%, 2.5% is at least 2 standard deviations below the mean and 2.5% is at least 2 standard deviations above the mean. Then:

Approximately 2.5% of cars are traveling at a speed greater than or equal to 80 mph.

(d) What percent of cars are traveling at a speed greater than 80 mph?

Same as item c, as in the normal distribution, the probability of an exact value is considered to be 0.

(e) 95% of cars are traveling between what two speeds?

Within two standard deviations of the mean.

70 - 2*5 = 60 mph

70 + 2*5 = 80 mph.

Between 60 mph and 80 mph.

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