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Elden [556K]
3 years ago
6

How can a translation and a reflection be used to map ΔHJK to ΔLMN? Translate K to N and reflect across the line containing HJ.

Translate K to N and reflect across the line containing JK. Translate H to L and reflect across the line containing JK. Translate K to L and reflect across the line containing HJ.

Mathematics
2 answers:
aliya0001 [1]3 years ago
9 0

Answer:

Its B on ED2020

Step-by-step explanation:

Svet_ta [14]3 years ago
3 0

Answer:

<u>Translate K to N and reflect across the line containing JK. </u>

Step-by-step explanation:

The rest of the question is the attached figure.

From the figure, we can deduce the following:

∠K = ∠N

JK = MN

HK = LN

So, N will be the image of K

By translating K to N, The segment JK will over-lap the segment MN,

Then, we need to reflect the point H across the the line containing JK to get the point L

So, the translation and a reflection that will be used to map ΔHJK to ΔLMN:

<u>Translate K to N and reflect across the line containing JK. </u>

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The price of paper increases by a lot. Book producers respond by supplying ____ books.
lubasha [3.4K]
If the price of paper increases by a lot, then book producers would respond by supplying less books.
5 0
2 years ago
Find the volume of the square pyramid, given its slant height​
Reptile [31]

Answer:

Step-by-step explanation:

base length b = 32 cm

slant height L = 30 cm

 

height h = √(L²-(½b)²)

= √(30²-0.25·32²)

≅ 25.3771550809 cm

 

volume V = (⅓)b²h

= (⅓)32²·25.38

≅ 8662.07 cm³

7 0
2 years ago
The result of rounding the whole number 2,746,052 to the nearest hundred thousands place is:
Rasek [7]
The answer would be 2,700,000.
This is because of the rules of rounding.
If a digit is 4 or smaller, it rounds down.
If a digit is 5 or more, it rounds up.
The number is 2, 746, 052 so it stays at 700,000.

TLDR: 2,700,000
6 0
2 years ago
The average amount of time that students use computers at a university computer center is 36 minutes with a standard deviation o
frosja888 [35]

Answer:

2119 students use the computer for more than 40 minutes. This number is higher than the threshold estabilished of 2000, so yes, the computer center should purchase the new computers.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 36, \sigma = 5

The first step to solve this question is finding the proportion of students which use the computer more than 40 minutes, which is 1 subtracted by the pvalue of Z when X = 40. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{40 - 36}{5}

Z = 0.8

Z = 0.8 has a pvalue of 0.7881.

1 - 0.7881 = 0.2119

So 21.19% of the students use the computer for longer than 40 minutes.

Out of 10000

0.2119*10000 = 2119

2119 students use the computer for more than 40 minutes. This number is higher than the threshold estabilished of 2000, so yes, the computer center should purchase the new computers.

8 0
3 years ago
Anyone know this answer?
damaskus [11]
La of sine:

sinC/c = sinB/b==> sin  37°/8 = sin B/12 ==> sin B = 0.903


arcsinB or sin⁻¹ B = 64.5°, & sin (B°) = sin (180° - B°), then

sin(64.5) = sin(180°-64.5°) ==> B = 64.5° or 115.5°
6 0
2 years ago
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