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2 months ago

ΔMNO was dilated, then _____________, to create ΔYHQ.

2 answers:

8 0

The answer is rotated. When dilating figures, we either increase or decrease their dimensions. Knowing the scale factor is crucial as it indicates whether the figure enlarges or reduces; a positive scale factor suggests expansion, while a negative one indicates reduction. The center of dilation is where lines connecting the original and dilated shapes intersect. Conversely, to rotate a shape involves turning it around a designated fixed point known as the center of rotation.

Inessa [12.5K]2 months ago

3 0

The answer is A.

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If mJI = (3x+2)°, mHLK = (15x-36)°, and m∠HML = (8x-1)°, find mHLK

lawyer [12517]

Response:

The measure of mHLK is "(204)°".

Step-by-step breakdown:

Given values include:

mJI = (3x+2)°

mHLK = (15x-36)°

and,

m∠HML = (8x-1)°

then,

What is mHLK?

Now,

Utilizing the chord-chord angle formula, we find

$mHMK=\frac{1}{2}(mJL+mHLK)$

Inserting the known values into the equation gives us

⇒ $(8x-1)=\frac{1}{2}(15x-36+3x+2)$

By carrying out cross-multiplication, we arrive at

⇒ $2(8x-1)=18x-34$

⇒ $16x-2=18x-34$

By subtracting "18x" from both sides, we obtain

⇒ $16x-2-18x=18x-34-18x$

⇒ $-2x-2=-34$

Upon adding "2" to both sides, we end up with

⇒ $-2x=-34+2$

⇒ $-2x=-32$

⇒ $x=\frac{32}{2}$

⇒ $x=16$

By substituting the value of "x" into mHLK = (15x-36)°, we calculate

⇒ (15x-36)° = (15×16-36)°

⇒ = (240-36)°

⇒ = (204)°

Thus, mHLK = (204)°

4 0

2 months ago

A batch of 20 semiconductor chips is inspected by choosing a sample of 3 chips. Assume 10 of the chips do not conform to custome

Svet_ta [12734]

Answer:

To inspect a batch consisting of 20 semiconductor chips, a sample of 3 is selected. Out of these, 10 chips fail to meet customer specifications.

a) Total distinct samples possible = 20C3 = $\frac{20!}{3!(17!)}$ =1140

b) For exactly 2 good chips and 1 bad chip

Total samples = 10C2 * 10C1 = 45 * 10 =450

c) Combinations of 2 good 1 bad, 1 good 2 bad, and 3 bad chips

Total samples = 10C2 * 10C1 + 10C1 * 10C2 + 10C3

= $450+450+120=1020$

6 0

2 months ago

What is the nearest ten thousand, the population of Vermont was estimated to be about 620,000 in 2008 . What might have been the

AnnZ [12381]

Answer:

The exact population of Vermont in 2008 could have been around 618,000

Step-by-step explanation:

* Here's how rounding to the nearest ten thousand works:

- Numbers ending with the last four digits between 0001 and 4999 are rounded down to the nearest lower multiple of ten thousand

- Example: 83,525 rounds down to 80,000.

- If the last four digits are 5000 or above, round up to the next higher ten thousand

- Example: 58,988 rounds up to 60,000

* Applying this rule to the problem given:

Since the rounded population is 620,000 to the nearest ten thousand, the actual population could be any value with the last four digits from 0001 to 4999 (like 618,000) or from 5000 upwards (like 624,000).

Therefore, 618,000 might represent the actual population of Vermont in 2008

6 0

3 months ago