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Dvinal [7]
2 years ago
6

An object's __________ is simply the data that is stored in the object's fields at any given moment. A. value B. assessment C. r

ecord D. state
Mathematics
1 answer:
Veseljchak [2.6K]2 years ago
8 0
I believe the answer is c.
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Triangle OMG has vertices at O(−3,3), M(4,2), and G(−3,1). Point G is rotated 90° counterclockwise around the origin, forming G'
Irina18 [472]

Answer:

O'(-3,-3) M'(-2,4) G'(-1-3)

Step-by-step explanation:

Rule: rotation of (x,y) counterclockwise around the origin gives (-y,x)

Thus

O(-3,3) will be O'(-3,-3)

M(4,2) will be M'(-2,4)

G(-3,1) will be G'(-1,-3)

Therefore O'(-3,-3),M'(-2,4) and G'(-1,-3) would be the answer.

3 0
2 years ago
On a coordinate plane, a line is drawn from point K to point J. Point K is at (160, 120) and point J is at (negative 40, 80).
Nadusha1986 [10]

Answer:

the answer is (40,96)

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
the jellybean jar has a radius of 4.8 cm and a height of 11.8 cm. what would be a reasonable lower limit for the number of jelly
azamat

Answer:

           \large\boxed{\large\boxed{100}}

Explanation:

This question comes with four answer choices:

  • 10,000
  • 100,000
  • 100
  • 1

You just need to find a rough estimate of the volume of the jar and the volume of one jelly bean. This is called magnitude order.

The <em>jellybean ja</em>r is cylindrical, Thus, its volume is \pi \times radius^2\times height

To find the order of magnitude, you just use the numbers rounded to one signficant figure: round π to 3, the radius to 5, cm, and the height to 10:

            Volume\approx 3cm\times (5cm)^2\times10cm=750cm^3\approx1,000cm^3

The order of magnitude for the radius of a jellybean is 1 cm. And the order of magnitude of the volume of a sphere with a radius of 1 cm is the cube of the diameter (2cm):

             (2cm)^3=8cm^3\approx10cm^3

Hence, a reasonable lower limit for the number of jellybeans in the jar is:

                     1,000cm^3/10cm^3=100

7 0
2 years ago
Prajna is listing squares of two-digit numbers in such a way that the digit in their units place is 6.
dem82 [27]

Answer:

2 (option B)

Question:

Prajna is listing squares of two-digit numbers in such a way that the digit in their units place is 6.

How many numbers should he have listed?

a) 9 b) 2 c)18 d) 20

Step-by-step explanation:

We would find the perfect squares that are two digit numbers

For two digit perfect squares:

Largest two digit number= 81 = 9²

Smallest two digit number= 16 = 4²

Total Two digit number perfect squares = Highest number - lowest number + 1

= 9-4+1 = 6

There are 6 two digit perfect squares

Next, we have to find how many of them have 6 in their units place.

The rule to be applied: If a number has 4 or 6 in the unit place, then its square has 6 in the units place.

Following this, the possibilities are 4², 6².

We have two of such numbers.

Therefore, he would have listed 2 numbers.

6 0
2 years ago
Which ratio is also equal to StartFraction R T Over R X EndFraction and StartFraction R S Over R Y EndFraction? StartFraction X
jekas [21]

Answer:

△RST ~ △RYX by the SSS similarity theorem. Which ratio is also equal to RT/RX and RS/RY ?

A.XY/TS

B.SY/RY

C.RX/XT

D.ST/YX

Step-by-step explanation:

Check attachment for solution

5 0
2 years ago
Read 2 more answers
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