Please help I need it ASAP x

Sam rotated parallelogram ABCD 90° clockwise around the origin. If angle A is 130° and angle B is 50°, what is the degree measur

irakobra [83]

Answer-

The measurement of angle A' will be 130° .

Solution-

Transformations like - rotations, reflections, and translations are isometric. That means that these transformations do not change the size of the figure. If the size and shape of the figure is not changed, then the figures are congruent.

It doesn't matter the order or how much degree of rotation has taken place, the final image will be congruent to the original image.

So, even after the rotation of 90° clockwise around the origin, the measurement of angle A will be 130° and so do all the rest of the angles.

8 0

3 years ago

Read 2 more answers

PLEASE HELP MEEEEE <3

Svetradugi [14.3K]

Answer:

The second choice is correct.

Step-by-step explanation:

The softball cannot be anything under 180 grams because that would go past the ten gram limit. As well as the ball cannot go any higher than 200 grams.

8 0

3 years ago

What's the greatest common factor of 36,48,60 plz explain I don't understand

vova2212 [387]

The factors of 36 are: 1, 2, 3, 4, 6, 9, 12, 18, 36

The factors of 48 are: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48

The factors of 60 are: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

Then the greatest common factor is 12.

6 0

3 years ago

Read 2 more answers

Please help, you don't have to show your work. Like at all.

elixir [45]

The slope of the line is -2.

Solution:

Given points are (-3, -7) and (2, -17).

Here $x_1=-3, y_1=-7, x_2=2, y_2=-17$

Slope of the line:

$$m=\frac{y_2-y_1}{x_2-x_1}$

$$m=\frac{-17-(-7)}{2-(-3)}$

$$m=\frac{-17+7}{2+3}$

$$m=\frac{-10}{5}$

m = -2

The slope of the line that passes through the pair of points is -2.

3 0

3 years ago

a) How many three-digit numbers can be formed from the digits 0, 1, 2, 3, 4, 5, and 6?6x7x7=294 b) How many three-digit numbers

love history [14]

Answer:

a) 294

b) 180

c) 75

d) 168

e) 105

Step-by-step explanation:

Given the numbers 0, 1, 2, 3, 4, 5 and 6.

Part A)

How many 3 digit numbers can be formed ?

Solution:

Here we have 3 spaces for the digits.

Unit's place, ten's place and hundred's place.

For unit's place, any of the numbers can be used i.e. 7 options.

For ten's place, any of the numbers can be used i.e. 7 options.

For hundred's place, 0 can not be used (because if 0 is used here, the number will become 2 digit) i.e. 6 options.

Total number of ways = $7 \times 7 \times 6$ = 294

Part B:

How many 3 digit numbers can be formed if repetition not allowed?

Solution:

Here we have 3 spaces for the digits.

Unit's place, ten's place and hundred's place.

For hundred's place, 0 can not be used (because if 0 is used here, the number will become 2 digit) i.e. 6 options.

Now, one digit used, So For unit's place, any of the numbers can be used i.e. 6 options.

Now, 2 digits used, so For ten's place, any of the numbers can be used i.e. 5 options.

Total number of ways = $6 \times 6 \times 5$ = 180

Part C)

How many odd numbers if each digit used only once ?

Solution:

For a number to be odd, the last digit must be odd i.e. unit's place can have only one of the digits from 1, 3 and 5.

Number of options for unit's place = 3

Now, one digit used and 0 can not be at hundred's place So For hundred's place, any of the numbers can be used i.e. 5 options.

Now, 2 digits used, so For ten's place, any of the numbers can be used i.e. 5 options.

Total number of ways = $3 \times 5 \times 5$ = 75

Part d)

How many numbers greater than 330 ?

Case 1: 4, 5 or 6 at hundred's place

Number of options for hundred's place = 3

Number of options for ten's place = 7

Number of options for unit's place = 7

Total number of ways = $3 \times 7 \times 7$ = 147

Case 2: 3 at hundred's place

Number of options for hundred's place = 1

Number of options for ten's place = 3 (4, 5, 6)

Number of options for unit's place = 7

Total number of ways = $1 \times 3 \times 7$ = 21

Total number of required ways = 147 + 21 = 168

Part e)

Case 1: 4, 5 or 6 at hundred's place

Number of options for hundred's place = 3

Number of options for ten's place = 6

Number of options for unit's place = 5

Total number of ways = $3 \times 6 \times 5$ = 90

Case 2: 3 at hundred's place

Number of options for hundred's place = 1

Number of options for ten's place = 3 (4, 5, 6)

Number of options for unit's place = 5

Total number of ways = $1 \times 3 \times 5$ = 15

Total number of required ways = 90 + 15 = 105

7 0

3 years ago