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2 years ago

14

what is the scale factor of a triangle with a vertex of a -6 for that is been dilated with a center of dilation at the origin wi

th vertex of its image is a prime negative 2416

1 answer:

Sedaia [141]2 years ago

3 0

Answer:

im just tryna get points so hopefully sum on else helps and u pass

Step-by-step explanation:

You might be interested in

Rewrite (123 + 456) + 789 using the Associative Law of Addition

adelina 88 [10]

Answer:

123 + (456 + 789)

Step-by-step explanation:

since this is an addition problem, the parenthesis don't matter, and they can be put wherever you want.

(123 + 456) + 789

123 + 456 + 789

(123 + 456 + 789)

123 + (456 + 789)

technically, any of these would work, but I think the kind of answer the problem is looking for is;

123 + (456 + 789)

5 0

3 years ago

The equation below represents Function A and the graph represents Function B:

kozerog [31]

The slope of function A is the coefficient of x, which is 6.

The slope of function B is the rise divided by the run, which is 3/1 = 3.

The appropriate choice is ...

... c) Slope of Function A = 2×Slope of Function B

8 0

3 years ago

Suppose you pick two cards from a deck of 52 playing cards. What is the probability that they are both queens?

photoshop1234 [79]

Answer:

0.45% probability that they are both queens.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes

The combinations formula is important in this problem:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

Desired outcomes

You want 2 queens. Four cards are queens. I am going to call then A,B,C,D. A and B is the same outcome as B and A. That is, the order is not important, so this is why we use the combinations formula.

The number of desired outcomes is a combinations of 2 cards from a set of 4(queens). So

$D = C_{4,2} = \frac{4!}{2!(4-2)!} = 6$

Total outcomes

Combinations of 2 from a set of 52(number of playing cards). So

$T = C_{52,2} = \frac{52!}{2!(52-2)!} = 1326$

What is the probability that they are both queens?

$P = \frac{D}{T} = \frac{6}{1326} = 0.0045$

0.45% probability that they are both queens.

4 0

2 years ago

Is a three-digt whole number

Rudiy27

Answer:

693

Step-by-step explanation:

7*9*11*3=693

5 0

2 years ago

What’s the answer for 12?

Lisa [10]

Answer:

$- 3 \sqrt{4 \times 4 \times 3} \\ - 3 \times 2 \times 2 \sqrt{3 } \\ - 12 \sqrt{3}$

6 0

2 years ago