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2 years ago

URGENT!!! Help fast!!!

Mathematics

1 answer:

Mashutka [201]2 years ago

6 0

second graph from the left

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13s - r^2; if r = 9 and s = 6

FromTheMoon [43]

Answer:

Step-by-step explanation:

r = 9

s = 6

r^2

= r x r

= 9 x 9

= 81

13s

= 13 x 6

= 108

13s - r^2

= 108 - 81

= 27

8 0

2 years ago

Read 2 more answers

What is the relationship between lines d and b?

docker41 [41]

Answer:

the are perpendicular to each other (option B)

5 0

2 years ago

A circle with the equation (x + 6)2 + (y + 3)2 = 16 is reflected over the line x = 2.

fomenos

Answer: $(x-10)^{2}+(y+3)^{2}=16$

Step-by-step explanation:

We know the center of the original circle is (-6, -3). After reflecting this over the line x=2, the new center is now (10, -3).

Thus, the equation of the circle is $(x-10)^{2}+(y+3)^{2}=16$ .

6 0

1 year ago

25 POINTS HURRY Brainliest!!!

Alborosie

I think the answer is B

7 0

2 years ago

Read 2 more answers

The numbers of teams remaining in each round of a single-elimination tennis tournament represent a geometric sequence where an i

Anit [1.1K]

Answer:

$a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}$

Step-by-step explanation:

We are given the following in the question:

The numbers of teams remaining in each round follows a geometric sequence.

Let a be the first the of the geometric sequence and r be the common ration.

The $n^{th}$ term of geometric sequence is given by:

$a_n = ar^{n-1}$

$a_4 = 16 = ar^3\\a_6 = 4 = ar^5$

Dividing the two equations, we get,

$\dfrac{16}{4} = \dfrac{ar^3}{ar^5}\\\\4}=\dfrac{1}{r^2}\\\\\Rightarrow r^2 = \dfrac{1}{4}\\\Rightarrow r = \dfrac{1}{2}$

the first term can be calculated as:

$16=a(\dfrac{1}{2})^3\\\\a = 16\times 6\\a = 128$

Thus, the required geometric sequence is

$a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}$

4 0

3 years ago