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3 years ago

Solve for Q Q/B + A = Y

Mathematics

1 answer:

Otrada [13]3 years ago

6 0

What are the numbers that are supposed to be plugged in?

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What is the SURFACE AREA of the pyramid?

Tamiku [17]

Answer:

85in²

Step-by-step explanation:

s = side

triangle: ((b x h) ÷ 2) x 4

((5 x 6) ÷ 2) x 4 = 60in

square: s2(squared not 2)

5² = 25

total sa: 25 + 60 = 85in²

5 0

3 years ago

Find the surface area of the composite figure.

Serggg [28]

Fhfhfhejdhfhebcbfbdbcbrdhfhfbfhfhfhffhfhchfhdhchfhdhhchfjdhf

5 0

1 year ago

34. The product of two algebraic

S_A_V [24]

Answer:

3a²b

Step-by-step explanation:

The product of two algebraic

terms is 6a3b2. If one of the terms is

2ab, find the other term.

Let us represent

First term = a

Other term = b

a × b = 6a³b²

a = 2ab

b = ?

b = 6a³b²/a

b = 6a³b²/2ab

b = 6a³b²/2a¹b¹

b = (6 ÷ 2) × a^3 - 1 × b ^2 - 1

b = 3a²b

Therefore, the other term = 3a²b

7 0

3 years ago

Hall and Mindy are playing a guessing game. Hall tells Mindy: ""The difference between 17 and the square root of my mystery numb

cestrela7 [59]

Answer:

The possible number that Hall could be thinking of is 144.

Step-by-step explanation:

Difference between 17 and the square root of my mystery number

This is represented by:

$17 - \sqrt{x}$

is 5

Then:

$17 - \sqrt{x} = 5$

What are two possible numbers that Hall could be thinking of?

We have to solve the equation, for x. So

$17 - \sqrt{x} = 5$

$\sqrt{x} = 12$

$(\sqrt{x})^2 = (12)^2$

$x = 144$

The possible number that Hall could be thinking of is 144.

5 0

3 years ago

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