Each participant in a certain study was assigned a sequence of 3 different letters from the set {A, B, C, D, E, F, G, H}. If no sequence was assigned to more than one participant and if 36 of the possible sequences were not assigned, what was the number of participants in the study? (Note, for example, that the sequence A, B, C is different from the sequence C, B, A.)

Answer:

B

Step-by-step explanation:

8 yards = 3 * 8 = 24 ft^2

1 mile = 5280

3*3 = 9 square feet

9 square feet holds 9 people.

1 square foot holds 1 person

8*3 * 5280 people could stand in an area of 8 yards * 1 mile

Though it's not quite correct, the answer is B

Answer:

2=124  124/2

4=248 248/4

5=310 310/5

8=496 496/8

Step-by-step explanation:

40 + 22 = 62

62 x 2 = 124

62 x 4 = 248

62 x 5 = 310

62 x 8 = 496

i think

Answer:

13.9

Step-by-step explanation:

For the 41° angle, x is the opposite leg, and 16 is the adjacent leg. The trig ratio that relates the opposite leg and the adjacent leg is the tangent.

tan A = opp/adj

tan 41° = x/16

x = 16 * tan 41°

x = 13.9085...

Answer: 13.9

Segment addition postulate states that given points X, and Z, on a line, a point Y, can be located between X, and Z, ony if we have;

XZ = XY + YZ

The length of the segment BD is 36  cm

The reason the above value is correct is as follows:

Known:

The ratio in which the points A, B, C, and D divide the line segment = 4:3:1

The length of segment AD = 72 cm

Required:

The length of BD

Method:

Calculate the length of BC and CD and add their values to get BD

Solution:

Let the ratios be given unit proportions of the segment AD such that we have;

AB = 4 units

BC = 3 units

CD = 1 unit

By segment addition postulate, we have;

AD = AB + BC + CD

∴ AD = 4 units + 3 units + 1 unit = 8 units = 72 cm

∴ 1 unit = 72 cm/8 = 9 cm

1 unit = 9 cm

BD = BC + CD by segment addition postulate

BC = 3 units = 3 × 1 unit

∴ BC = 3 × 9 cm = 27 cm

BC = 27 cm

CD = 1 unit

∴ CD = 9 cm

∴ BD = 27 cm + 9 cm = 36 cm

The length of segment BD = 36 cm

Learn more about segment addition postulate here:

https://brainly.com/question/17015321

Answer:

I think E

Step-by-step explanation:

You know the shortest building is 25 m.

to find the rest, use trigo so Tan(20)=opposite/adjacent.

Adjacent is 50. Do the math and add the answer with 25.

Answer:

The answer would be E. 43.2

According to TOA, The opposite side is tan(20) x adjacent side( 50m)

the answer is 18.2( to 1 dp). Add the height of the second building together with 18.2 and you will get ur answer. HOpe this helps:)

Answer:

8 cm

Step-by-step explanation:

Answer:

4

Step-by-step explanation:

So first let's write the information we got

Angle BAC = 90 degrees

Midpoint of BC = M

AC = 6cm

Am= 5cm

Also I found MC = BM, since it has a line that represents both lines are same

so to find it we have to Pythagoras theorem (A^2 + B^2 = C^2), well it is question we have the 'A value and C Value', also we need to find the value of B to find the length of MC and BM

A = 5

C = 6

so therefore to find B^2, we have to do the reverse, we don't add but subtract

C^2 - A^2 = B^2

___________________________________________________________

Moving on to Calculation

6^2 - 5^2 = B^2

36 - 25 = B^2

B^2 = 9

B = √9

B = 3

MC and BM Length =  3 cm

____________________________

Now, we know the length we again need to use Pythagoras theorem to solve this.

Since we know

A = 5

B = 3

So..

A^2 + B^2 = C^2

5^2 + 3^2 = C^2

25 - 9 = C^2

C^2 = 16

C = √16

C = 4

Answer:

Step-by-step explanation:

Cos^-1(1/2) = 60

To get from a degree to radian, simply multiply by pi then divide by 180. So the final answer is 1/3pi or approximately 1.0472

I hope this helped! :D

Count how many multiples of 8, 12, or 18 there are in the range {1, 2, 3, …, 300}:

⌊300/8⌋ = 37

⌊300/12⌋ = 25

⌊300/18⌋ = 16

(where ⌊n⌋ denotes the floor of n, i.e. the largest integer that is smaller than n; for instance, 300/8 = 37.5, so ⌊300/8⌋ = ⌊37.5⌋ = 37)

Take pairwise LCMs, as well as the LCM of all three numbers:

LCM(8, 12) = LCM(2³, 2²×3) = 2³×3 = 24

LCM(8, 18) = LCM(2³, 2×3²) = 2³×3² = 72

LCM(12, 18) = LCM(2²×3, 2×3²) = 2²×3² = 36

LCM(8, 12, 18) = LCM(2³, 2²×3, 2×3²) = 2³×3² = 72

Count how many multiples there are of each of these LCMs that are less than 300:

⌊300/24⌋ = 12

⌊300/72⌋ = 4

⌊300/36⌋ = 8

Then, using the inclusion/exclusion principle, the number of numbers less than 300 that are exactly divisible by 8, 12, or 18 is

{multiples of 8} + {multiples of 12} + {multiples of 18}

- {multiples of 24} - {multiples of 72} - {multiples of 36}

+ {multiples of 72}

= 37 + 25 + 16 - 12 - 4 - 8 + 4 = 58

Answer: See explanation

Step-by-step explanation:

Since Smart Dot Company costs $12 per month plus $0.50 for each hour of Internet used, the equation to represent the cost of using the internet with them will be:

C = 12 + 0.5t

Since Communication Plus costs $2.50 for every hour of Internet used, the equation to represent the cost of using the internet with them will be:

C = 2.5t

Answer:

B. Never

Step-by-step explanation:

When a number is irrational, it means that it cannot be written as a fraction.

I hope this helps!

pls ❤ and mark brainliest pls!

Answer:

c) when it is improper fraction

Answer:

Equation of line is y = -8x + 21

Step-by-step explanation:

Slope, m = -8

General equation of line:

[tex]{ \boxed{ \bf{y = mx + c}}}[/tex]

At point (2, 5), y = 5 and x = 2:

[tex]{ \tt{5 = ( - 8 \times 2) + c}} \\ { \tt{c = 21}}[/tex]

Therefore:

[tex]{ \sf{y = - 8x + 21}}[/tex]

[tex]{ \underline{ \blue{ \sf{christ \:† \: alone }}}}[/tex]

Answer:

height = 21.5 ft

Step-by-step explanation:

Substitute t = 1.5 into h(t) and evaluate

h(1.5) = - 16(1.5)² + 35(1.5) + 5

        = - 16(2.25) + 52.5 + 5

       = - 36 + 57.5

      = 21.5 ft

Answer:

21.5 feet.

Step-by-step explanation:

Let t = 1.5

[tex]h(1.5)=-16(1.5)^2+35(1.5)+5\\h(1.5)=-36+52.5+5\\h(1.5)=21.5[/tex]

Therefore, at 1.5 seconds, the basketball is 21.5 feet in the air.

Answer:

see explanation

Step-by-step explanation:

Using x and y to represent the 2 numbers with x > y , then

x + y = 15 → (1)

x - y = 11 → (2)

Adding the 2 equations term by term to eliminate y

2x = 26 ( divide both sides by 2 )

x = 13

Substitute x = 13 into (1)

13 + y = 15 ( subtract 13 from both sides )

y = 2

The 2 numbers are 13 and 2

Part (a)

This is the empty set

------------------

Explanation:

It doesn't matter what set A is composed of. Intersecting any set with the empty set Ø will always result in the empty set.

This is because we're asking the question: "What does some set A and the empty set have in common?". The answer of course being "nothing" because there's nothing in Ø. Not even the value zero is in this set.

We can write Ø as { } which is a set of curly braces with nothing inside it.

=========================================================

Part (b)

-----------------

Explanation:

When you union the universal set with any other set, you'll get the universal set.

The rule is [tex]A \cup B = B[/tex] where I've made B the universal set to avoid confusion of the letter U and the union symbol [tex]\cup[/tex] which looks nearly identical.

Why does this rule work? Well if an item is in set [tex]\overline{A}[/tex], then it's automatically in set U (everything is in set U; it's the universe). So we're not adding anything to the universe when applying a union involving this largest set.

It's like saying

we can see that [tex]\overline{A} \cup U[/tex] will basically form the set of every item, aka the universal set.

Answer:

Klog earned $135.45 altogether.

Step-by-step explanation:

Hours

Monday - Friday : 5 days / 3.5 hours

Saturday : 1 day / 4 hours

3.5 · 5 + 4

= 17.5 + 4

= 21.5

Money

$6.30 per hours / 21.5 hours

6.30 · 21.5

= $135.45

Step-by-step explanation:

(+4)+(-7)

=4-7

=-3

Hope it will help you..

Answer:

[tex]P(x \ge 5) = 0.60[/tex]

Step-by-step explanation:

Given

[tex]S = \{5, 2, 7, 2, 3, 4, 10, 6,4,6,3, 6, 6, 4, 8,5,7,7,1,5\}[/tex]

[tex]n(S) = 20[/tex]

Required

[tex]P(x \ge 5)[/tex]

First, we count the number of trials that are at least 5

[tex]x = \{5, 7, 10, 6,6, 6, 6, 8,5,7,7,5\}[/tex]

So, we have:

[tex]n(x \ge 5) = 12[/tex]

So, we have:

[tex]P(x \ge 5) = \frac{n(x \ge 5)}{n(S)}[/tex]

This gives

[tex]P(x \ge 5) = \frac{12}{20}[/tex]

[tex]P(x \ge 5) = 0.60[/tex]

Answer:

(2x+3y)^2

= (2x)^2 + 2(2x)(3y) + (3y)^2

= 4x^2 + 12xy + 9y^2

Answer:

4x^2 12xy +9y^2

Step-by-step explanation:

(2x+3y)^2

(2x+3y)(2x+3y)

FOIL

4x^2 + 6xy+6xy + 9y^2

4x^2 12xy +9y^2

Answer:

-4

Step-by-step explanation:

a = 3

b = -4

c = -2

2a + 3b - c

= 2(3) + 3(-4) - (-2)

= 6 + (-12) + 2

= -4

Answer:

sorry

Step-by-step explanation:

polynomial is not given

question is incomplete

The statistical question is; "What are the ages of the students in this class?

A statistical question is always aimed at data collection which can subsequently used for analysis and decision making. Statistical questions are asked in the course of research.

Among the two questions, the statistical question is; "What are the ages of the students in this class?

Learn more about statistics:https://brainly.com/question/8058700

#SPJ1

Answer:

(i) a = 2

   b = -1

   c = -1

(ii) x= 1

Step-by-step explanation:

Step 1: Factorise

[tex]f(x) = 2(x^{2} -2x+\frac{1}{2})[/tex]  

Step 2: Use the complete square method.

[tex]f(x)=2(x^{2} -2x+(\frac{-2}{2})^{2} + \frac{1}{2} - (\frac{-2}{2})^{2} )[/tex]  

Step 3: Have it to [tex]a(x+b)^{2} +c[/tex]

[tex]f(x)=2((x-1)^{2} -\frac{1}{2} )\\ = 2(x-1)^{2} -1[/tex]

Line of symmetry:  

To find line of symmetry, we use -b/2a formula.

Based from [tex]2x^{2} -4x+1[/tex]:

a=2, b=-4

-b/2a = -(-4)/2(2) = 1

Answer:

- 2 x y^2+3 x^2 y -xy

-120

Step-by-step explanation:

xy – 2xy + 3x^2 y — 4x y^2 + 2xy^2

Combine like terms

xy – 2xy + 3x^2 y — 4x y^2 + 2xy^2

- x y+3 x^2 y - 2 x y^2

Let x = 4 y = -2

-(4)(-2) +3(4)^2 (-2) -2(4)(-2)^2

Exponents first

-(4)(-2) +3(16) (-2) -2(4)(4)

Multiply

+8 -96-32

Add and subtract

-120

Answer:

5m

Step-by-step explanation:

5^(k-1)+5^(k-1)=m

Going to combine like terms on left:

2×5^(k-1)=m

Law of exponents applied:

2×5^k×5^(-1)=m

Reciprocal:

2×5^k×1/5=m

Multiply 5 on both sides to obtain the requested:

2×5^k=5m

[tex]\\ \sf\longmapsto (-42)+15+(-63)[/tex]

[tex]\\ \sf\longmapsto -42+15+(-63)[/tex]

[tex]\\ \sf\longmapsto -27+(-63)[/tex]

[tex]\\ \sf\longmapsto -27-63[/tex]

[tex]\\ \sf\longmapsto -90[/tex]

Answer:

42 + 15 + (-63) = -90

Step-by-step explanation:

If you like my answer than please mark me brainliest thanks

Answer:

[tex]\displaystyle \displaystyle \cos A = \frac{a^2-b^2}{a^2 + b^2}\text{ and } \sin A = \frac{2ab}{a^2 + b^2}[/tex]

Step-by-step explanation:

We are given that:

[tex]\displaystyle \tan A = \frac{2ab}{a^2 - b^2}[/tex]

And we want to find the value of cos(A) and sin(A).

Recall that tangent is the ratio of the opposite side to the adjacent side.

Therefore, the opposite side measures 2ab, and the adjacent side measures a² - b².

Using the Pythagorean Theorem, solve for the hypotenuse:

[tex]\displaystyle \begin{aligned} c^2 &= a^2 + b^2 \\ \\ c&= \sqrt{(2ab)^2 + (a^2-b^2)} \\ \\ &= \sqrt{(4a^2b^2)+(a^4-2a^2b^2+b^4)} \\ \\ &= \sqrt{a^4 + 2a^2b^2 + b^4 } \\ \\ &= \sqrt{(a^2 +b^2)^2} \\ \\ &= a^2 + b^2\end{aligned}[/tex]

Thus, our hypotenuse is given by a² + b².

Cosine is the ratio between the adjacent side and the hypotenuse. Thus:

[tex]\displaystyle \cos A = \frac{a^2-b^2}{a^2 + b^2}[/tex]

And sine is the ratio between the opposite side and the hypotenuse. Thus:

[tex]\displaystyle \sin A = \frac{2ab}{a^2 + b^2}[/tex]

In conclusion:

[tex]\displaystyle \displaystyle \cos A = \frac{a^2-b^2}{a^2 + b^2}\text{ and } \sin A = \frac{2ab}{a^2 + b^2}[/tex]

Answer:

Step-by-step explanation:

[tex]sec^2A-tan^2A=1\\sec^2A=1+tan^2A=1+\frac{4a^2b^2}{(a^2-b^2)^2} =\frac{(a^2-b^2)^2+4a^2b^2}{(a^2-b^2)^2} =\frac{(a^2+b^2)^2}{(a^2-b^2)^2} \\cos^2A=\frac{(a^2-b^2)^2}{(a^2+b^2)^2} \\cos A=\frac{a^2-b^2}{a^2+b^2} \\sin A=\sqrt{1-cos^2A} =\sqrt{1-(\frac{a^2-b^2}{a^2+b^2} )^2} =\sqrt{\frac{(a^2+b^2)^2-(a^2-b^2)^2}{(a^2+b^2)^2} } =\sqrt{\frac{4a^2b^2}{(a^2+b^2)^2} }=\frac{2ab}{a^2+b^2}[/tex]