Find the exact value of sin A in simplest radical form.

3/7 of the apples in a box are red apples the rest are green apples there are 24 green apples how many apples are together

hjlf

4/7 of the box must be green apples. If there are 24 green apples, the numerator is multiplied by 6. Since it must stay proportional, multiply the denominator, 7 by six to find there are a total of 42 apples

4 0

3 years ago

I need help. I’ve been stuck

solniwko [45]

Ok RYU? Isn’t it 110*?

3 0

3 years ago

Using the 45°-45°-90° triangle theorem, find the value of h, the height of the wall.

SIZIF [17.4K]

Answer:

Step-by-step explanation:

Using the 45°-45°-90° triangle theorem, find the value of h, the height of the wall.

4 0

3 years ago

Four universities - 1, 2, 3, and 4 - are participating in a holiday basketball tournament. In first round, 1 will play 2 and 3 w

tia_tia [17]

Answer:

Part A. {S} = {1324, 1342, 1423, 1432, 2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241, 4123, 4132, 4213, 4231}

Part B. {A} = {1324, 1342, 1423. 1432}

Part C. {B} = {2314, 2341, 2413, 2431, 3214, 3241, 4213, 4231}

Part D.

(A∪B) = {1324, 1342, 1423, 1432, 2314, 2341, 2413, 2431, 3214, 3241, 4213, 4231}

(A∩B) = ∅

{A'} = {2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241, 4123, 4132, 4213, 4231}

Step-by-step explanation:

A. All possible outcomes

There are four teams, each play a semi final where 1 and 2 plays against each other while 3 and 4 plays against each other. Winner of the first semi final can be either 1 or 2 therefore they both can not be in the championship game or in the losers game at the same time same goes for the other semi final.

Using this explanation (1324 denotes: 1 beats 2 and 3 beats 4 in first-round games and then 1 beats 3 and 2 beats 4), All possible outcomes are

{S} = {1324, 1342, 1423, 1432, 2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241, 4123, 4132, 4213, 4231}

B. Event A in which 1 wins the tournament

From {S} we only have to write the outcomes in which 1 is the first number in 4digit combinations given in part A

{A} = {1324, 1342, 1423. 1432}

C. Event B in which 2 gets into championship game

From {S} we only have to write the outcomes in which 2 is either the first or second digit in 4digit combinations given in part A

{B} = {2314, 2341, 2413, 2431, 3214, 3241, 4213, 4231}

D. Outcomes in (A∪B), (A∩B) and A'

I. (A∪B)

(A∪B) means A union B therefore all we have to do is combine all the members of A and B

(A∪B) = {1324, 1342, 1423, 1432} ∪{2314, 2341, 2413, 2431, 3214, 3241, 4213, 4231}

(A∪B) = {1324, 1342, 1423, 1432, 2314, 2341, 2413, 2431, 3214, 3241, 4213, 4231}

II. (A∩B)

(A∩B) means A intersection B in which we have to find the common members of A and B. If there are no common members then the result of (A∩B) is a null set.

(A∩B) = {1324, 1342, 1423, 1432} ∩ {2314, 2341, 2413, 2431, 3214, 3241, 4213, 4231}

(A∩B) = ∅

III. {A'}

A' means A compliment, in other words it can be described as all the possible outcomes that are not part of A. So all we do is to subtract outcomes of A from the total possible outcomes S

{A'} = {S} - {A}

{A'} = {1324, 1342, 1423, 1432, 2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241, 4123, 4132, 4213, 4231} - {1324, 1342, 1423. 1432}

{A'} = {2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241, 4123, 4132, 4213, 4231}

7 0

3 years ago

Can someone please help me with my Question #29 of The Quadratic Relations for me please?

lara [203]

Answer:

Calculate the first differences between the y-values:

$\sf 3 \underset{+1}{\longrightarrow} 4 \underset{+3}{\longrightarrow} 7 \underset{+5}{\longrightarrow} 12 \underset{+7}{\longrightarrow} 19$

As the first differences are not the same, we need to calculate the second differences:

$\sf 1 \underset{+2}{\longrightarrow} 3 \underset{+2}{\longrightarrow} 5 \underset{+2}{\longrightarrow} 7$

As the second differences are the same, the relationship between the variable is quadratic and will contain an $x^2$ term.

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To determine the quadratic equation

The coefficient of $x^2$ is always half of the second difference.

As the second difference is 2, and half of 2 is 1, the coefficient of $x^2$ is 1.

The standard form of a quadratic equation is: $y=ax^2+bx+c$

(where a, b and c are constants to be found).

We have already determined that the coefficient of $x^2$ is 1.

Therefore, a = 1

From the given table, when $x=0$ , $y=12$ .

$\implies a(0)^2+b(0)+c=12$

$\implies c=12$

Finally, to find b, substitute the found values of a and c into the equation, then substitute one of the ordered pairs from the given table:

$\begin{aligned}\implies x^2+bx+12 & = y\\ \textsf{at }(1,19) \implies (1)^2+b(1)+12 & = 19\\ 1+b+12 & = 19\\b+13 & =19\\b&=6\end{aligned}$

Therefore, the quadratic equation for the given ordered pairs is:

$y=x^2+6x+12$

7 0

2 years ago