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svlad2 [7]
2 years ago
15

Alina, Robert, and Makayla are on the student council at their high school. They all want to go to a dinner where student counci

l representatives from other schools will participate. However, only one student per school can go. How can a fair decision be made about who goes to dinner? Select all that apply.
If the answer is correct I will mark as brainliest.

A. Write each students name on a slip of paper and put the slips in a hat. Randomly draw a slip of paper. The Student whose name is on the paper wins.

B. Flip a coin twice. If both tossed are heads, Alina wins. If both are tails, Makayla wins. If one is heads and one is tails, Robert wins.

C. Roll a number cube. If it lands on 1 or 2, Alina wins. If it lands on 3 or 5, Robert wins. If it lands on any other number, Makayla wins.

D. Roll a number cube. If the number is odd, Alina wins. If the number is even, Robert wins. If it's any other number, Makayla wins​
Mathematics
1 answer:
Gnesinka [82]2 years ago
7 0

Answer:

A. Write each students name on a slip of paper and put the slips in a hat. Randomly draw a slip of paper. The Student whose name is on the paper wins.

C. Roll a number cube. If it lands on 1 or 2, Alina wins. If it lands on 3 or 5, Robert wins. If it lands on any other number, Makayla wins.

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Derive these identities using the addition or subtraction formulas for sine or cosine: sinacosb=(sin(a+b)+sin(a-b))/2
Sergeu [11.5K]

Answer:

The work is in the explanation.

Step-by-step explanation:

The sine addition identity is:

\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b).

The sine difference identity is:

\sin(a-b)=\sin(a)\cos(b)-\cos(a)\sin(a).

The cosine addition identity is:

\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b).

The cosine difference identity is:

\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b).

We need to find a way to put some or all of these together to get:

\sin(a)\cos(b)=\frac{\sin(a+b)+\sin(a-b)}{2}.

So I do notice on the right hand side the \sin(a+b) and the \sin(a-b).

Let's start there then.

There is a plus sign in between them so let's add those together:

\sin(a+b)+\sin(a-b)

=[\sin(a+b)]+[\sin(a-b)]

=[\sin(a)\cos(b)+\cos(a)\sin(b)]+[\sin(a)\cos(b)-\cos(a)\sin(b)]

There are two pairs of like terms. I will gather them together so you can see it more clearly:

=[\sin(a)\cos(b)+\sin(a)\cos(b)]+[\cos(a)\sin(b)-\cos(a)\sin(b)]

=2\sin(a)\cos(b)+0

=2\sin(a)\cos(b)

So this implies:

\sin(a+b)+\sin(a-b)=2\sin(a)\cos(b)

Divide both sides by 2:

\frac{\sin(a+b)+\sin(a-b)}{2}=\sin(a)\cos(b)

By the symmetric property we can write:

\sin(a)\cos(b)=\frac{\sin(a+b)+\sin(a-b)}{2}

3 0
3 years ago
The product of 2 numbers is -24 and their sum is -5. What are the two numbers and why are they the two numbers?
lana [24]

The numbers are (-8, 3) or (3, -8).

Step-by-step explanation:

  • Step 1: Given the product of the numbers are -24 and their sum are -5. Let the numbers be a and b. Form equations out of these details.

⇒ a + b = -5 ⇒ b = -5 - a

⇒ a × b = -24 -------- (1)

  • Step 2: Substitute the value of b in eq(1)

⇒ a × (-5 - a) = - 24

⇒ -5a - a² = -24

⇒ a² + 5a - 24 = 0

  • Step 3: Solve the quadratic equation for a.

a = (-5 ± √25 - 4 × 1 × -24)/2

  = (-5 ± √121)/2

  = - 16/2, 6/2 = -8 or 3

  • Step 4: For each value of a, find b.

When a = -8, b = -5 +8 = 3

When a = 3, b = -5 - 3 = -8.

6 0
2 years ago
5.
enyata [817]

Answer:

  (E) 90°

Step-by-step explanation:

The diagonals of a parallelogram are equal in length if and only if the parallelogram is a rectangle.

Any angle of a rectangle is 90°.

_____

Adjacent angles of a parallelogram are supplementary. If the diagonals are the same length, the triangles formed by either diagonal are congruent with the triangles formed by the other diagonal. This means adjacent angles are congruent. Congruent supplementary angles are 90°.

7 0
2 years ago
Find z such that 7.5% of the standard normal curve lies to the left of z.
sergeinik [125]
A probability of 0.075 corresponds to a z-score of about z\approx-1.44, i.e.

\mathbb P(Z
3 0
3 years ago
Choose the appropriate pattern and use it to find the product: (p4−q4)(p4+q4).
sp2606 [1]

The expression can be solved by expanding the bracket and multiplying out the terms

(p^4-q^4)(p^4+q^4)\begin{gathered} =p^4(p^4+q^4)-q^4(p^4+q^4) \\ =p^8+p^4q^4-p^4q^4-q^8 \\ =p^8-q^8 \end{gathered}

Therefore, the expression can be simplified as;

p^8-q^8

Alternatively, using the theorem of difference of two squares, which is

a^2-b^2=(a-b)(a+b)

Hence,

p^8-q^8=(p^4)^2-(q^4)^2

7 0
1 year ago
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