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3 years ago

PLEASE HELP THE QUESTION IS BELOW!!

Mathematics

2 answers:

Mamont248 [21]3 years ago

7 0

Answer:

Step-by-step explanation:

Hello,

What the question requires of us is just the property of complimentary angles and which is

The sum of their angles must be equal to 90°.

When two angles combine together to form angle 90°, it is said to be complimentary and they're referred to as a right angled triangle.

From the option, none of the angles can be up to a right angle triangle (90°)

The sum of their angles can be an acute triangle I.e 45°

45° + 45° = 90°

Knowing the fact that the angles can both be an acute angle (45°), it would be wrong to say both angles cannot be equal.

Hence option D is is wrong and it's the right answer in this case

Ne4ueva [31]3 years ago

6 0

Answer:

The answer is option D.

Their angles measures cannot be equal.

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3 years ago

A donut store has 11 different types of donuts. You can only buy a bag of 3 of them, where each donut has to be of a different t

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$165$ .

Step-by-step explanation:

Since repetition isn't allowed, there would be $11$ choices for the first donut, $(11 - 1) = 10$ choices for the second donut, and $(11 - 2) = 9$ choices for the third donut. If the order in which donuts are placed in the bag matters, there would be $11 \times 10 \times 9$ unique ways to choose a bag of these donuts.

In practice, donuts in the bag are mixed, and the ordering of donuts doesn't matter. The same way of counting would then count every possible mix of three donuts type $3 \times 2 \times 1 = 6$ times.

For example, if a bag includes donut of type $x$ , $y$ , and $z$ , the count $11 \times 10 \times 9$ would include the following $3 \times 2 \times 1$ arrangements:

$xyz$ .
$xzy$ .
$yxz$ .
$yzx$ .
$zxy$ .
$zyx$ .

Thus, when the order of donuts in the bag doesn't matter, it would be necessary to divide the count $11 \times 10 \times 9$ by $3 \times 2 \times 1 = 6$ to find the actual number of donut combinations:

$\begin{aligned} \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}$ .

Using combinatorics notations, the answer to this question is the same as the number of ways to choose an unordered set of $3$ objects from a set of $11$ distinct objects:

$\begin{aligned}\begin{pmatrix}11 \\ 3\end{pmatrix} &= \frac{11 !}{(11 - 3)! \times 3 !} \\ &= \frac{11 !}{8 ! \times 3 !} \\ &= \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}$ .