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3 years ago

Solve these 3 separately, please!:

Mathematics

2 answers:

MAVERICK [17]3 years ago

6 0

Answer:

4-6w=16 is w=-2

-9-5r=-19 is r=2

5k-7=3k+3 is k=5

eimsori [14]3 years ago

6 0

Answers:
• w = -2
• r = 2
• k = 5

Explanation:
• Problem #1

Step 1 - Subtract four from both sides of the equation

4 - 6w = 16
4 - 4 - 6w = 16 - 4
-6w = 12

Step 2 - Divide both sides of the equation by negative six

-6w = 12
-6w/ -6 = 12/ -6
w = -2

• Problem #2

Step 1 - Add nine to both sides of the equation

-9 - 5r = -19
-9 + 9 - 5r = -19 + 9
-5r = -10

Step 2 - Divide both sides of the equation by negative five

-5r = -10
-5r/ -5 = -10/ -5
r = 2

• Problem #3

Step 1 - Add seven to both sides of the equation

5k - 7 = 3k + 3
5k - 7 + 7 = 3k + 3 + 7
5k = 3k + 10

Step 2 - Subtract 3k from both sides

5k = 3k + 10
5k - 3k = 3k - 3k + 10
2k = 10

Step 3 - Divide both sides of the equation by 2

2k = 10
2k/ 2 = 10/ 2
k = 5

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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

MakcuM [25]

Answer:

$165$ .

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:

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