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2 years ago

Can you please answer this

Mathematics

2 answers:

Crazy boy [7]2 years ago

5 0

13. 71.5=5.5u divide each side by 5.5 to solve for u
71.5/5.5=13 so u=13

14. 5/6x=5/14 to solve for x you divide both sides by 5/6. But to divide fractions you multiply the reciprical. So:
x=5/14*6/5=30/70 or 3/7

15. 44=t/11 multiply both sides by 11 to solve for t
484=t

16. 8.3=b/14 multiply both sides by 14 to solve for b
116.2=b

17. x/3= 15 multiply both sides by 3 to solve for x
x=45

18. 7.8x=23.4 divide both sides by 7.8 to solve for x
x=3

notka56 [123]2 years ago

4 0

13) divide 71.5 by 5.5 to find the answer for u.
14) find the common denominator then divide like in the last one
15) multiply 11 by 44 to find t
16) multiply 14 by 8.3 to find b
17) multiply 3 by 15 to find x
18) divide 23.4 by 7.8 to find x

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Flauer [41]

Answer:

Y1 intercept = 2

(0,2) (3,-7)

m= (-7 - 2) / (3-0) =-9/3 = - 3

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2 years ago

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2 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:

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

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1 year ago