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2 years ago

Someone please help! Tysm if you answer this for me! Have a great day:)

Mathematics

2 answers:

AnnZ [28]2 years ago

5 0

Only 1 student watched 6 movies over the break!!

Lena [83]2 years ago

5 0

Answer:

One

Step-by-step explanation:

Only one student watch six movies

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4x^2-24x+4y^2+72y=76

podryga [215]

<span>Simplifying 4x2 + -24x + 4y2 + 72y = 76
Reorder the terms: -24x + 4x2 + 72y + 4y2 = 76
Solving -24x + 4x2 + 72y + 4y2 = 76
Solving for variable 'x'.
Reorder the terms: -76 + -24x + 4x2 + 72y + 4y2 = 76 + -76
Combine like terms: 76 + -76 = 0 -76 + -24x + 4x2 + 72y + 4y2 = 0
Factor out the Greatest Common Factor (GCF), '4'. 4(-19 + -6x + x2 + 18y + y2) = 0
Ignore the factor 4.

</span><span>Subproblem 1
Set the factor '(-19 + -6x + x2 + 18y + y2)' equal to zero and attempt to solve: Simplifying -19 + -6x + x2 + 18y + y2 = 0 Solving -19 + -6x + x2 + 18y + y2 = 0 The solution to this equation could not be determined. This subproblem is being ignored because a solution could not be determined.
The solution to this equation could not be determined.</span>

7 0

2 years ago

The fourth grade wants to observe Martin Luther king day. They can show a video about his life or have a guest speaker. They can

nalin [4]

The anwser is 6 possible combinations.

8 0

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:

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

5 0

1 year ago

Round the following to the nearest cent: $16.997

Paul [167]

$\$16.99\underbrace{7}_{>5}\approx\$17.00$

7 0

2 years ago

Mrs. Kidd took her two children to a waterpark for summer fun. An adult ticket cost $32 and the total price of admission for Mrs

Leona [35]

32 + 2x = 84
2x = 52
x = 26
child's ticket is $26

4 0

3 years ago