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Nastasia [14]
3 years ago
11

IF YOU ANSWER CORRECTLY I WILL MAKE ANOTHER QUESTION ADN GIVE YOU 20 POINTS

Mathematics
2 answers:
Montano1993 [528]3 years ago
6 0

Answer:

Im not sure what the question is but if its what the relation is between the boxes the first one would be y = x+5     the second one would be  y=x-4    the third would be  y= x+3    and the fourth would be  y= x times 2

Step-by-step explanation:

Korolek [52]3 years ago
4 0

Step-by-step explanation:

Pretty sure it is Linear Function, just graphed it.

Linear Function: y=mx+b when m is slope and b is y-intercept.

From the graph that I drew, the first one. y-intercept is at 5 and the slope is 1.

So the first one is y=x+5.

For the second one, x-intercept is at 4. That means y-intercept is at -4.

So the second one is y=x-4

If you draw both lines all at once, you will notice that both lines don't have/share interception. That means there are no answers for the equation.

y=x+5 — 1

y=x-4 — 2

Substitute 1 in 2

x+5=x-4

x-x+5+4=0

9=0

The equation doesn't have answers.

And I cant see the third one anyways, because I am on phone.

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In how many ways can we seat 3 pairs of siblings in a row of 7 chairs, so that nobody sits next to their sibling
monitta

Answer:

1,968

Step-by-step explanation:

Let x₁ and x₂, y₁ and y₂, and z₁ and z₂ represent the 3 pairs of siblings, and let;

Set X represent the set where the siblings x₁ and x₂ sit together

Set Y represent the set where the siblings y₁ and y₂ sit together

Set Z represent the set where the siblings z₁ and z₂ sit together

We have;

Where the three siblings don't sit together given as X^c∩Y^c∩Z^c

By set theory, we have;

\left | X^c \cap Y^c \cap Z^c  \right | = \left | X^c \cup Y^c \cup Z^c  \right | =  \left | U  \right | - \left | X \cup Y \cup Z  \right |

\left | U  \right | - \left | X \cup Y \cup Z  \right | = \left | U  \right | - \left (\left | X \right | +  \left | Y\right | +  \left | Z\right | -  \left | X \cap Y\right | -  \left | X \cap Z\right | -  \left | Y\cap Z\right | +  \left | X \cap Y \cap Z\right | \right)

Therefore;

\left | X^c \cap Y^c \cap Z^c  \right | = \left | U  \right | - \left (\left | X \right | +  \left | Y\right | +  \left | Z\right | -  \left | X \cap Y\right | -  \left | X \cap Z\right | -  \left | Y\cap Z\right | +  \left | X \cap Y \cap Z\right | \right)

Where;

\left | U\right | = The number of ways the 3 pairs of siblings can sit on the 7 chairs = 7!

\left | X\right | = The number of ways x₁ and x₂ can sit together on the 7 chairs = 2 × 6!

\left | Y\right | = The number of ways y₁ and y₂ can sit together on the 7 chairs = 2 × 6!

\left | Z\right | = The number of ways z₁ and z₂ can sit together on the 7 chairs = 2 × 6!

\left | X \cap Y\right | = The number of ways x₁ and x₂ and y₁ and y₂ can sit together on the 7 chairs = 2 × 2 × 5!

\left | X \cap Z\right | = The number of ways x₁ and x₂ and z₁ and z₂ can sit together on the 7 chairs = 2 × 2 × 5!

\left | Y \cap Z\right | = The number of ways y₁ and y₂ and z₁ and z₂ can sit together on the 7 chairs = 2 × 2 × 5!

\left | X \cap Y \cap Z\right | = The number of ways x₁ and x₂,  y₁ and y₂ and z₁ and z₂ can sit together on the 7 chairs = 2 × 2 × 2 × 4!

Therefore, we get;

\left | X^c \cap Y^c \cap Z^c  \right | = 7! - (2×6! + 2×6! + 2×6! - 2 × 2 × 5! - 2 × 2 × 5! - 2 × 2 × 5! + 2 × 2 × 2 × 4!)

\left | X^c \cap Y^c \cap Z^c  \right | = 5,040 - 3072 = 1,968

The number of ways where the three siblings don't sit together given as \left | X^c \cap Y^c \cap Z^c  \right |  = 1,968

5 0
3 years ago
which of the three sets could not be the lengths of the sides of a triangle {6,10,12} {5,7,10} {4,4,9} {2,3,3}
Phoenix [80]

So for this, we will be applying the Triangle Inequality Theorem, which states that the sum of 2 sides must be greater than the third side for it to be a triangle. If any inequality turns out to be false, then the set cannot be a triangle.

A+B>C\\A+C>B\\B+C>A

<h2>First Option: {6, 10, 12}</h2>

Let A = 6, B = 10, and C = 12:

6+10>12\\16>12\ \textsf{true}\\\\6+12>10\\18>10\ \textsf{true}\\\\12+10>6\\22>6\ \textsf{true}

<h2>Second Option: {5, 7, 10}</h2>

Let A = 5, B = 7, and C = 10

5+7>10\\12>10\ \textsf{true}\\\\5+10>7\\15>7\ \textsf{true}\\\\10+7>5\\17>5\ \textsf{true}

<h2>Third Option: {4, 4, 9}</h2>

Let A = 4, B = 4, and C = 9

4+4>9\\8>9\ \textsf{false}\\\\4+9>4\\13>4\ \textsf{true}\\\\4+9>4\\13>4\ \textsf{true}

<h2>Fourth Option: {2, 3, 3}</h2>

Let A = 2, B = 3, and C = 3

2+3>3\\5>3\ \textsf{true}\\\\2+3>3\\5>3\ \textsf{true}\\\\3+3>2\\6>2\ \textsf{true}

<h2>Conclusion:</h2>

Since the third option had an inequality that was false, <u>the third option cannot be a triangle.</u>

5 0
2 years ago
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Help asap find the area of this trapazoid
Contact [7]

Step-by-step explanation:

here you have the answers

8 0
3 years ago
Read 2 more answers
For the real-valued functions f(x)=2x+3 and g(x)=√x-5, find the composition fog and specify its domain using interval notation.
Burka [1]

Find fog

  • fog(x)
  • f(g(x))
  • f(√x-5)
  • 2√(x-5)+3

For real range the domain

  • x-5≥0
  • x≥5

Domain is [5,oo)

7 0
1 year ago
10th grade level LT1
Slav-nsk [51]

Answer:

<h2>? = 3.</h2>

Hope this will help you.

5 0
2 years ago
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