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melisa1 [442]
3 years ago
9

How does this rule of one output for each input justify the use of the vertical line test for graphs

Mathematics
1 answer:
Alisiya [41]3 years ago
7 0

Answer:

The vertical line test can be used to determine whether a graph represents a function. If we can draw any vertical line that intersects a graph more than once, then the graph does not define a function because a function has only one output value for each input value.

Step-by-step explanation:

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What is 3(x-1)<12 or x+7>10
Otrada [13]

Simplifying

3[x + -1] = 12

Reorder the terms:

3[-1 + x] = 12

[-1 * 3 + x * 3] = 12

[-3 + 3x] = 12

Solving

-3 + 3x = 12

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '3' to each side of the equation.

-3 + 3 + 3x = 12 + 3

Combine like terms: -3 + 3 = 0

0 + 3x = 12 + 3

3x = 12 + 3

Combine like terms: 12 + 3 = 15

3x = 15

Divide each side by '3'.

x = 5

Simplifying

x = 5


6 0
3 years ago
10 less than the sum of two consecutive numbers
damaskus [11]
The consecutive numbers 5and 6 equal 11 is that what yr asking
3 0
4 years ago
Read 2 more answers
Suppose the time a child spends waiting at for the bus as a school bus stop is exponentially distributed with mean 7 minutes. De
Gala2k [10]

Answer:

The probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning is 0.148.

Step-by-step explanation:

Let the random variable <em>X</em> represent the time a child spends waiting at for the bus as a school bus stop.

The random variable <em>X</em> is exponentially distributed with mean 7 minutes.

Then the parameter of the distribution is,\lambda=\frac{1}{\mu}=\frac{1}{7}.

The probability density function of <em>X</em> is:

f_{X}(x)=\lambda\cdot e^{-\lambda x};\ x>0,\ \lambda>0

Compute the probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning as follows:

P(6\leq X\leq 9)=\int\limits^{9}_{6} {\lambda\cdot e^{-\lambda x}} \, dx

                      =\int\limits^{9}_{6} {\frac{1}{7}\cdot e^{-\frac{1}{7} \cdot x}} \, dx \\\\=\frac{1}{7}\cdot \int\limits^{9}_{6} {e^{-\frac{1}{7} \cdot x}} \, dx \\\\=[-e^{-\frac{1}{7} \cdot x}]^{9}_{6}\\\\=e^{-\frac{1}{7} \cdot 6}-e^{-\frac{1}{7} \cdot 9}\\\\=0.424373-0.276453\\\\=0.14792\\\\\approx 0.148

Thus, the probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning is 0.148.

6 0
3 years ago
Suppose that 600 yards of fencing material are available to fence in
postnew [5]

Answer:

Length = 150 yards

Width = 100 yards

Step-by-step explanation:

We want 600 yards of fencing that will result in the largest 2 fenced corrals, sharing a common border.

It will take the shape of a rectangle, with a dividing fence down the center.

Let W and L,  Width and Length of the larger enclosure.

See attachment.

W= Area of the larger enclosure.

The perimeter is 2W + 2L.

The dividing fence is 1W

We know that we only have 600 yards of fence, so:

2W + 2L + 1W = 600 yards

Area = W x L

---

3W + 2L  = 600    (yards)

2L  = 600 -3W

L = (600-3W)/2

L = 300 -(3/2)W

---

Use this expression in the Area calculation:

Area = W x L

Area = W x (300 -(3/2)W)

Area = 300W -(3/2)W^2)

To find the maximum area, take the first derivative and set to zero to find the value of W that results in the greatest area.

Area' = 300 -2(3/2)W)

0 = 300 - 3W

3W = 300

W = 100 yards

Since 3W + 2L  = 600

L = (600 - 3W)/2

L = (600 - 3(100))/2

L = 150 yards

Area = 150*100 = 15,000 yards^2

 

8 0
3 years ago
List multiples to find the lcm 10<br> , 15
morpeh [17]
10-- 10,20,30,40,50,60,70,80,90
15-- 15,30,45,60,75,90,105
here are a few. The LCM would be 30.
I hope this helps!!
6 0
4 years ago
Read 2 more answers
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