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3 years ago

Kendra found this relationship between the quantities in the table.

Mathematics

2 answers:

slava [35]3 years ago

6 0

Answer:

Kendra should have multiplied the x-values by 75 to get the y-values

Step-by-step explanation:

Given

Table

X|| Y

1 || 75

2 || 150

3 || 225

4 || 300

5 || 375

Given that Kendra multiply x by 7.5 to get y

The relationship of x and y can be calculated as thus;

y = rx

Where y and x are the values at the y and x column respectively and r is the constant of proportionality

When y = 75, x = 1.

Plug in these values in the above formula

y = rx becomes

75 = r * 1

75 = r

r = 75

When y = 150, x = 2

150 = r * 2

Multiply both sides by ½

150 * ½ = r * 2 * ½

75 = r

r = 75

When y = 225, x = 3

225 = r * 3

Multiply both sides by ⅓

225 * ⅓ = r * 3 * ⅓

75 = r

r = 75

Notice that r remains 75 and the difference between y values is 75

If you apply these formula on when y = 300 or 375 and when x = 4 or 5, the constant of proportionality will remain The value of 75.

Hence, Kendra mistake is that; Kendra should have multiplied the x-values by 75 to get the y-values

inessss [21]3 years ago

4 0

Answer:

C-Kendra should have multiplied the x-values by 75 to get the y-values.

Step-by-step explanation:

took test

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Solve the following equation. Remember to check for extraneous solutions 1/(x-6) + (x/(x-2)) = (4/(x²-8x+12)).

xxMikexx [17]

Answer:

-1, 2, 6

Step-by-step explanation:

We have to solve the equation as follows: 1/(x-6) + (x/(x-2)) = (4/(x²-8x+12)).

Now, we have, $\frac{1}{x-6} +\frac{x}{x-2} = \frac{4}{x^{2}-8x+12 }$

⇒ $\frac{(x-2)+x(x-6)}{(x-2)(x-6)} = \frac{4}{x^{2}-8x+12 }$

⇒ $\frac{x-2+x^{2}-6x }{(x-2)(x-6)} =\frac{4}{(x-2)(x-6)}$

⇒ $\frac{(x-2)(x-6)}{x^{2}-5x-2 }=\frac{(x-2)(x-6)}{4}$

⇒ $(x-2)(x-6)[\frac{1}{x^{2} -5x-2} -\frac{1}{4} ]=0$

⇒ $(x-2)(x-6) =0$ or, $[\frac{1}{x^{2} -5x-2} -\frac{1}{4} ]=0$

If, (x-2)(x-6) =0, then x=2 or x=6

If, $[\frac{1}{x^{2} -5x-2} -\frac{1}{4} ]=0$ , then $x^{2} -5x-2=4$

and (x-6)(x+1) =0

Therefore, x=6 or -1

So the solutions for x are -1, 2 6. (Answer)

4 0

3 years ago

For what value of k does the equation 6(x + 1) + 2 = 3(k5x + 1) + 3 have no solution?

NARA [144]

Answer:

k = (6/15)

Step-by-step explanation:

The equation is:

6*(x + 1) + 2 = 3*(k*5*x + 1) + 3

To have no solutions, we need to have something like:

x + 7 = x + 4

where we can remove x in both sides and end with

7 = 4

So this equation is false, meaning that there is no value of x such that this equation is true, then the equation has no solutions.

First, let's try to simplify our equation:

6*(x + 1) + 2 = 3*(k*5*x + 1) + 3

6*x + 6 + 2 = 3*k*5*x + 3*1 + 3

6*x + 8 = 15*k*x + 6

if 15*k = 6, then the system clerly has no solution.

then:

k = 6/15

then we get:

6*x + 8 = (6/15)*15*x + 6

6*x + 8 = 6*x + 6

8 = 6

The system has no solutions.

8 0

3 years ago

Complete the pattern 274÷1=

Orlov [11]

The answer 274 because anything divided by one is itself

5 0

3 years ago

Find a 75% confidence interval for the population average weight μ of all adult mountain lions in the specified region. (Round y

kupik [55]

Answer:

75% confidence interval is 91.8±16.66. That is between 75.1 and 108.5 pounds.

Step-by-step explanation:

The question is missing. It is as follows:

Adult wild mountain lions (18 months or older) captured and released for the first time in the San Andres Mountains had the following weights (pounds): 69 104 125 129 60 64

Assume that the population of x values has an approximately normal distribution.

Find a 75% confidence interval for the population average weight μ of all adult mountain lions in the specified region. (Round your answers to one decimal place.)

75% Confidence Interval can be calculated using M±ME where