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padilas [110]
3 years ago
13

What are the restrictions for (x^2+10x+25)/(x^2-3x-28) 0,14 -4,7 0,7 4,7

Mathematics
1 answer:
nirvana33 [79]3 years ago
5 0
The answer is -4,7

The denominator can't equal zero. Factor the denominator:
x^2 - 3x - 28=
(X - 7)(x + 4); next set each set of parentheses equal to 0;
x - 7 = 0; so x=7 is one value
x + 4 = 0; so x=-4 is the other
Remember, x = 7 and x= -4 make the denominator zero, which is a "restriction" because you can't divide by zero.
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Suppose that A and B are points on the number line.
oksian1 [2.3K]

Answer:

Suppose that A and B are points on the number line.

If AB=11 and A lies at 6, where could B be located?

If there is more than one location, separate them with commas

Step-by-step explanation:

Suppose that A and B are points on the number line.

If AB=11 and A lies at 6, where could B be located?

If there is more than one location, separate them with commas

4 0
3 years ago
The sample consisted of 50 night students, with a sample mean GPA of 3.02 and a standard deviation of 0.08, and 25 day students,
bulgar [2K]

Answer: The test statistic is t= -0.90.

Step-by-step explanation:

Since we have given that

n₁ = 50

n₂ = 25

\bar{x_1}=3.02\\\\\bar{x_2}=3.04\\\\\sigma_1=0.08\\\\\sigma_2=0.06

So, s would be

s=\sqrt{\dfrac{n_1\sigma_1^2+n_2\sigma_2^2}{n_1+n_2-2}}\\\\s=\sqrt{\dfrac{50\times 0.08^2+25\times 0.06^2}{50+25-2}}\\\\s=0.075

So, the value of test statistic would be

t=\dfrac{\bar{x_1}-\bar{x_2}}{s(\dfrac{1}{n_1}+\dfrac{1}{n_2})}\\t=\dfrac{3.02-3.04}{0.074(\dfrac{1}{50}+\dfrac{1}{25})}\\\\t=\dfrac{-0.04}{0.074(0.02+0.04)}\\\\t=\dfrac{-0.04}{0.044}\\\\t=-0.90

Hence, the test statistic is t= -0.90.

7 0
3 years ago
How to find out 4 x 754 by using two methods
baherus [9]
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3 years ago
Pointsss here ya gooooo.,
alexandr1967 [171]

Answer:

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6 0
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Read 2 more answers
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