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3 years ago

True or false: any nonzero integers, the product and the quotient have the same sign. Why?

1 answer:

3 0

False. This is because no matter what the number is, the properties of multiplication and division stay the same. So, if one odd and even number multiply/divide, then the answer will be negative. But, if two numbers of the same sign (both either negative or positive) are divided/multiplied, they will both be positive. Therefore, the result is not always the same sign as the starting integers

Hope this helped! :D

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Chose parameters h and k such that the system has a) a unique solution, b) many solutions, and c) no solution. X1 + 3x2 = 4 2x1

AysviL [449]

Answer:

a) The system has a unique solution for $k\neq 6$ and any value of $h$ , and we say the system is consisted

b) The system has infinite solutions for $k=6$ and $h=8$

c) The system has no solution for $k=6$ and $h\neq 8$

Step-by-step explanation:

Since we need to base the solutions of the system on one of the independent terms ( $h$ ), the determinant method is not suitable and therefore we use the Gauss elimination method.

The first step is to write our system in the augmented matrix form:

$\left[\begin{array}{cc|c}1&3&4\\2&k&h\end{array}\right]$

The we can use the transformation $r_0\rightarrow r_0 -2r_1$ , obtaining:

$\left[\begin{array}{cc|c}1&3&4\\0&k-6&h-8\end{array}\right]$ .

Now we can start the analysis:

If $k\neq 6$ then, the system has a unique solution for any value of $k$ , meaning that the last row will transform back to the equation as:

$(k-6)x_2=h-8\\x_2=h-8/(k-6)$

from where we can see that only in the case of $k=6$ the value of $x_2$ can not be determined.

if $k=6$ and $h=8$ the system has infinite solutions: this is very simple to see by substituting these values in the equation resulting from the last row:

$(k-6)x_2=h-8\\0=0$ which means that the second equation is a linear combination of the first one. Therefore, we can solve the first equation to get $x_1$ as a function of $x_2$ o viceversa. Thus, $x_2$ ( $x_1$ ) is called a parameter since there are no constraints on what values they can take on.

if $k=6$ and $h\neq 8$ the system has no solution. Again by substituting in the equation resulting from the last row:

$(k-6)x_2=h-8\\0=h-8$ which is false for all values of $h\neq 8$ and since we have something that is not possible $(0\neq h-8,\ \forall \ h\neq 8)$ the system has no solution

6 0

4 years ago

Something about evaluating numbers or something. Question is in the picture.

Law Incorporation [45]

The answer is 10 - 3√2

:) hope this helps!

3 0

3 years ago

Aron has a bag of marbles that contains three red (r) marbles, four green (g) marbles, and three blue (b) marbles. which express

pshichka [43]

We know that

red marbies=3------> r=3
green marbies=4----> g=4
blue marbies=3-----> b=3
total marbles in the bag=r+g+b-----> 3+4+3-----> 10

a) probability of randomly selecting a green (g) marble
P(green)=g/total marbles------> 4/10-----> 2/5

b) probability of randomly selecting a red (r) marble
the marble is replaced so we have the same total marbles
P(red)=r/total marbles------> 3/10

c) Total probability of selecting this path is:
P(green)*P(red)-----> (2/5)*(3/10)-----> 6/50----> 3/25----> 0.12 (12%)

the answer is
0.12 (12%)

7 0

3 years ago

Read 2 more answers

X+2y = −4 I forgot how to do the problem

enot [183]

Solving for X it would be
X=-4-2y
Solving for Y it would be
Y=-2-x/2

7 0

4 years ago

Read 2 more answers

A rectangle has a length that is 2 greater then its width. The area is 15 square centimeters. what are the dimensions of the rec

GaryK [48]

9514 1404 393

Answer:

length 5 cm
width 3 cm

Step-by-step explanation:

Your knowledge of multiplication facts tells you 3×5 = 15. Your knowledge of addition facts tells you 3+2 = 5. So, the dimensions 3 cm and 5 cm are the width and length of the rectangle, respectively.

3 0

3 years ago