For what values of x is the rational expression below undefined?
Check all that apply.
x-4/3x^2-75
A. -3
B. -4
C. 3
D. -5
E. 4
F. 5

we have  

[tex]0.5-\left|x-12\right|=-0.25[/tex]  

we know that        

The absolute value has two solutions

Subtract  [tex]0.5[/tex] both sides

[tex]-\left|x-12\right|=-0.25-0.5[/tex]  

[tex]-\left|x-12\right|=-0.75[/tex]  

Step 1

Find the first solution (Case positive)

[tex]-[+(x-12)]=-0.75[/tex]

[tex]-x+12=-0.75[/tex]

Subtract  [tex]12[/tex] both sides

[tex]-x+12-12=-0.75-12[/tex]

[tex]-x=-12.75[/tex]

Multiply by [tex]-1[/tex] both sides

[tex]x=12.75[/tex]

Step 2

Find the second solution (Case negative)

[tex]-[-(x-12)]=-0.75[/tex]

[tex]x-12=-0.75[/tex]

Adds  [tex]12[/tex] both sides

[tex]x=-0.75+12[/tex]

[tex]x=11.25[/tex]

Statements

case A) The equation will have no solutions

The statement is False

Because the equation has two solutions------> See the procedure

case B) A good first step for solving the equation is to subtract 0.5 from both sides of the equation

The statement is True ----->  See the procedure

case C) A good first step for solving the equation is to split it into a positive case and a negative case

The statement is False ----->  See the procedure

case D) The positive case of this equation is 0.5 – |x – 12| = 0.25

The statement is False

Because the positive case is [tex]0.5-(x-12)=-0.25[/tex] -----> see the procedure

case E) The negative case of this equation is x – 12 = –0.75

The statement is True -----> see the procedure

case F) The equation will have only 1 solution

The statement is False

Because The equation has two solutions------> See the procedure

The equation 0.5 - |x - 12| = -0.25 has no solutions because an absolute value cannot be negative. Attempting to split the equation into positive and negative cases or solving for x is fruitless because the left side of the equation will always be at least 0.5.

When solving the equation 0.5 - |x - 12| = -0.25, we can immediately notice that it will have no solutions because the absolute value is always non-negative, and therefore the left-hand side cannot be less than 0.5. Hence, subtracting 0.5 from both sides is not a good first step. Instead, you would typically isolate the absolute value on one side, but given that the equation equals a negative number, we know it has no solutions without additional steps.

Additionally, splitting the equation into a positive case and a negative case isn't useful here, because no matter what's inside the absolute value, the output cannot lead to a negative result, thus making both cases moot.

The statements that say "The positive case of this equation is 0.5 - |x - 12| = 0.25" and "The negative case of this equation is x - 12 = -0.75" are incorrect as they misinterpret how the absolute value works. Lastly, the equation does not have any solution, so it cannot have only one solution.

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Whole numbers are sometimes integers. Correct option is b.

Integers include both positive and negative whole numbers, as well as zero. Whole numbers are a subset of integers, but they do not include negative numbers. So, while all whole numbers are integers, not all integers are whole numbers.

Relationship between Whole Numbers and Integers:

Every whole number is an integer: Since whole numbers include zero and all positive counting numbers, they are also part of the set of integers.

Not every integer is a whole number: Integers also include negative numbers, which are not part of the set of whole numbers.

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7x4 = 28

5 x z = 28

z = 28/5 = 5.6

They can differ only by a constant is true of f(x) and g(x) if both are antiderivatives of f(x) Hence, option D is correct.

When two functions, F(x)  and G(x), are antiderivatives of the same function f(x), it means that their derivatives are equal to f(x).

This relationship can be represented as:

F'(x) = G'(x) = f(x)

However, it's important to note that if  F(x) and G(x) are both antiderivatives of F(x), then their difference, F(x) - G(x), will have a derivative of zero.

Consequently, F(x) and G(x) can differ only by a constant.

So, the correct option is D.

Complete question:

What must be true of f(x) and G(x) if both of them are antiderivatives of f(x)?

A. They are the same function

B. They can differ by a factor of x

C. If is not possible for two functions to be antiderivatives of the same function

D. They can differ only by a constant

Answer: [tex]x-5=\pm 3[/tex] will be the next step of the given expression.

Step-by-step explanation:

Since, Given expression is |x-5|+2=5

On solving the above expression,

Step 1.  [tex]|x-5|+2=5[/tex]

step 2. [tex]|x-5| = 5-2[/tex]

Step 3. [tex]|x-5| = 3[/tex]

Step 4. [tex]x-5=\pm 3[/tex] (because mode takes both positive and negative values)

A = hours for plan A
B = hours for plan B

Monday: 6A + 5B = 7
Tuesday: 2A + 3B = 3

use elimination by multiplying the 2nd equation by 3.

Doing that we get 3(2A + 3B = 3) = 6A + 9B = 9

So the two equations are now:
6A + 9B = 9

6A + 5B = 7

Subtract and we have 4B = 2

B = 2/4 = 1/2 of an hour

Now put 1/2 back into either equation to solve for A

6A + 5(1/2) = 7
6A + 5/2 = 7
6A = 14/2 -5/2
6A = 9/2
divide by 6 to get A = 9/12 = ¾  hours

Plan A = 3/4 hour

 Plan B = 1/2 hour

Final answer:

By setting up and solving a system of equations, we find that Plan A lasts for 45 minutes per session and Plan B lasts for 30 minutes per session.

Explanation:

Solving for the Duration of Workout Plans

We have information regarding the total duration of workouts and the number of clients for two consecutive days. To find the duration of each workout plan, we use a system of equations. Let A represent the duration of Plan A and B represent the duration of Plan B. The equations based on the given information are:

6A + 5B = 420 minutes (7 hours on Monday)

2A + 3B = 180 minutes (3 hours on Tuesday)

Multiplying the second equation by 3 gives us:

6A + 9B = 540

Subtracting the first equation from this result gives us:

4B = 120 minutes, therefore, B = 30 minutes

Now we substitute B = 30 in the first equation:

6A + 150 = 420, which simplifies to 6A = 270, hence A = 45 minutes

Thus, Plan A lasts for 45 minutes and Plan B lasts for 30 minutes.

Answer:

Transitive

Step-by-step explanation:

Just took the test

Hope it helps :)

Answer: Option 'B' is correct.

Step-by-step explanation:

Since we have given that

Number of cards in a deck = 52

Number of heart = 13

Number of king = 4

But we know that heart contains one king too.

So, to avoid double counting we have to subtract 1 from it.

so, Number of king = 3

So, Probability that Sam picked a heart or a king is given by

[tex]\frac{13}{52}+\frac{3}{52}\\\\=\frac{16}{52}\\\\[/tex]

Hence, Option 'B' is correct.

Answer:

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Step-by-step explanation: I just took the test and got it right.

An underlined greater than sign in mathematics represents a strict inequality, indicating that one value is significantly greater than another.

In mathematics, an underlined greater than sign usually represents an inequality. When a greater than sign (>) is underlined, it indicates a strict inequality, meaning that the value on the left side is significantly greater than the value on the right side.

For example, if we have the underlined inequality 5 > 3, it means that 5 is larger than 3 and there is a clear distinction between the two values.

It's important to note that this is just one possible interpretation of an underlined greater than sign, as the context in which it is used can vary.

Zeros of the given equation [tex]3x^{2} -7x+ 1[/tex] are [tex]\frac{7+\sqrt{37} }{6} \ or \frac{7-\sqrt{37} }{6}[/tex].

The zeros of a quadratic equation f(x) are all the x-values that make the polynomial equal to zero.

The quadratic formula helps us solve any quadratic equation. First, we bring the equation to the form ax²+bx+c=0, where a, b, and c are coefficients. Then, we plug these coefficients in the formula:

[tex]x = \frac{-b \pm \sqrt{b^{2} -4ac} }{2a}[/tex]

According to the given question.

We have a function.

[tex]f(x) = 3x^{2} -7x+1[/tex]

To find the zeros of the function equate f(x) = 0.

[tex]3x^{2} -7x+1 = 0\\[/tex]

Solve the above equation by quadratic method.

[tex]x = \frac{7\pm\sqrt{(7)^{2} -4(3)(1)} }{2(3)}[/tex]

[tex]\implies x = \frac{7\pm\sqrt{49-12} }{6}[/tex]

[tex]\implies x = \frac{7\pm\sqrt{37} }{6}[/tex]

[tex]\implies x = \frac{7+\sqrt{37} }{6} \ or \frac{7-\sqrt{37} }{6}[/tex]

Hence, zeros of the given equation [tex]3x^{2} -7x+ 1[/tex] are [tex]\frac{7+\sqrt{37} }{6} \ or \frac{7-\sqrt{37} }{6}[/tex].

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Answer:

5.2

Step-by-step explanation:

The expression that satisfies the given condition is using if-else conditional statements to check the value of the variable 'month' and assign the corresponding month name.

The expression that satisfies the given condition is:

if (month == 1) {     answer = "jan"; } else if (month == 2) {     answer = "feb"; } else if (month == 3) {     answer = "mar"; } // ... continue this pattern for the remaining months

This code uses conditional statements (if-else) to check the value of the variable 'month' and assigns the corresponding month name to the variable 'answer'.

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