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12.04.2022

What is x in 15 = -3(x -2)

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24.06.2023, solved by verified expert

Bhaskar Singh Bora

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x = -3

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to Right

Equality Properties

Multiplication Property of Equality Division Property of Equality Addition Property of Equality Subtraction Property of Equality

Step-by-step explanation:

Step 1: Define

15 = -3(x - 2)

Step 2: Solve for x

[Division Property of Equality] Divide -3 on both sides: -5 = x - 2[Addition Property of Equality] Add 2 on both sides: -3 = xRewrite/Rearrange: x = -3

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Mathematics

What is x in 15 = -3(x -2)

Step-by-step answer

P Answered by PhD

1

x = -3

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to Right

Equality Properties

Multiplication Property of Equality Division Property of Equality Addition Property of Equality Subtraction Property of Equality

Step-by-step explanation:

Step 1: Define

15 = -3(x - 2)

Step 2: Solve for x

[Division Property of Equality] Divide -3 on both sides: -5 = x - 2[Addition Property of Equality] Add 2 on both sides: -3 = xRewrite/Rearrange: x = -3

Mathematics

The function fff is given in three equivalent forms. Which form most quickly reveals the zeros (or roots ) of the function? Choose 1 Choose 1 (Choice A) A f(x)=-3(x-2)^2+27f(x)=−3(x−2) 2 +27f, (, x, ), equals, minus, 3, (, x, minus, 2, ), squared, plus, 27 (Choice B) B f(x)=-3(x+1)(x-5)f(x)=−3(x+1)(x−5)f, (, x, ), equals, minus, 3, (, x, plus, 1, ), (, x, minus, 5, )(Choice C) C f(x)=-3x^2+12x+15f(x)=−3x 2 +12x+15f, (, x, ), equals, minus, 3, x, squared, plus, 12, x, plus, 15 Write one of the zeros. xxx =

Step-by-step answer

P Answered by PhD

9

(B) f(x)=-3(x+1)(x-5)

x=5

Step-by-step explanation:

Given the three equivalent forms of f(x):

$f(x)=-3(x-2)^2+27\\f(x)=-3(x+1)(x-5)\\f(x)=-3x^2+12x+15$

The form which most quickly reveals the zeros (or "roots") of f(x) is

(B) f(x)=-3(x+1)(x-5)

This is as a result of the fact that on equating to zero, the roots becomes immediately evident.

$f(x)=-3(x+1)(x-5)=0\\-3\neq 0\\Therefore:\\x+1=0$ or x-5=0\\The zeros are x=-1 or x=5$

Therefore, one of the zeros, x=5

Mathematics

The function fff is given in three equivalent forms. Which form most quickly reveals the zeros (or roots ) of the function? Choose 1 Choose 1 (Choice A) A f(x)=-3(x-2)^2+27f(x)=−3(x−2) 2 +27f, (, x, ), equals, minus, 3, (, x, minus, 2, ), squared, plus, 27 (Choice B) B f(x)=-3(x+1)(x-5)f(x)=−3(x+1)(x−5)f, (, x, ), equals, minus, 3, (, x, plus, 1, ), (, x, minus, 5, )(Choice C) C f(x)=-3x^2+12x+15f(x)=−3x 2 +12x+15f, (, x, ), equals, minus, 3, x, squared, plus, 12, x, plus, 15 Write one of the zeros. xxx =

Step-by-step answer

P Answered by PhD

9

(B) f(x)=-3(x+1)(x-5)

x=5

Step-by-step explanation:

Given the three equivalent forms of f(x):

$f(x)=-3(x-2)^2+27\\f(x)=-3(x+1)(x-5)\\f(x)=-3x^2+12x+15$

The form which most quickly reveals the zeros (or "roots") of f(x) is

(B) f(x)=-3(x+1)(x-5)

This is as a result of the fact that on equating to zero, the roots becomes immediately evident.

$f(x)=-3(x+1)(x-5)=0\\-3\neq 0\\Therefore:\\x+1=0$ or x-5=0\\The zeros are x=-1 or x=5$

Therefore, one of the zeros, x=5

Mathematics

Can anyone tell me why I got this wrong? -3(x+2) 15 or x-3 ≤ -1 my answer was ( -7, 2 ) The options are A. ( - infinity, 2) B. (-7, -2) C. [ - 7, infinity) D. ( -7, 2)

Step-by-step answer

P Answered by PhD

1

The answer should be A. $x \in (-\infty,\, 2)$ .

Step-by-step explanation:

Start with the second inequality, which is more straightforward than the first. Add to both sides of this equation to separate from the numbers:

$x - 3 + 3 \le -1 + 3$ . The direction of the inequality sign shall stay the same.

$x \le 2$ .

On the other hand, simplifying the first inequality takes a bit more work. Start by dividing both sides by (-3) . However, keep in mind that dividing or multiplying both sides of an inequality with a negative number will invert the direction of the inequality sign. For example, the left-hand side of an inequality might have been greater than the right-hand side. When both sides are multiplied with a number less than zero, the new left-hand side would become smaller than the new right-hand side.

Because $(-3)\!$ is a negative number, dividing both sides of the first inequality inequality by (-3) will turn the "greater-than" sign into a "less-than" sign.

The inequality used to be:

$-3\, (x + 2) 15$ .

After dividing both sides by the negative number (-3) , it becomes:

$\displaystyle \frac{(-3)\, (x + 2 )}{(-3)} < \frac{15}{(-3)}$ .

$\displaystyle (x + 2) < -5$ .

x < -7 .

The question connected the two original inequalities with "or". Therefore, the conditions will be satisfied as long as either of at least one of the two inequalities is true. Note, that x < 2 is true whenever is true. Therefore, any real number that is smaller than will meet the requirements- not just those that are smaller than (-7) .

The corresponding interval will be $(-\infty,\, 2)$ .

Mathematics

Can anyone tell me why I got this wrong? -3(x+2) 15 or x-3 ≤ -1 my answer was ( -7, 2 ) The options are A. ( - infinity, 2) B. (-7, -2) C. [ - 7, infinity) D. ( -7, 2)

Step-by-step answer

P Answered by PhD

1

The answer should be A. $x \in (-\infty,\, 2)$ .

Step-by-step explanation:

Start with the second inequality, which is more straightforward than the first. Add to both sides of this equation to separate from the numbers:

$x - 3 + 3 \le -1 + 3$ . The direction of the inequality sign shall stay the same.

$x \le 2$ .

On the other hand, simplifying the first inequality takes a bit more work. Start by dividing both sides by (-3) . However, keep in mind that dividing or multiplying both sides of an inequality with a negative number will invert the direction of the inequality sign. For example, the left-hand side of an inequality might have been greater than the right-hand side. When both sides are multiplied with a number less than zero, the new left-hand side would become smaller than the new right-hand side.

Because $(-3)\!$ is a negative number, dividing both sides of the first inequality inequality by (-3) will turn the "greater-than" sign into a "less-than" sign.

The inequality used to be:

$-3\, (x + 2) 15$ .

After dividing both sides by the negative number (-3) , it becomes:

$\displaystyle \frac{(-3)\, (x + 2 )}{(-3)} < \frac{15}{(-3)}$ .

$\displaystyle (x + 2) < -5$ .

x < -7 .

The question connected the two original inequalities with "or". Therefore, the conditions will be satisfied as long as either of at least one of the two inequalities is true. Note, that x < 2 is true whenever is true. Therefore, any real number that is smaller than will meet the requirements- not just those that are smaller than (-7) .

The corresponding interval will be $(-\infty,\, 2)$ .

Mathematics

A new school randomly assigns a six-character student code to every prospective student. The school uses the characters 1, 2, 3, 4, 5, X, Y, and Z. If none of the student codes contain a repeated character, what is the total number of codes that can be randomly assigned?

Step-by-step answer

P Answered by PhD

The total nom of code that can be used is equal to 5+3 = 8

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Step-by-step answer

P Answered by PhD

F=ma

where F=force

m=mass

a=acceleration

Here,

F=4300

a=3.3m/s2

m=F/a

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