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Let’s check out your problem:

(x-4)(x-5)=0
If
x=s and
x=t are the solutions to the given equation, which of the following is equal to the value of
|s-t| ?
Choose 1 answer:
(A) -9
(B) -1
C 1
(D) 9

$(x-4)(x-5)=0$ $\newline$ If $x=s$ and $x=t$ are the solutions to the given equation, which of the following is equal to the value of $|s-t|$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $-9$ $\newline$ (B) $-1$ $\newline$ C $1$ $\newline$ (D) $9$

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Transformations of functions

Full solution

Q.

(x-4)(x-5)=0

\newline

If

x=s

and

x=t

are the solutions to the given equation, which of the following is equal to the value of

|s-t|

?

\newline

Choose

1

answer:

\newline

(A)

-9

\newline

(B)

-1

\newline

C

1

\newline

(D)

9

Solve Equation: Solve the equation $(x-4)(x-5)=0$ to find the values of $x$ that satisfy the equation. $\newline$ The equation is in factored form, so we can use the zero product property, which states that if a product of factors equals zero, then at least one of the factors must be zero. $\newline$ Set each factor equal to zero and solve for $x$ : $\newline$ $x - 4 = 0$ or $x - 5 = 0$ $\newline$ $x = 4$ or $x = 5$
Use Zero Product Property: Assign the solutions to $s$ and $t$ . According to the problem, $x = s$ and $x = t$ are the solutions to the equation. Therefore, we can assign $s = 4$ and $t = 5$ without loss of generality.
Assign Solutions: Calculate the absolute value of the difference between $s$ and $t$ . We need to find $|s - t|$ , which is the absolute value of the difference between the two solutions. $|s - t| = |4 - 5| = |-1|$ The absolute value of $-1$ is $1$ .

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\text{translates it } 6 \text{ units down}

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What kind of transformation converts the graph of

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\newline

\newline

\text{[A]translation 10 units left}

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\text{[B]translation 10 units right}

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Question

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\newline

\newline

g(x) = 6(x - 7)^2 - 2

\newline

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Question

\lim _{x \rightarrow \frac{\pi}{4}} \cos (x)=?

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1

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1

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Question

Tess tried to solve the differential equation

\frac{d y}{d x}=2 x y-6 x

. This is her work:

\newline

\frac{d y}{d x}=2 x y-6 x

\newline

1

\quad \frac{d y}{d x}=(y-3) 2 x

\newline

2

\quad \int \frac{1}{y-3} d y=\int 2 x d x

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3

\quad \ln |y-3|=x^{2}+C_{1}

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4

\quad e^{\ln |y-3|}=e^{x^{2}}+C_{1}

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5

\quad|y-3|=e^{x^{2}}+C_{1}

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7

\quad y= \pm e^{x^{2}}+C

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Is Tess's work correct? If not, what is her mistake?

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1

\newline

(A) Tess's work is correct.

\newline

1

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\newline

4

is incorrect. The right-hand side of the equation should be

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\newline

7

is incorrect. Tess didn't account for adding

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Question

Aditya tried to solve the differential equation

\frac{d y}{d x}=6 x-30

. This is his work:

\newline

\frac{d y}{d x}=6 x-30

\newline

1

\quad \int 30 d y=\int 6 x d x

\newline

2

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3

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1

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1

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2

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\newline

3

is incorrect. The right-hand side of the equation should be

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Question

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\newline

2

of the following expressions represents the distance between the sandbag and the target?

\newline

2

\newline

|-250+75|

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|250+75|

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Question

f(x) = -(x+1)(x+7)

\newline

The function represents a parabola in the

xy

-plane. Which of the following is an equivalent form of

f

y

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f

appears as a constant or coefficient?

\newline

1

\newline

f(x) = -(x+4)^{2}+9

\newline

f(x) = -(x+4)^{2}-9

\newline

f(x) = -x^{2}+8x+7

\newline

f(x) = -x^{2}-8x-7

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Question

Determine whether the function

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\newline

f(x)=\left\{\begin{array}{ll} 18-x^{2}, & x \leq-3 \\ 15+3 x, & x>-3 \end{array}\right.

\newline

f(x)

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\newline

f(x)

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