Resources

Resources

Bytelearn - cat image with glasses

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Eva sees a dolphin 3.2 meters below sea level and a bird (47)/(10) meters above sea level.
Which 2 of the following expressions represents the vertical distance between the dolphin and the bird?
Choose 2 answers:
(A) |(47)/(10)+3.2|
(B) |-3.2-(47)/(10)|
(C)|3.2-(47)/(10)|

Eva sees a dolphin $3.2$ meters below sea level and a bird $\frac{47}{10}$ meters above sea level. $\newline$ Which $2$ of the following expressions represents the vertical distance between the dolphin and the bird? $\newline$ Choose $2$ answers: $\newline$ (A) |\(\frac{47}{10}+3.2|\) $\newline$ (B) $|-3.2$ \(-\frac{47}{10}|\) $\newline$ (C) $|3.2$ \(-\frac{47}{10}|\)

View step-by-step help

View step-by-step help

Transformations of functions

Full solution

Q. Eva sees a dolphin

3.2

meters below sea level and a bird

\frac{47}{10}

meters above sea level.

\newline

Which

2

of the following expressions represents the vertical distance between the dolphin and the bird?

\newline

Choose

2

answers:

\newline

(A) |\(\frac{47}{10}+3.2|\)

\newline

(B)

|-3.2

\(-\frac{47}{10}|\)

\newline

(C)

|3.2

\(-\frac{47}{10}|\)

Find Vertical Distance: To find the vertical distance between two points, we subtract the lower point's position from the higher point's position. Since the dolphin is below sea level, its position is $-3.2$ meters. The bird is above sea level, so its position is $+\frac{47}{10}$ meters. The vertical distance is the absolute value of the difference between these two positions. $\newline$ Calculation: $|(\frac{47}{10}) - (-3.2)| = |(\frac{47}{10}) + 3.2|$
Perform Calculation: Now, let's perform the calculation inside the absolute value. $\newline$ $(\frac{47}{10}) + 3.2 = 4.7 + 3.2 = 7.9$ $\newline$ So, the expression becomes $|7.9|$ , which is simply $7.9$ since the absolute value of a positive number is the number itself.
Express Vertical Distance: We can also express the vertical distance by starting with the dolphin's position and subtracting the bird's position, taking the absolute value of the result. $\newline$ Calculation: $|-3.2 - (\frac{47}{10})| = |-3.2 - 4.7| = |-7.9|$
Confirm Result: The absolute value of $-7.9$ is also $7.9$ , which confirms that the vertical distance between the dolphin and the bird is $7.9$ meters.

More problems from Transformations of functions

Question

What does the transformation

f(x) \mapsto f(x) - 6

do to the graph of

f(x)

\newline

\newline

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

\newline

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

\newline

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

\newline

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

Posted 5 months ago

Question

g(x)

g(x)

is the reflection across the

y

f(x) = -4x + 1

\newline

\newline

\text{[g(x) = 4x - 1]}

\newline

\text{[g(x) = -4x + 1]}

\newline

\text{[g(x) = 4x + 1]}

\newline

\text{[g(x) = -4x - 1]}

Posted 5 months ago

Question

What kind of transformation converts the graph of

f(x) = -8(x - 4)^2 + 6

into the graph of

g(x) = -8(x + 6)^2 + 6

\newline

\newline

\text{[A]translation 10 units left}

\newline

\text{[B]translation 10 units right}

\newline

\text{[C]translation 10 units up}

\newline

\text{[D]translation 10 units down}

Posted 5 months ago

Question

g(x)

g(x)

is the reflection across the x-axis of

f(x) = -6(x - 7)^2 + 2

\newline

\newline

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

\newline

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

\newline

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

\newline

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

Posted 5 months ago

Question

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

\newline

1

\newline

\frac{1}{2}

\newline

1

\newline

\frac{\sqrt{2}}{2}

\newline

(D) The limit doesn't exist.

Posted 10 months ago

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

\newline

3

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

\newline

4

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

\newline

5

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

\newline

6

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

\newline

7

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

\newline

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

\newline

1

\newline

(A) Tess's work is correct.

\newline

1

is incorrect. We're not allowed to factor an expression when solving differential equations.

\newline

4

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

e^{x^{2}+C_{1}}

\newline

7

is incorrect. Tess didn't account for adding

3

to the right side.

Posted 8 months ago

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

\quad 30 y=3 x^{2}+C

\newline

3

\quad y=\frac{x^{2}}{10}+C

\newline

Is Aditya's work correct? If not, what is his mistake?

\newline

1

\newline

(A) Aditya's work is correct.

\newline

1

is incorrect. The separation of variables wasn't done correctly.

\newline

2

is incorrect. Aditya didn't integrate

30

\newline

3

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

\frac{30}{3 x^{2}+C}

Posted 9 months ago

Question

Jonael dropped a sandbag from a hot air balloon at a target that is

250

meters below the balloon. At this moment, the sandbag is

75

meters below the balloon.

\newline

2

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

\newline

2

\newline

|-250+75|

\newline

|250+75|

\newline

|-75+250|

Posted 9 months ago

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

-intercept of the graph of

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

Posted 9 months ago

Question

Determine whether the function

f(x)

is continuous at

x=-3

\newline

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

\newline

f(x)

is discontinuous at

x=-3

\newline

f(x)

is continuous at

x=-3

Posted 9 months ago

Related topics

Algebra - Order of Operations Algebra - Distributive Property `X` and `Y` Axes Geometry - Scalene Triangle Common Multiple Geometry - Quadrant