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3 years ago

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A flea 1/16 of an inch tall can jump 9 3/8 inches high. If a person 6 feet tall had the ability to jump like a flea, how high co

uld he jump?
The First-person to answer both questions gets the brainiest

The scale on a globe is 1/8 inch equals 125 miles. How many inches is the circumference of that globe? (Use 25,000 miles for the circumference of the earth.)

2 answers:

rodikova [14]3 years ago

8 0

Answer:

first you.. then you have to.. and after you do that then... you have your answer. Hope that helps, have a good day!

Step-by-step explanation:

Daniel [21]3 years ago

4 0

^^^^^^^^^^^^^^^^^^^^^

You might be interested in

I need the answers to this please and thank you:)

Liono4ka [1.6K]

Q # 1

Explanation

Given the parabola

$f\left(x\right)=\left(x-3\right)^2-1$

Openness

It OPENS UP, as 'a=1' is positive.

Finding Vertex

The vertex of an up-down facing parabola of the form

$y=ax^2+bx+c\:\mathrm{is}\:x_v=-\frac{b}{2a}$

$\mathrm{Rewrite}\:y=\left(x-3\right)^2-1\:\mathrm{in\:the\:form}\:y=ax^2+bx+c$

$y=x^2-6x+8$

$a=1,\:b=-6,\:c=8$

$x_v=-\frac{\left(-6\right)}{2\cdot \:1}$

$x_v=3$

Finding $y_v$

$y_v=3^2-6\cdot \:3+8$

$y_v=-1$

So vertex is:

$\left(3,\:-1\right)$

Horizontal Translation

$y=\left(x-3\right)^2$ moves the graph RIGHT 3 units.

Vertical Translation

$f\left(x\right)=\left(x-3\right)^2-1$ moves the graph DOWN 1 unit.

Stretch or Compress Vertically

As $a = 1$ , so it does not affect the stretchiness or compression.

Q # 2

Explanation:

$f\left(x\right)=-\left(x+1\right)^2-2$

Openness

It OPENS DOWN, as 'a=-1' is negative.

Vertex

$\mathrm{Rewrite}\:y=-\left(x+1\right)^2-2\:\mathrm{in\:the\:form}\:y=ax^2+bx+c$

$y=-x^2-2x-3$

$a=-1,\:b=-2,\:c=-3$

$x_v=-\frac{\left(-2\right)}{2\left(-1\right)}$

$x_v=-1$

$\mathrm{Plug\:in}\:\:x_v=-1\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}$

$y_v=-2$

So vertex is:

$\left(-1,\:-2\right)$

Horizontal Translation

$y=\left(x+1\right)^2$ moves the graph LEFT 1 unit.

Vertical Translation

$f\left(x\right)=\left(x+1\right)^2-2$ moves the graph DOWN 2 unit.

Stretch or Compress Vertically

As $a = -1$ < 0, so it is either stretched or compressed.

Q # 3

Explanation:

$f\left(x\right)=\frac{1}{3}\left(x-4\right)^2+6$

It OPENS UP, as 'a=1/3' is positive.

Vertex

$\mathrm{Rewrite}\:y=\frac{1}{3}\left(x-4\right)^2+6\:\mathrm{in\:the\:form}\:y=ax^2+bx+c$

$y=\frac{1\cdot \:x^2}{3}-\frac{8x}{3}+\frac{34}{3}$

$a=\frac{1}{3},\:b=-\frac{8}{3},\:c=\frac{34}{3}$

$x_v=-\frac{\left(-\frac{8}{3}\right)}{2\left(\frac{1}{3}\right)}$

$x_v=4$

Finding $y_v$

$y_v=\frac{1\cdot \:4^2}{3}-\frac{8\cdot \:4}{3}+\frac{34}{3}$

$y_v=6$

So vertex is:

$\left(4,\:6\right)$

Horizontal Translation

$f\left(x\right)=\left(x-4\right)^2$ moves the graph RIGHT 4 units.

Vertical Translation

$f\left(x\right)=}\left(x-4\right)^2+6$ moves the graph UP 6 unit.

Stretch or Compress Vertically

As $a=\frac{1}{3}$ , so it the graph is vertically compressed by a factor of 1/3.

Check the attached comparison graphs.

Q # 4

Explanation:

Given the function

$f\left(x\right)=-\left(x+3\right)^2$

It OPENS DOWN, as 'a=-1' is negative.

Vertex

The vertex of an up-down facing parabola of the form $y=a\left(x-m\right)\left(x-n\right)$

is the average of the zeros $x_v=\frac{m+n}{2}$

$y=-\left(x+3\right)^2$

$a=-1,\:m=-3,\:n=-3$

$x_v=\frac{m+n}{2}$

$x_v=\frac{\left(-3\right)+\left(-3\right)}{2}$

$x_v=-3$

Finding $y_v$

$y_v=-\left(-3+3\right)^2$

$y_v=0$

So vertex is:

$\left(-3,\:0\right)$

Horizontal Translation

$y=\left(x+3\right)^2$ moves the graph LEFT 3 units.

Vertical Translation

$y=\left(x+3\right)^2$ does not move the graph vertically.

Stretch or Compress Vertically

As $a=-1$ , so it the graph is either vertically stretched or compressed.

Q # 5

Explanation:

$f\left(x\right)=\left(x+5\right)^2-3$

Openness

It OPENS UP, as 'a=1' is positive.

Vertex

$\mathrm{Rewrite}\:y=\left(x+5\right)^2-3\:\mathrm{in\:the\:form}\:y=ax^2+bx+c$

$y=x^2+10x+22$

$a=1,\:b=10,\:c=22$

$x_v=-\frac{10}{2\cdot \:1}$

$x_v=-5$

Finding $y_v$

$y_v=\left(-5\right)^2+10\left(-5\right)+22$

So vertex is:

$\left(-5,\:-3\right)$

Horizontal Translation

$f\left(x\right)=\left(x+5\right)^2$ moves the graph LEFT 5 units.

Vertical Translation

$f\left(x\right)=\left(x+5\right)^2-3$ moves the graph DOWN 3 unit.

Stretch or Compress Vertically

As $a = 1$ , so it does not affect the stretchiness or compression.

Check the attached comparison graphs.

Q # 6

THE DETAILS OF COMPLETE SOLUTION OF QUESTION 6 IS ATTACHED IN THE DIAGRAM AS THE 5000 CHARACTERS WERE ALREADY FILLED. SO, I solved via the attached figure.

SO, PLEASE CHECK THE LAST FIGURE TO FIND THE COMPLETE SOLUTION OF THE Q#6.

6 0

3 years ago

Graph the following piecewise function on a separate piece of graph paper and upload your graph below.

Alecsey [184]

ANSWER

To graph the function

$f(x)=\left \{ {{x+4\:\:if\:\:-4\leq x$

follow the steps below.

1. Find y- intercept by plugging in $x=0$ .

$x=0$ is on the interval, $-4\leq x$ , so we substitute in to

$f(x)=x+4$

$\Rightarrow f(0)=0+4$

$\Rightarrow f(0)=4$

Hence the y-intercept is $(0,4)$

2. Find x-intercept by setting $f(x)=0$

This implies that

$x+4=0,$ on $-4\leq x$

or

$2x-1=0$ on $3\leq x$

We now solve for $x$ on each interval,

$x=-4,$ on $-4\leq x$

or

$x=\frac{1}{2}$ on $3\leq x$

But observe that

$x=\frac{1}{2}$ does not belong to $3\leq x$

This means it can never be an intercept for this piece-wise function.

Hence our x-intercept is $(-4,0)$

3. Plotting the boundaries of the interval.

For $f(x)=x+4$ on $-4\leq x$

$f(-4)=-4+4$

$\Rightarrow f(-4)=0$ .

This point $(-4,0)$ coincides with the x-intercept.

$f(3)=3+4$

$f(3)=7$

So we have the point $(3,7)$ . But note that $x=3$ does not belong to this interval so we plot this point as a hole.

For $f(x)=2x-1$ on $3\leq x$

$f(3)=2(3)-1$

$\Rightarrow f(3)=5$

So we plot $(3,5)$

$f(6)=2(6)-1$

$\Rightarrow f(6)=11$

So we plot $(6,11)$ also as a hole.

Plotting all these points we can now graph the function,

$f(x)=\left \{ {{x+4\:\:if\:\:-4\leq x$

See attachment for graph.

7 0

3 years ago

Read 2 more answers

A crate made of wood measures 50 cm wide, 70 cm long, and 40 cm tall. Which unit of measurement should be used to figure out the

saveliy_v [14]

Surface area is the area of all the sides added up. If you were to unfold all the areas and make it flat (2-D), area is $cm^{2}$ .

Volume would be cubed because it's 3-D

8 0

3 years ago

5.) To help with a fundraiser, Ms. Rummel baked a total of 23 pies. Each apple pie

madam [21]

Answer:

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Area of a triangle with verticals of (-4,1),(-7,5),(0,1)

Lynna [10]

Answer:

$8 ft^2$

Step-by-step explanation:

I assume that the three points given are the vertices of the triangle. So the three points are:

A (-4,1)

B (-7,5)

C (0,1)

The area of a triangle is given by:

$A=\frac{1}{2}bh$

where b is the base and h is the height.

Here we can take as follows:

- We can take the base to be the side AC, so that its length is equal to the distance between the two points:

$AC=\sqrt{(0-(-4))^2+(1-1)^2}=4$

- The height of the triangle is therefore the perpendicular distance between AC and point B. This is equivalent the difference in the x-coordinate between any point on AC and B, so:

$h=(5-1)=4 ft$

Therefore, the area of the triangle is:

$A=\frac{1}{2}(4)(4)=8 ft^2$

7 0

3 years ago