A statue is mounted on top of a 21 foot hill. From the base of the hill to where you are standing is 57feet and the statue subte

Point T is the midpoint of RS, W is the midpoint of RT, and Z is the midpoint of WS. If the length of TZ is x, find the lengths

PIT_PIT [12445]

Let RS be denoted as y.

Given that T is the midpoint of RS, this results in RT = TS = $\frac{y}{2}$

And W acting as the midpoint of RT, leads to

thus RW = WT = $\frac{\frac{y}{2} }{2} = \frac{y}{4}$

Given that Z is the midpoint of WS.

Meaning WZ = ZS = $\frac{WS}{2} = \frac{WT+TS}{2} = \frac{(\frac{y}{4} +\frac{y}{2} )}{2} =\frac{3y}{8}$

Consequently, TZ = TS - ZS = $\frac{y}{2} -\frac{3y}{8} = \frac{4y-3y}{8} = \frac{y}{8}$

However, TZ is defined as x.

Thus, $\frac{y}{8} = x$

y = 8x = RS.

A) Length of RW = $\frac{y}{4} = \frac{8x}{4} = 2x$

B) Length of WZ = $\frac{3y}{8} =\frac{3*8x}{8} = 3x$

C) Length of RS = y = 8x

D) Length of ZS = WZ = 3x.

An image is provided for clarification.

8 0

2 months ago

Cullen is 10 years younger than Ada. The product of their ages 2 years ago was 39.

Inessa [12570]

Answer:

15

Step-by-step explanation:

8 0

2 months ago

Construct a scatter diagram using the data table to the right. This data is from a study comparing the amount of tar and carbon

PIT_PIT [12445]

Thank you for bringing your question here. I hope you find the answer useful. Don't hesitate to ask additional questions. Yes, as the level of tar rises, the concentration of carbon monoxide also increases.

7 0

1 month ago

If 30,000 cm2 of material is available to make a box with a square base and an open top, what is the largest possible volume (in

Inessa [12570]

Answer:

The highest achievable volume for the box is 2000000 cubic meters.

Step-by-step explanation:

Below is an outline of the volume ( $V$ ), measured in cubic centimeters, and surface area ( $A_{s}$ ), measured in square centimeters, for a box featuring a square base:

$A_{s} = l^{2}+h\cdot l$ (1)

$V = l^{2}\cdot h$ (2)

Where:

$l$ - The length of the base's side, in centimeters.

$h$ - The height of the box, in centimeters.

Using (2), we isolate $h$ in the formula:

$h = \frac{V}{l^{2}}$

Then, we substitute into (1) and simplify the outcome:

$A_{s} = l^{2}+ \frac{V}{l}$

$A_{s}\cdot l = l^{3}+V$

$V = A_{s}\cdot l -l^{3}$ (3)

Next, we calculate the first and second derivatives of this expression:

$V' = A_{s}-3\cdot l^{2}$ (4)

$V'' = -6\cdot l$ (5)

If $V' = 0$ and $A_{s} = 30000\,cm^{2}$ , then we find that the critical value for the base's side length is:

$30000-3\cdot l^{2} = 0$

$3\cdot l^{2} = 30000$

$l = 100\,cm$

Subsequently, we assess this outcome using the second derivative's expression:

$V'' = -600$

According to Second Derivative Test, this critical value signifies an absolute maximum. Consequently, the largest volume obtainable for the box is:

$V = 30000\cdot l - l^{3}$

$V = 2000000\,cm^{3}$

The highest achievable volume for the box is 2000000 cubic meters.

4 0

1 month ago