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3 years ago

Point O is on line segment \overline{NP}

Mathematics

1 answer:

Kobotan [32]3 years ago

4 0

Answer:

$OP = 11$

Step-by-step explanation:

Given

OP = 4x - 5

NP = 3x + 4

NO = 2x - 3

Required

Determine the value of OP

Since O lines n the segment, them

$NP = NO+ OP\\$

$3x + 4 = 2x - 3 + 4x - 5$

Collect Like Terms

$3x - 2x - 4x = -3 -5 - 4$

$-3x=-12$

$x = \frac{-12}{-3}$

$x = 4$

Substitute 4 for x in OP = 4x - 5

$OP = 4 * 4 - 5$

$OP = 16 - 5$

$OP = 11$

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A pyramid has a square base of length 8cm and a total surface area of 144cm². Find the volume of the pyramid. (Please use Pythag

Sveta_85 [38]

Answer:

$\displaystyle V_{ \text{pyramid}}= 64 \: {cm}^{3}$

Step-by-step explanation:

we are given surface area and the length of the square base

we want to figure out the Volume

to do so

we need to figure out slant length first

recall the formula of surface area

$\displaystyle A_{\text{surface}}=B+\dfrac{1}{2}\times P \times s$

where B stands for Base area

and P for Base Parimeter

$\sf\displaystyle \: 144=(8 \times 8)+\dfrac{1}{2}\times (8 \times 4) \times s$

now we need our algebraic skills to figure out s

simplify parentheses:

$\sf\displaystyle \: 64+\dfrac{1}{2}\times32\times s = 144$

reduce fraction:

$\sf\displaystyle \: 64+\dfrac{1}{ \cancel{ \: 2}}\times \cancel{32} \: ^{16} \times s = 144 \\ 64 + 16 \times s = 144$

simplify multiplication:

$\displaystyle \: 16s + 64 = 144$

cancel 64 from both sides;

$\displaystyle \: 16s = 80$

divide both sides by 16:

$\displaystyle \: \therefore \: s = 5$

now we'll use Pythagoras theorem to figure out height

according to the theorem

$\displaystyle \: {h}^{2} + (\frac{l}{2} {)}^{2} = {s}^{2}$

substitute the value of l and s:

$\displaystyle \: {h}^{2} + (\frac{8}{2} {)}^{2} = {5}^{2}$

simplify parentheses:

$\displaystyle \: {h}^{2} + (4 {)}^{2} = {5}^{2}$

simplify squares:

$\displaystyle \: {h}^{2} + 16 = 25$

cancel 16 from both sides:

$\displaystyle \: {h}^{2} = 9$

square root both sides:

$\displaystyle \: \therefore \: {h}^{} = 3$

recall the formula of a square pyramid

$\displaystyle V_{pyramid}=\dfrac{1}{3}\times A\times h$

where A stands for Base area (l²)

substitute the value of h and l:

$\sf\displaystyle V_{ \text{pyramid}}=\dfrac{1}{3}\times \{8 \times 8 \}\times 3$

simplify multiplication:

$\sf\displaystyle V_{ \text{pyramid}}=\dfrac{1}{3}\times 64\times 3$

reduce fraction:

$\sf\displaystyle V_{ \text{pyramid}}=\dfrac{1}{ \cancel{ 3 \: }}\times 64\times \cancel{ \: 3}$

hence,

$\sf\displaystyle V_{ \text{pyramid}}= 64 \: {cm}^{3}$

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Nina owns a vacation home valued at $75,500. She owes $32,126 on her mortgage. How much are the asset and liability values for N

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Answer:

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