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Usimov [2.4K]
4 years ago
9

Are all congruent polygons similar?

Mathematics
2 answers:
marshall27 [118]4 years ago
8 0
Polygons whose corresponding angles are congruent are proportion aka similar polygons tip: check the angles and side and see if they are the same
Leya [2.2K]4 years ago
4 0
Yes. because congruent means that all the parts of the graphs are the same, i.e. congruent is a special kind of similar
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Tara and Levi are trying to decide between homeowners insurance policies offered by two different agencies. AAA Insurance has of
nlexa [21]
Given:
House valued at $ 425,000
AAA Insurance: $0.38 per 100 with a $500 deductible
425,000 / 100 = 4,250
4,250 x 0.38 = 1,615 annual premium

Thompson’s insurance: $0.26 per 100 with a $1000 deductible
425,000 / 100 = 4,250
4,250 x 0.26 = 1,105 annual premium
If an incident occurs that results to damage or loss to their home, the couple shall shell out 500 before AA Insurance can take over payment for expenses. On the other hand, the couple will shoulder 1,000 before Thompson’s insurance take over payment for expenses in the event there is an incident that results to damage or loss to their home.

c. Thompson’s Insurance is cheaper even if Tara and Levi experience an incident that results in severe damage or loss to their home.

6 0
3 years ago
Read 2 more answers
Can someone help me
Hitman42 [59]
Median as there is an outlier (anomaly)
5 0
4 years ago
What is the derivative of ln(lnx^3)
chubhunter [2.5K]

the derivative of ln(lnx^3) is \frac{1}{x(lnx)} .

<u>Step-by-step explanation:</u>

Here we have to find the derivative of ln(lnx^3) , Let's find out:

We have , ln(lnx^3) , Let's differentiate it w.r.t x :

⇒ \frac{d(ln(lnx^3))}{dx}

Let lnx^3 = u

⇒ \frac{d(ln(u))}{dx}

⇒ \frac{1}{u} (\frac{d(u)}{dx} )

⇒ \frac{1}{u} (\frac{d(ln(x^3))}{dx} )

Let x^3=v

⇒ \frac{1}{u} (\frac{d(ln(v))}{dx} )

⇒ \frac{1}{u}(\frac{1}{v} ) (\frac{d(v)}{dx} )

⇒ \frac{1}{u}(\frac{1}{v} ) (\frac{d(x^3)}{dx} )            

⇒ 3x^2(\frac{1}{u})(\frac{1}{v} )

Putting value of u & v we get:

⇒ 3x^2(\frac{1}{lnx^3})(\frac{1}{x^3} )

⇒ \frac{3}{x(lnx^3)}    { lnx^n = n(lnx)  }

⇒ \frac{3}{3x(lnx)}

⇒ \frac{1}{x(lnx)}

Therefore , the derivative of ln(lnx^3) is \frac{1}{x(lnx)} .

6 0
3 years ago
#13 please help solve this
rusak2 [61]

Answer:

The change in the total cost for each book printed is $10

The cost to get started is $1250

Step-by-step explanation:

In the linear equation y = m x + b, where

  • m is the rate of change per unit
  • b is the initial amount (value y at x = 0)

∵ y represents the total cost of publishing a book in dollars

∵ x represents the number of copies of the book printed

∵ y = 1250 + 10 x

- Compare it with the linear equation y = m x + b

∴ m = 10

∴ b = 1250

∵ m = Δy/Δx

- That means m is the change in the total cost per book

∴ The change in the total cost for each book printed is $10

∵ b is value y at x = 0

- That means b is the cost to get started before print any book)

∴ The cost to get started is $1250

5 0
4 years ago
Kathryn is cooking pancakes. The recipe calls for 3 2/9 cups of flour. She accidentally put in 5 4/5 cups. how many extra cups d
atroni [7]
Subtract
29/5 - 29/9
261/45 - 145/45
116/45
4 0
3 years ago
Read 2 more answers
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