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Aleks04 [339]
3 years ago
9

A pool company is creating a blueprint for a family pool and a similar dog pool for a new client. Which statement explains how t

he company can determine whether pool ABCD is similar to pool EFGH? quadrilaterals ABCD and EFGH Translate EFGH so that point F of EFGH lies on point B of ABCD, then dilate EFGH by the ratio segment EF over segment AB. Translate EFGH so that point E of EFGH lies on point A of ABCD, then translate EFGH so that point F of EFGH lies on point B of ABCD. Translate EFGH so that point E of EFGH lies on point A of ABCD, then dilate EFGH by the ratio segment AB over segment EF. Translate EFGH so that point F of EFGH lies on point B of ABCD, then translate EFGH so that point E of EFGH lies on point A of ABCD.
Mathematics
2 answers:
Xelga [282]3 years ago
5 0

Answer: B.

Step-by-step explanation: These are the multiples of 15 but what you are probably asking or what are the factors of 15. If ABCD and EFGH are similar, then you should be able to line up corresponding points (for example, A corresponds to E, B corresponds to F). if they're different sizes, you should then dilate one of them so that they are the exact same size.

Annette [7]3 years ago
5 0

Answer: B

Step-by-step explanation:

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Ainat [17]

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7\frac{1}{12} or \frac{85}{12}

Step-by-step explanation:

total boxes = Jills boxes + Mikes boxes

total boxes = 5\frac{1}{3} +1\frac{3}{4}

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3 0
2 years ago
For his birthday, Tyrone's parents gave him $7,790.00 which they put into a savings account that earns 15% interest compounded m
torisob [31]

Answer:

5 years and 5 months

Step-by-step explanation:

<u />

<u>Compound Interest Formula</u>

\large \text{$ \sf A=P(1+\frac{r}{n})^{nt} $}

where:

  • A = final amount
  • P = principal amount
  • r = interest rate (in decimal form)
  • n = number of times interest applied per time period
  • t = number of time periods elapsed

Given:

  • A = $17,474.00
  • P = $7,790.00
  • r = 15% = 0.15
  • n = 12
  • t = number of years

Substitute the given values into the formula and solve for t:

\implies \sf 17474=7790\left(1+\dfrac{0.15}{12}\right)^{12t}

\implies \sf \dfrac{17474}{7790}=\left(1.0125}\right)^{12t}

\implies \sf \ln\left(\dfrac{17474}{7790}\right)=\ln \left(1.0125}\right)^{12t}

\implies \sf \ln\left(\dfrac{17474}{7790}\right)=12t \ln \left(1.0125}\right)

\implies \sf t=\dfrac{\ln\left(\frac{17474}{7790}\right)}{12 \ln \left(1.0125}\right)}

\implies \sf t=5.419413037...\:years

Therefore, the money was in the account for 5 years and 5 months (to the nearest month).

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