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Zinaida [17]
3 years ago
5

Pagina 32 de libri de tercero de secundaria

Mathematics
1 answer:
irga5000 [103]3 years ago
4 0

Answer:

Assignment: 06.05 Data Observation Assignment: 06.05 Data Observation

Step-by-step explanation:

Assignment: 06.05 Data Observation Assignment: 06.05 Data Observation Assignment: 06.05 Data Observation Assignment: 06.05 Data Observation Assignment: 06.05 Data Observation Assignment: 06.05 Data Observation Assignment: 06.05 Data Observation hi

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PLEASE ANSWER 90 POINTS!!
denpristay [2]

Answer:

unlikey

Step-by-step explanation:

6 0
2 years ago
PLS HELP IF YOU CAN!THIS IS DUE IN 30 MINUTES :(
ExtremeBDS [4]

Answer:

Length: 20 or 40 feet

Width: 40 or 20 feet

Step-by-step explanation:

Area = 800 = length*width = x*(60-x)= 60x - x^2

- x^2 + 60x - 800 = 0

x = 20

Or x = 40

So length can be either 20 feet or 40 feet and width can be 40 (60-20) or 20 (60-40)

6 0
3 years ago
As a river guide on the gauja river, antra earned $55 per passenger and a $25 tip. if she earned $410 altogether, how many passe
Ket [755]

410 - 25 = 386

385 ÷ 55 = 7 passengers


5 0
3 years ago
Read 2 more answers
√₂
snow_lady [41]

Answer:

$2000 was invested at 5% and $5000 was invested at 8%.

Step-by-step explanation:

Assuming the interest is simple interest.

<u>Simple Interest Formula</u>

I = Prt

where:

  • I = interest earned.
  • P = principal invested.
  • r = interest rate (in decimal form).
  • t = time (in years).

Given:

  • Total P = $7000
  • P₁ = principal invested at 5%
  • P₂ = principal invested at 8%
  • Total interest = $500
  • r₁ = 5% = 0.05
  • r₂ = 8% = 0.08
  • t = 1 year

Create two equations from the given information:

\textsf{Equation 1}: \quad \sf P_1+P_2=7000

\textsf{Equation 2}: \quad \sf P_1r_1t+P_2r_2t=I\implies 0.05P_1+0.08P_2=500

Rewrite Equation 1 to make P₁ the subject:

\implies \sf P_1=7000-P_2

Substitute this into Equation 2 and solve for P₂:

\implies \sf 0.05(7000-P_2)+0.08P_2=500

\implies \sf 350-0.05P_2+0.08P_2=500

\implies \sf 0.03P_2=150

\implies \sf P_2=\dfrac{150}{0.03}

\implies \sf P_2=5000

Substitute the found value of P₂ into Equation 1 and solve for P₁:

\implies \sf P_1+5000=7000

\implies \sf P_1=7000-5000

\implies \sf P_1 = 2000

$2000 was invested at 5% and $5000 was invested at 8%.

Learn more about simple interest here:

brainly.com/question/27743947

brainly.com/question/28350785

5 0
1 year ago
An asset is purchased on April 1st and has an annual depreciation amount of $3,600. Using the straight-line method of depreciati
Free_Kalibri [48]
From April to December =9 months

the depreciation expense
3,600×(9÷12)=2,700
5 0
3 years ago
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