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QveST [7]
3 years ago
13

5.6=I+4.09 what is I? I need help I just need answers in the 10 minutes please helpp​

Mathematics
2 answers:
serious [3.7K]3 years ago
3 0

Answer:

the answer should be I= 1.51

Dahasolnce [82]3 years ago
3 0

Answer:

i=1+0.51/1

Step-by-step explanation:

5.6=i+4.09

We move all terms to the left:

5.6-(i+4.09)=0

We get rid of parentheses

-i-4.09+5.6=0

We add all the numbers together, and all the variables

-1i+1.51=0

We move all terms containing i to the left, all other terms to the right

-i=-1.51

i=-1.51/-1

i=1+0.51/1

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VashaNatasha [74]

Answer:

a. Ameribank-$15,157.50

b. Capital Two-$4,646.25

Step-by-step explanation:

a. Tad's savings is $15,000, we calculate his total amount at the end of the year for each bank:

#Ameribank

A=P+I=P+PRT\\\\=15000+15000\times 0.0105\times 1\\\\=\$15,157.50

#Huffington( we use the effective rate to calculate the compound amount):

i_m=(1+i/m)^m-1\\\\=(1+0.0095/12)^[12}-1=0.009541\\\\A=P(1+i_m)^n\\\\=15000(1.009541)^1\\\\=\$15,143.12

#Sixth-Third, Take 1 yrs=52 weeks:

i_m=(1+i/m)^m-1\\\\=(1+0.01/52)^{52}-1=0.01005\\\\A=15000(1.01005)^1\\\\=\$15,150.74

#Hence, Ameribank is the best option as his money grows to $15,157.50 which is greater than all the remaining two options.

b. We use the compound interest formula A=P(1+r/n)^{nt} to determine which bank gives the best option:

#Capital Two. r=3.75%, n=12,t=4

A=P(1+r/n)^{nt}\\\\=4000(1+0.0375/12)^{12\times4}\\\\=\$4,646.25

#J.C Morgan, t=2, r=3.55% n=12

A=P(1+r/n)^{nt}\\\\=4000(1+0.0355/12)^{12\times 2}\\\\=\$4,293.87

#Silverman Slacks, n=12,t=3, r=3.65%

A=P(1+r/n)^{nt}\\\\=4000(1+0.0365/12)^{12\times3}\\\\=\$4,462.14

We compare the investment amounts after t years:

Capital>Silver>Morgan=4646.25>4462.14>4293.87

Hence, Capital two is the best option with an investment amount of $4,646.25

8 0
3 years ago
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SIMPLIFY 1/4 × 1/3 - 1/5× 1/4 + 1/4 × 1/6
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Answer:

3/40

Step-by-step explanation:

1/4 x 1/3 - 1/5 x 1/4 +1/4 x 1/6

(1/4 x 1/3) - (1/5 x 1/4) + (1/4 x 1/6)

1/12 - 1/20 + 1/24

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3/40

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