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Fofino [41]
2 years ago
8

Need help on this math problem of parallelograms.

Mathematics
1 answer:
tamaranim1 [39]2 years ago
6 0

Answer:

z=124 because it opposite

Step-by-step explanation:

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emmasim [6.3K]
The equation is y=-3/2+3
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I mainly need help with #10, thank you!
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Answer:

10) 40

Step-by-step explanation:

just subtract 95 and 55

8 0
3 years ago
Factor completely. 25w² + 60w + 36 Express the answer in the form (aw+b) 2 .  
Lana71 [14]
<span>25w² + 60w + 36
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6 0
3 years ago
Help plzzz thank you !
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
Read 2 more answers
Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false:
kondaur [170]
\text{Proof by induction:}
\text{Test that the statement holds or n = 1}

LHS = (3 - 2)^{2} = 1
RHS = \frac{6 - 4}{2} = \frac{2}{2} = 1 = LHS
\text{Thus, the statement holds for the base case.}

\text{Assume the statement holds for some arbitrary term, n= k}
1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2} = \frac{k(6k^{2} - 3k - 1)}{2}

\text{Prove it is true for n = k + 1}
RTP: 1^{2} + 4^{2} + 7^{2} + ... + [3(k + 1) - 2]^{2} = \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2} = \frac{(k + 1)[6k^{2} + 9k + 2]}{2}

LHS = \underbrace{1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2}}_{\frac{k(6k^{2} - 3k - 1)}{2}} + [3(k + 1) - 2]^{2}
= \frac{k(6k^{2} - 3k - 1)}{2} + [3(k + 1) - 2]^{2}
= \frac{k(6k^{2} - 3k - 1) + 2[3(k + 1) - 2]^{2}}{2}
= \frac{k(6k^{2} - 3k - 1) + 2(3k + 1)^{2}}{2}
= \frac{k(6k^{2} - 3k - 1) + 18k^{2} + 12k + 2}{2}
= \frac{k(6k^{2} - 3k - 1 + 18k + 12) + 2}{2}
= \frac{k(6k^{2} + 15k + 11) + 2}{}
= \frac{(k + 1)[6k^{2} + 9k + 2]}{2}
= \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2}
= RHS

Since it is true for n = 1, n = k, and n = k + 1, by the principles of mathematical induction, it is true for all positive values of n.
3 0
3 years ago
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