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2 years ago

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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Find the equation to the line , Please help !!!!

emmasim [6.3K]

The equation is y=-3/2+3

5 0

2 years ago

Read 2 more answers

I mainly need help with #10, thank you!

Masja [62]

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
= (5w + 6)^2

hope it helps</span>

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