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motikmotik
3 years ago
6

Need help with this question ​

Mathematics
2 answers:
mrs_skeptik [129]3 years ago
4 0

Answer:

<h2>- 2 {x}^{5}  {y}^{7}</h2>

Last option is correct.

Step-by-step explanation:

- 2 {x}^{3}  {y}^{4}  {x}^{2}  {y}^{3}

Multiply the terms with the same base by adding their exponents

- 2 {x}^{3 + 2}  {y}^{4 + 3}

Add the numbers

- 2 {x}^{5}  {y}^{7}

Hope this helps..

Best regards!

romanna [79]3 years ago
3 0

- 2 {x}^{5} {y}^{7}

<u>Solution</u>:

- 2 {x}^{3}  {y}^{4}  {x}^{2}  {y}^{3}

= 2 {x}^{(3 + 2)} {y}^{(4 + 3)}

=  - 2 {x}^{5} {y}^{7}

{\boxed{\blue{\textsf{Some Important Laws of Indices}}}}

{a}^{n}.{a}^{m}={a}^{(n + m)}

{a}^{-1}=\dfrac{1}{a}

\dfrac{{a}^{n}}{ {a}^{m}}={a}^{(n-m)}

{({a}^{c})}^{b}={a}^{b\times c}={a}^{bc}

{a}^{\frac{1}{x}}=\sqrt[x]{a}

a^0 = 1

[\text{Where all variables are real and greater than 0}]

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To minimize this function d^2 subject to the constraint, <span>2x+y−10=0
</span>If we substitute, the y-values the distance function can take will be related to the x-values by the line:<span>y=10−2x 
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<span>d=sqrt [<span><span><span>x2</span>+(10−2x<span>)^2]
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d′=1/2 (5x2−40x+100)^(−1/2)   (10x−40)<span>
</span>Setting the derivative to zero to find optimal x,
<span><span>d′</span>=0→10x−40=0→x=4
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Answer:

You should invest $820 in account A and $940 in account B

Step-by-step explanation:

* Lets use the system of linear equations to solve the problem

- Simple Interest Equation I = Prt , Where:

# P = Invested Amount

# I = Interest Amount

# r = Rate of Interest per year in decimal; r = R/100

# t = Time Period involved in months or years

* Lets solve the problem

- The total money invested is $1760

- Account A pays 7​% annual interest

- Account B pays 4​% annual interest

- Let A represent the amount of money invested in the account A

- Let B represent the amount of money invested in the account B

- You would like to earn $ 95 at the end of one year

∴ The interest from both accounts at the end of one year is $95

- Lets write the equations

# Account A :

∵ Account A has $A invested

∴ P = $A

∵ Account A pays 7​% annual interest

∴ r = 7/100 = 0.07

∵ t = 1 year

∵ I = Prt

∴ I = A(0.07)(1) = 0.07A

# Account B :

∵ Account B has $B invested

∴ P = $B

∵ Account A pays 4​% annual interest

∴ r = 4/100 = 0.04

∵ t = 1 year

∵ I = Prt

∴ I = B(0.04)(1) = 0.04B

- The total amount of interest from both accounts at the end of one

  year is $95

∴ I from A + I from B = 95

∴ 0.07A + 0.04B = 95 ⇒ multiply both sides by 100

∴ 7A + 4B = 9500 ⇒ (1)

- The total money to invest in both accounts is $1760

∵ Account A has $A invested

∵ Account B has $B invested

∴ A + B = 1760 ⇒ (2)

* Lets solve the system of equations to find the amount of money

  invested in each account

- Multiply equation (2) by -4 to eliminate B

∵ A + B = 1760 ⇒ × -4

∴ -4A - 4B = -7040 ⇒ (3)

- Add equation (1) and (3)

∵ 7A + 4B = 9500 ⇒ (1)

∵ -4A - 4B = -7040 ⇒ (3)

∴ 7A - 4A = 9500 - 7040

∴ 3A = 2460 ⇒ divide both side by 3

∴ A = 820

- Substitute the value of A in equation (1) or (2)

∵ A + B = 1760 ⇒ (2)

∴ 820 + B = 1760 ⇒ subtract 820 from both sides

∴ B = 940

- From all above

* You should invest $820 in account A and $940 in account B

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