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Tresset [83]
2 years ago
8

Perform the indicated operation. Be sure the answer is reduced.

Mathematics
1 answer:
avanturin [10]2 years ago
8 0
<h3>Given Equation:-</h3>

\boxed{ \rm  \frac{4x^{2}y^{3}z}{9} \times  \frac{45y}{8 {x}^{5} {z}^{5} }}

<h3>Step by step expansion:</h3>

\dashrightarrow \sf\dfrac{4x^{2}y^{3}z}{9} \times  \dfrac{45y}{8 {x}^{5} {z}^{3} }

\\  \\

\dashrightarrow \sf\dfrac{ \cancel4x^{2}y^{3}z}{9} \times  \dfrac{45y}{ \cancel8 {x}^{5} {z}^{3} }

\\  \\

\dashrightarrow \sf\dfrac{x^{2}y^{3}z}{9} \times  \dfrac{45y}{2{x}^{5} {z}^{3} }

\\  \\

\dashrightarrow \sf\dfrac{x^{2}y^{3}z}{ \cancel9} \times  \dfrac{ \cancel{45}y}{2{x}^{5} {z}^{3} }

\\  \\

\dashrightarrow \sf\dfrac{x^{2}y^{3}z}{1} \times  \dfrac{5y}{2{x}^{5} {z}^{3} }

\\  \\

\dashrightarrow \sf\dfrac{x^{0}y^{3}z}{1} \times  \dfrac{5y}{2{x}^{5 - 2} {z}^{3} }

\\  \\

\dashrightarrow \sf\dfrac{y^{3}z}{1} \times  \dfrac{5y}{2{x}^{3} {z}^{3} }

\\  \\

\dashrightarrow \sf\dfrac{y^{3}z {}^{0} }{1} \times  \dfrac{5y}{2{x}^{3} {z}^{3 - 1} }

\\  \\

\dashrightarrow \sf\dfrac{y^{3}}{1} \times  \dfrac{5y}{2{x}^{3} {z}^{2} }

\\  \\

\dashrightarrow \sf  \dfrac{5y \times  {y}^{3} }{2{x}^{3} {z}^{2} }

\\  \\

\dashrightarrow \sf  \dfrac{5y {}^{0}  \times  {y}^{3 + 1} }{2{x}^{3} {z}^{2} }

\\  \\

\dashrightarrow \sf  \dfrac{5 \times  {y}^{4} }{2{x}^{3} {z}^{2} }

\\  \\

\dashrightarrow \bf  \dfrac{5 {y}^{4} }{2{x}^{3} {z}^{2} }

\\  \\

\therefore \underline{ \textbf{ \textsf{option \red c \: is \: correct}}}

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enot [183]
A
Because she deposits the same amount but got 60 dollars more for the birthday money.
5 0
3 years ago
<img src="https://tex.z-dn.net/?f=8a%283b%2B6c-7c%29" id="TexFormula1" title="8a(3b+6c-7c)" alt="8a(3b+6c-7c)" align="absmiddle"
tino4ka555 [31]

Answer:

24ab-8ac

Step-by-step explanation:

Assuming you want to simplify it.

Original equation: 8a(3b+6c-7c)

Apply 8a to each variable in the paranthesis: (8a)(3b)+(8a)(6c)+(8a)(-7c)

After multiplication: 24ab+48ac-56ac

Combine like terms:  24ab-8ac

8 0
2 years ago
Solve for 6y+12=3y-3
kherson [118]
6y + 12 = 3y - 3

Subtract 3y on both sides

3y + 12 = -3

Subtract 12 on both sides

3y = -15

Divide by 3 on both sides

y = -5
5 0
3 years ago
Read 2 more answers
List 3 values that would make this inequality true. ​
3241004551 [841]

Answer:

5, 6, 7

Step-by-step explanation:

You can put any number for n as long as it is equal to or greater than 12 when added to 7.

5+7=12 its equal to 12 so its good

6+7=13 it adds to be greater than 12 so this one is good too

7+7=14 same with this one, so they all check out.

4 0
2 years ago
Give a third degree polynomial that has zeros of 13, 5i, and −5i, and has a value of −680 when x=3. Write the polynomial in stan
Vikki [24]

Answer:

\dfrac{17x^3-221x^2+425x-5525}{9}\\

Step-by-step explanation:

hello

we need to find something like, a being real

a(x-13)(x-5i)(x+5i)

with a so that

a(3-13)(3-5i)(3+5i)=-680\\\\a(-10)(3^2-(5i)^2)= -10(9+25)a=-360a=-680\\ a = \dfrac{680}{360}=\dfrac{17*40}{9*40}=\dfrac{17}{9}

so the third degree polynomial that we are looking for is

\dfrac{17}{9}(x-13)(x-5i)(x+5i)\\=\dfrac{17}{9}(x-13)(x^2+25)\\\\=\dfrac{17}{9}(x^3+25x-13x^2-13*25)\\\\\\=\dfrac{17}{9}(x^3-13x^2+25x-325)\\\\\\\\=\dfrac{17}{9}x^3-\dfrac{17*13}{9}x^2+\dfrac{17*25}{9}x-\dfrac{17*325}{9}\\=\dfrac{17}{9}x^3-\dfrac{221}{9}x^2+\dfrac{425}{9}x-\dfrac{5525}{9}\\

hope this helps

3 0
3 years ago
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