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uranmaximum [27]
3 years ago
9

Simplify the expression:

D%29%5E%7B2%7D%20" id="TexFormula1" title=" ({2x + 5y})^{2} - ({3x + 4y})^{2} " alt=" ({2x + 5y})^{2} - ({3x + 4y})^{2} " align="absmiddle" class="latex-formula">
​
Mathematics
2 answers:
Naya [18.7K]3 years ago
8 0

Answer:

3x+18y

Step-by-step explanation:

2x+5y+2x+5y-3x+4y+3x+4y

2x+2x-3x+3x+5y+5y+4y+4y

4x-x+10y+8y

3x+18y

Rasek [7]3 years ago
5 0

Step-by-step explanation:

the answer is in the picture

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father left home at 7:20.After 1 hour and 25 minutes he arrived and chaguanas and immediately got a taxi to go port-of-spain at
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It took him 1 hour and 13 minutes to go from changuanas to port-of-spain 
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The red figure is congruent to the blue figure. Which choice describes a sequence of transformations in which the blue figure is
Colt1911 [192]

Answer:

  • 2. Rotate the triangle 90º clockwise about the origin and then translate it 10 units left and 9 units down.

Step-by-step explanation:

  • <em>Easy way to take one of the vertices and apply the transformations</em>

1. Rotate the triangle 90º counterclockwise about the origin and then translate it 10 units left and 9 units down.

  • False

2. Rotate the triangle 90º clockwise about the origin and then translate it 10 units left and 9 units down.

  • True
  • (-3, 3) →  (3, 3) → (3 - 10, 3 - 9) = (-7, -6)

3. Rotate the triangle 90º counterclockwise about the origin then translate it 1 unit up.

  • False

4. Rotate the triangle 90º clockwise about the origin then translate it 1 unit up.

  • False
4 0
3 years ago
Alfonso earned a score of 87.5% on his test. he did 56 problems correctly. how many problems were on the​ test?
Yuri [45]
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5 0
4 years ago
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What are all the solutions to the following<br> equation?<br> |10x + 2| =48
Greeley [361]

\huge \boxed{\mathfrak{Question} \downarrow}

What are all the solutions to the following

equation?

|10x + 2| =48

\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}

| 10 x + 2 | = 48

Combine like terms and use the properties of equality to get the variable on one side of the equals sign and the numbers on the other side. Remember to follow the order of operations.

|10x+2|=48

Use the definition of absolute value.

10x+2=48  \\ 10x+2=-48

Subtract 2 from both sides of the equation.

10x=46 \\  10x=-50

Divide both sides by 10.

\boxed{ \boxed{ \bf \: x=\frac{23}{5} = 46}} \\    \boxed{\boxed{ \bf \: x=-5 }}

7 0
3 years ago
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Can anyone help me out with this?​
bogdanovich [222]

{\large{\textsf{\textbf{\underline{\underline{Question \: 1 :}}}}}}

\star\:{\underline{\underline{\sf{\purple{Solution:}}}}}

\bullet \sf \:   {(a + b)}^{ab}

<u>Putting value of a as 3 and b as -2, we get</u><u> </u><u>:</u>

\longrightarrow \sf \:   {( 3 +  (- 2))}^{3 \times  - 2}

\longrightarrow \sf \:   {( 3 - 2)}^{3 \times  - 2}

\longrightarrow \sf \:   {( 1)}^{ - 6}

• <u>Using negative Exponents Law</u>

\longrightarrow \sf   \dfrac{1}{ {1}^{6} }

\longrightarrow \sf   \dfrac{1}{ 1 \times 1 \times 1 \times 1 \times 1 \times 1 }

\longrightarrow \sf   \dfrac{1}{  1 }

\longrightarrow \sf   \purple{1}

{\large{\textsf{\textbf{\underline{\underline{Question \: 2 :}}}}}}

\star\:{\underline{\underline{\sf{\red{Solution:}}}}}

\bullet  \sf \:  \dfrac{ {8}^{ - 1} \times   {5}^{3} }{ {2}^{ - 4}}

\longrightarrow  \sf \:  {8}^{ - 1} \times   {5}^{3}  \times  \dfrac{1}{{2}^{ - 4}}

<u>• Using negative Exponents Law</u>

\longrightarrow  \sf \:  {8}^{ - 1} \times   {5}^{3}  \times   {2}^{4}

\longrightarrow  \sf \:  {8}^{ - 1} \times   5 \times 5 \times 5  \times   {2}^{4}

\longrightarrow  \sf \:  {8}^{ - 1} \times 125  \times   {2}^{4}

\longrightarrow  \sf \:  {8}^{ - 1} \times 125  \times   2 \times 2 \times 2 \times 2

<u>• Using negative Exponents Law</u>

\longrightarrow  \sf \:   \dfrac{1}{ \cancel{8}_{4}} \times 125  \times   \cancel{2}_{1} \times 2 \times 2 \times 2

\longrightarrow  \sf \:   \dfrac{1}{ \cancel4_{2}} \times 125  \times   \cancel{2}_{1}  \times 2 \times 2

\longrightarrow  \sf \:   \dfrac{1}{ \cancel2} \times 125  \times   \cancel{2}   \times 2

\longrightarrow  \sf \:    125  \times 2

\longrightarrow  \sf \red{  250}

{\large{\textsf{\textbf{\underline{\underline{Question \: 3 :}}}}}}

\star\:{\underline{\underline{\sf{\green{Solution(1):}}}}}

\bullet \sf  \dfrac{ \sqrt{32} +  \sqrt{48}  }{ \sqrt{8} +  \sqrt{12}  }

\longrightarrow \sf  \dfrac{ \sqrt{4 \times 4 \times 2} +  \sqrt{4 \times 4 \times 3}  }{ \sqrt{2 \times 2 \times 2} +  \sqrt{2 \times 2 \times 3}  }

\longrightarrow \sf  \dfrac{ \sqrt{  {4}^{2}   \times 2} +  \sqrt{ {4}^{2}  \times 3}  }{ \sqrt{ {2}^{2}  \times 2} +  \sqrt{ {2}^{2}  \times 3}  }

\longrightarrow \sf  \dfrac{ 4\sqrt{    2} + 4 \sqrt{  3}  }{ 2\sqrt{  2} +2  \sqrt{  3}  }

\longrightarrow \sf  \dfrac{ \cancel{ 4}_{2}(\sqrt{    2} +  \sqrt{  3})  }{  \cancel{2}(\sqrt{  2} + \sqrt{  3})  }

\longrightarrow \sf  \dfrac{ 2  \: \cancel{(\sqrt{    2} +  \sqrt{  3}) } }{  \cancel{(\sqrt{  2} + \sqrt{  3})}  }

\longrightarrow \sf   \green{2}

\star\:{\underline{\underline{\sf{\blue{Solution(2):}}}}}

\bullet  \sf \dfrac{ \sqrt{5}  +  \sqrt{3} }{ \sqrt{80} +  \sqrt{48}  - \sqrt{45}  -  \sqrt{27}   }

\begin{gathered}  \longrightarrow \sf  \dfrac{ \sqrt{5}  +  \sqrt{3} }{ \sqrt{4 \times 4 \times 5} +  \sqrt{4 \times 4 \times 3}  - \sqrt{3 \times 3 \times 5}  -  \sqrt{3 \times 3 \times 3}   } \end{gathered}

\begin{gathered}\longrightarrow  \sf \dfrac{ \sqrt{5}  +  \sqrt{3} }{ \sqrt{ {4}^{2}  \times 5} +  \sqrt{ {4}^{2}  \times 3}  - \sqrt{ {3}^{2}  \times 5}  -  \sqrt{ {3}^{2}  \times 3}   } \end{gathered}

\longrightarrow \sf  \dfrac{ \sqrt{5}  +  \sqrt{3} }{4 \sqrt{  5} + 4 \sqrt{   3}  - 3\sqrt{    5}  -  3\sqrt{  3}   }

\longrightarrow \sf  \dfrac{ \sqrt{5}  +  \sqrt{3} }{4 \sqrt{  5}   - 3\sqrt{    5} + 4 \sqrt{   3} -  3\sqrt{  3}   }

\longrightarrow  \sf \dfrac{ \cancel{ \sqrt{5}  +  \sqrt{3}} }{ \cancel{\sqrt{    5}  +   \sqrt{  3}   } }

\longrightarrow   \blue{1}

{\large{\textsf{\textbf{\underline{\underline{Answers :}}}}}}

• Question 1 - \purple{1}

• Question 2 - \red{250}

• Question 3(1) - \green{2}

• Question 3(2) - \blue{1}

{\large{\textsf{\textbf{\underline{\underline{ Concept \: :}}}}}}

<u>★</u><u> </u><u>Negative</u><u> Exponents Law -</u>

\bullet  \sf \:  {a}^{ - m}  =  \dfrac{1}{ {a}^{m} }

★ \sqrt{32} can be written as 4 \sqrt{2}

‣ \sqrt{48} can be written as 4 \sqrt{3}

‣ \sqrt{8} can be written as 2 \sqrt{2}

‣ \sqrt{12} can be written as 2 \sqrt{3}

‣ \sqrt{80} can be written as 4 \sqrt{5}

‣ \sqrt{48} can be written as 4 \sqrt{3}

‣ \sqrt{45} can be written as 3 \sqrt{5}

‣ \sqrt{27} can be written as 3 \sqrt{3}

★ <u>During Addition and Subtraction</u>

• minus (-) minus (-) gives plus (+)

• minus (-) plus (+) gives minus (-)

• plus (+) minus (-) gives minus (-)

• plus (+) plus (+) gives plus (+)

• Also the sign of the resultant term depends upon the sign of the largest number.

{\large{\textsf{\textbf{\underline{\underline{ Note \: :}}}}}}

• Swipe to see the full answer.

\begin{gathered} {\underline{\rule{330pt}{3pt}}} \end{gathered}

5 0
2 years ago
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