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Alla [95]
2 years ago
8

Help please asap just the answers​

Mathematics
1 answer:
Stella [2.4K]2 years ago
4 0

\underline \bold{ \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  }

\huge\underline{\sf{\red{Problem:}}}

7.) Detemine the value of \sf{ {3a}^{2}  -b.}

\huge\underline{\sf{\red{Given:}}}

\quad\quad\quad\quad\sf{a = 2}

\quad\quad\quad\quad\sf{b =  - 1}

\quad\quad\quad\quad\sf{c =  - 3}

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

\quad\quad\quad\quad\sf{⟶{3a}^{2}  - b}

\quad \quad \quad \quad   \sf{⟶{(3)( 2)}^{2}  -( - 1)}

\quad \quad \quad \quad   \sf{⟶3(4)-( - 1)}

\quad \quad \quad \quad   \sf{⟶12-( - 1)}

\quad \quad \quad \quad   \sf{⟶12 +  1 }

\quad \quad \quad \quad ⟶ \boxed{ \sf{ 13}}

\huge\underline{\sf{\red{Answer:}}}

\huge\quad \quad \underline{ \boxed{ \sf{ \red{7.)\:13}}}}

8.) Find the value of \sf{ {a}^{3}  {b}^{3}  - abc.}

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

\quad\quad\quad\quad\sf{⟶ {a}^{3}  {b}^{3}  - abc}

\quad\quad\quad\quad\sf{⟶{(2)}^{3}  {( - 1)}^{3}  - (2)( - 1)( - 3)}

\quad\quad\quad\quad\sf{⟶{(8)}{( - 1)}  - ( - 2)( - 3)}

\quad\quad\quad\quad\sf{⟶{( - 8)}  - ( 6)}

\quad\quad\quad\quad ⟶\boxed{\sf{  - 14}}

\huge\underline{\sf{\red{Answer:}}}

\huge\quad \quad \underline{ \boxed{ \sf{ \red{8.)-14}}}}

\underline \bold{ \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  }

#CarryOnLearning

\sf{\red{✍︎ C.Rose❀}}

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PLEASE HELP SOLVE! FIRST TO SOLVE RIGHT WILL GET BRAINIEST
IrinaVladis [17]

Answer:

  40π/3 cm^2

Step-by-step explanation:

The centerline of the shaded region has a radius of 3 +4/2 = 5 cm. Its length is 1/3 of a circle with that radius, so is ...

  length of centerline = (1/3)(2π·5 cm) = (10/3)π cm

The shaded region is 4 cm wide, so the area is the product of that width and the centerline length:

  (4 cm)(10/3 π cm) = 40π/3 cm^2

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Given f(x) and g(x) = f(x) + k, use the graph to determine the value of k.
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<h3>Answer: C) 4</h3>

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Step-by-step explanation:

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He determines that 6 is an extraneous solution because the difference of the numerators is 6, so the 6s cancel to 0. which best
seropon [69]

The question is incomplete. Completed question is given below the answer.

His solution for x is correct, but in order for 6 to be an extraneous solution, one denominator has to result in 0 when 6 is substituted for x.

Given  

Then the solution follows thus:

Step 1: 8(x – 4) = 2(x + 2)

Step 2: 4(x – 4) = (x + 2)

Step 3: 4x – 16 = x + 2

Step 4: 3x = 18

Step 5: x = 6

It can be seen that his solution is correct. But 6 is not an extraneous solution.

An extraneous solution is a solution to an equation that emerges from the process of solving the problem but is not a valid solution to the original problem.

When 6 is substituted into the original equation, the original equation holds.

Therefore, his solution for x is correct, but in order for 6 to be an extraneous solution, one denominator has to result in 0 when 6 is substituted for x.

Learn more about extraneous solution here: brainly.com/question/3751209

#SPJ4

Completed question:-

A student solves the following equation for all possible values of x:His solution is as follows:

Step 1: 8(x – 4) = 2(x + 2)

Step 2: 4(x – 4) = (x + 2)

Step 3: 4x – 16 = x + 2

Step 4: 3x = 18

Step 5: x = 6

He determines that 6 is an extraneous solution because the difference of the numerators is 6, so the 6s cancel to 0.

Which best describes the reasonableness of the student’s solution?

His solution for x is correct and his explanation of the extraneous solution is reasonable.

His solution for x is correct, but in order for 6 to be an extraneous solution, both denominators have to result in 0 when 6 is substituted for x.

His solution for x is correct, but in order for 6 to be an extraneous solution, one denominator has to result in 0 when 6 is substituted for x.

His solution for x is incorrect. When solved correctly, there are no extraneous solutions.

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