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topjm [15]
3 years ago
12

Evaluate the following expression. 1/5^-2 Evaluate 6^-3

Mathematics
2 answers:
Marizza181 [45]3 years ago
7 0
1/5^-2 = 5^2 so this equals

 = 25
melisa1 [442]3 years ago
6 0
( \frac{1}{5}) ^{-2}  =  ( \frac{5}{1} )^{2}  =  5^{2}  = 25 \\  \\  6^{-3} = ( \frac{6}{1} )^{-3} =  ( \frac{1}{6} )^{3} =  \frac{ 1^{3} }{ 6^{3} } =  \frac{1}{216}
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Emily thinks the perfect tomato sauce has 8 cloves of garlic in every 500{ mL} of sauce. Raphael's tomato sauce has 12 cloves of
igomit [66]

For perfect solution we need to have :-  8 colves   in 500ml    solution

500   ml     has   8 cloves

1 ml             has     8 / 500 cloves

100ml         has   8 / 500  * 100  =    8/5  = 1.6 cloves

Raphael's mixture  

    900 ml  has 12 cloves

      1 ml     has   12 / 900 cloves

   100 ml has   12 / 900 *  100   =    12 / 9  =   1.33 cloves

so concentration of garlic in 100 ml solution of Raphael's solution is less than Emily's solution    so it is not garlicky enough.     ( option B)

5 0
3 years ago
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6(x² + 2x + 7)pls ans​
Kazeer [188]

Answer:

Use distributive property

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Easton bought 8 chicken wings for $8.80. How much would it cost for 17 wings?
olga2289 [7]
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Because they are 1.10 each
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3 years ago
What is the value of -9 1/5 - 10 + 2/5 + 2(14 1/2 - 7)?​
Art [367]

Answer:

-9 1/5 - 10 + 2/5 + 2(14 1/2 - 7) = -3.8

Step-by-step explanation:

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3 years ago
Find a polynomial P(x) with real coefficients having a degree 4, leading coefficient 4, and zeros 3-i and 4i.
Alisiya [41]

now, this polynomial has roots of 3-i and 4i, namely 3 - i and 0 + 4i.

let's bear in mind that a complex root never comes all by her lonesome, her sibling is always with her, the conjugate, so if 3 - i is there, 3 + i is also coming along, likewise if 0 + 4i is there, her sibling 0 - 4i is also there.

\bf \begin{cases} x=3-i\implies &x-3+i=0\\ x=3+i\implies &x-3-i=0\\ x=4i\implies &x-4i=0\\ x=-4i\implies &x+4i=0 \end{cases}\\\\[-0.35em] ~\dotfill\\\\ (x-3+i)(x-3-i)(x-4i)(x+4i)=\stackrel{y}{0} \\[2em] \underset{\textit{difference of squares}}{[(x-3)+i][(x-3)-i]}\underset{\textit{difference of squares}}{[x-4i][x+4i]}=0

\bf [(x-3)^2-i^2][x^2-(4i)^2]=y\implies [(x-3)^2-(-1)][x^2-(4^2i^2)]=0 \\[2em] [(x-3)^2-(-1)][x^2-(16(-1))]=0\implies [(x-3)^2+1][x^2+16]=0 \\[2em] [(x^2-6x+9)+1][x^2+16]=y\implies (x^2-6x+10)(x^2+16)=0 \\\\\\ x^4-6x^3+10x^2+16x^2-96x+160=0 \\\\\\ x^4-6x^3+26x^2-96x+160=0 \\\\\\ \stackrel{\textit{multiplying both sides by 4}}{4(x^4-6x^3+26x^2-96x+160)=4(0)} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill 4x^4-24x^3+104x^2-384x+640=y~\hfill

3 0
3 years ago
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