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daser333 [38]
3 years ago
10

PLEASE, SOMEONE, I CAN'T FAIL THIS CLASS

Mathematics
1 answer:
Alenkasestr [34]3 years ago
8 0

\text{You're finding the applicable expression}\\\\\text{One way we could solve this is by plugging in the values in the left}\\\text{column to the expressions}\\\\\text{Lets plug in 2 to the first expression:}\\\\2^2+2+3\\\\\text{Solve:}\\\\4+2+3\\\\6+3=9\\\\\text{You would see that it gave you the right value, lets try it on another one:}\\\\3^2+3+3\\\\9+3+3\\\\12+3=15\\\\\text{This expression works}\\\\\boxed{x^2+x+3}

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Adam went on a shopping trip to Somerville. He purchased a set of dishes originally priced at
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Answer:187.2

Step-by-step explanation:

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After 60% discount price will be 156

And Sales tax is 8% . Then the price of sales tax is $31.2.

156+31.2= $187.2

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What is 5x²+x³-2x+1 ?​
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Simplified: x^3+5x^2-2x+1
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3 years ago
Use the Quadratic formula to find all real zeros of the 2nd degree polynomial
schepotkina [342]

Answer:

  x ∈ {-5, -1}

Step-by-step explanation:

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x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-24\pm\sqrt{24^2-4\cdot 4\cdot 20}}{2\cdot 4}\\\\=\dfrac{-24\pm\sqrt{576-320}}{8}=\dfrac{-24\pm\sqrt{256}}{8}\\\\=\dfrac{-24\pm 16}{8}=-3\pm 2=\{-5,-1\}

The real zeros are -5 and -1.

_____

There are many ways to check your answer. One of them is to look at the given quadratic, which has no changes of sign in its coefficients. (They are all positive.) That means there can be no positive real roots, so already you know that x=0.5 won't work.

Also, the constant in the quadratic is the product of the roots, For your roots, their product is -7/4, so even multiplying by 4 (the leading coefficient in the given quadratic), you don't get anything like 20.

4 0
2 years ago
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