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Lera25 [3.4K]
3 years ago
12

The function below is written in vertex form or intercept form. Rewrite them in standard form and show your work.

Mathematics
2 answers:
lora16 [44]3 years ago
8 0

Answer:

The standard form as y=5x^2+30x+41

Step-by-step explanation:

Given: A function which is written in vertex form or intercept form.

We have to re-write it  in standard form that in terms of

Given y = 5(x+3)^2-4

Squaring  using (a+b)^2=a^2+b^2+2ab , we get,  

y=5(x^2+9+6x)-4

Multiply 5 inside , we get,

y=5x^2+45+30x-4

Solving further , we get,

y=5x^2+30x+41

Thus , we have obtained the standard form as y=5x^2+30x+41







olga55 [171]3 years ago
5 0

Answer: The standard form of equation will be

f(x)=5x^2+30x+41

Step-by-step explanation:

Since we have given that

The vertex form of equation is given by

y=5(x+3)^2-4

We need to find the standard form :

Standard form is written as :

f(x)=ax^2+bx+c

So, our equation becomes,

y =5(x+3)^2-4\\\\y=5(x^2+9+6x)-4\\\\y=5x^2+45+30x-4\\\\y=5x^2+30x+41

Hence, the standard form of equation will be

f(x)=5x^2+30x+41

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Write 14/√2 + √50 in the form b√2 where b is an integer.
kifflom [539]

Answer:

12 \sqrt{2}

Step-by-step explanation:

Taking the terms one at a time,

\frac{14}{ \sqrt{2} }  \\   \frac{14 *  \sqrt{2} }{ \sqrt{2} *  \sqrt{2}  }  \\   \frac{14  \sqrt{2} }{2}  \\  7 \sqrt{2}

\sqrt{50} \\  \sqrt{25 * 2} \\  \sqrt{25}*  \sqrt{2}    \\ 5 *  \sqrt{2}   \\ 5 \sqrt{2}

Adding the terms, you have;

7 \sqrt{2} + 5 \sqrt{2} =(7 + 5) \sqrt{2} = 12 \sqrt{2}

Hope it helps!

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Answer:

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Step-by-step explanation:

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