2 years ago

Can u guys help me with the visual question please

1 answer:

stepan [7]2 years ago

3 0

Which problem is it?

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Damm [24]

C. 35 because c. Is the right answer just trust me ok

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Korolek [52]

Given the function $f(x)=3x^2-18x+23$ . The above function can be written as

$f(x)=3x^2-18x+23\\ f(x)=3(x^2-6x)+23\\ f(x)=3(x^2-6x+9-9)+23\\ f(x)=3(x-3)^2-27+23\\ f(x)=3(x-3)^2-4$

a)Now, the function $f(x)=3(x-3)^2-4$ has minimum value since the coefficient of $(x-3)^2$ is $3>0$ .

b) The minimum value of the function occurs at $x=3$ and its value is

$f(3)=3(3-3)^2-4 =-4$

c)The minimum value of the function occurs at $x=3$ .

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2 years ago

elena-14-01-66 [18.8K]

The perimeter is 10+10+4+4=28

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Bond [772]

Answer:24 cm

Step-by-step explanation:

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1 year ago

Dmitry [639]

Answer:

(D) t^4

Step-by-step explanation:

You have defined ...

a3 = a2·a1

a4 = a3·(a2·a1) = a3²

a5 = a4·(a3·a2·a1) = a4² = (a3²)² = a3⁴

Then if a3 = t, a5 = t⁴

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2 years ago