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Tpy6a [65]
3 years ago
15

Whats the extreme value?y=3(x+5)^(2)-4

Mathematics
1 answer:
natta225 [31]3 years ago
6 0

Answer:

-4

Step-by-step explanation:

  • y = 3(x+5)²-4
  • y= 3(x²+10x+25)-4
  • y= 3x²+30x+75-4
  • y= 3x²+30x+71

Derivate the function

  • y' = 6x+30

solve the equation y'= 0

  • y'= 6x+30
  • 6x+30= 0
  • 6x= -30
  • x = -5

The extrme value of y is reached in -5

y = 3(-5+5)²-4= -4

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Factor out the greatest common factor. 10a3b−6a2b2
podryga [215]

Answer:

2a^2b(5a-3b)

If not then the other way is

5a-3b

Step-by-step explanation:

The GCF of 10a^3b and -6a^2b^2 is 2a^2b

So divide by that you will get

2a^2b(5a-3b)

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What is the value of 4 x 3 divided by (12 divided by2)^2
Digiron [165]

Answer:0.333

Step-by-step explanation:

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3 years ago
Read 2 more answers
Let g(x)=2x and h(x)=x^2+4 find the value. (hog)(a)
lisov135 [29]
In short, (h o g)(a) is just h(    g(a)    ).

so what we can do is simply get g(a) first and then plug that in h(x).

\bf \begin{cases}
g(x)&=2x\\
h(x)&=x^2+4\\
(h\circ g)(a)&=h(~~g(a)~~)
\end{cases}
\\\\\\
g(a)=2(a)\implies g(a)=2a
\\\\\\
h(~~g(a)~~)\implies h(~~2a~~)=(2a)^2+4
\\\\\\
h(~~2a~~)=(2^2a^2)+4\implies  h(~~2a~~)=4a^2+4
8 0
3 years ago
If a student was randomly guessing while completing a 20 question multiple choice exam, what is the probability they would score
mario62 [17]

Answer:

you want 4 correct and 16 incorrect

there are 20 questions

each question has four answers, so

P(right answer) = 1/4

P(wrong answer) = 3/4

----

Since you want 4 correct of 20 we have a combination of 20C4

This is a binomial problem where p = 1/4, q = 3/4 and we get

(20 "choose" 4)*(probability correct)^(number correct)*(probability incorrect)^(number incorrect)

putting numbers in we get

(20c4)*(1/4)^4*(3/4)^16

This gives us

~ .189685

Step-by-step explanation:

7 0
3 years ago
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