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natta225 [31]
3 years ago
9

I need help ASAP ........

Mathematics
1 answer:
Gennadij [26K]3 years ago
8 0

Answer:

What do they want you to do? do they want you to add up each one orrrr

Step-by-step explanation:

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G(x) = -2x^3 – 15x^2 + 36x
shusha [124]

Consider the function G(x) = -2x^3 - 15x^2 + 36x. First, factor it:

G(x) = -2x^3 - 15x^2 + 36x=-x(2x^2+15x-36)=\\ \\=-x\cdot 2\cdot \left(x-\dfrac{-15-\sqrt{513}}{4}\right)\cdot \left(x-\dfrac{-15+\sqrt{513}}{4}\right).

The x-intercepts are at points \left(\dfrac{-15-\sqrt{513} }{4},0\right),\ (0,0),\ \left(\dfrac{-15+\sqrt{513} }{4},0\right).

1. From the attached graph you can see that

  • function is positive for x\in \left(-\infrty, \dfrac{-15-\sqrt{513} }{4}\right)\cup \left(0,\dfrac{-15+\sqrt{513} }{4}\right);
  • function is negative for x\in \left(\dfrac{-15-\sqrt{513} }{4},0\right)\cup \left(\dfrac{-15+\sqrt{513} }{4},\infty\right).

2. Since

G(-x) = -2(-x)^3 - 15(-x)^2 + 36(-x)=2x^3-15x^2-36x\neq G(x)\ \text{and }\neq -G(x) the function is neither even nor odd.

3. The domain is x\in (-\infty,\infty), the range is y\in (-\infty,\infty).

8 0
3 years ago
Read 2 more answers
Joe has 2/10 of a dollar. Ashley has 20/100 of a dollar. Billy has 3/100 of a dollar. Together, what fraction of a dollar do the
Vinvika [58]

Answer:

they have 43/100 or 43% out of 100

Step-by-step explanation:

Because joe has 2/10| 2/10=20% or 2

Joe=20

Ashley also has the same thing as joe so 20.

Ashley=20

Billy has 3/100 that = 3 so what's 20+20+3=43 and there's your answer.

Have a great day:)

6 0
3 years ago
How do you round 25,386 to the nearest 1000
NeTakaya
You would round down to 25,000
3 0
3 years ago
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Find the hypotenuse
Anika [276]
A^2+b^2=c^2
11^2+4^2=c^2
121+16=c^2
137=c^2
\sqrt{137}  =  \sqrt{c}
11.70=c

8 0
3 years ago
Read 2 more answers
Carly paid $17.50 for 7 gallons of gas and Jade paid $45 for 15 gallons of gas. Which of the following statements is true?
KatRina [158]

Answer:

Jade paid more per gallon than Carly

Step-by-step explanation:

3 0
3 years ago
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