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Molodets [167]
2 years ago
9

Domain and range of y=-2x^2+12x-15

Mathematics
1 answer:
Dominik [7]2 years ago
8 0
Answer:
Domain is (-infinity,infinity)
Range is (-infinity,3)

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x = (0.456 x (0.01)^2)/(0.0401)^2 = 0.0284 ohms
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X / -9 equal -4 minus parentheses -11 answer x equal
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Do you mean x/-9=(-4)-(-11)x?

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Y to the 3rd power times y to the negative 5th power
lina2011 [118]
Y^3 * y^5 = y^8. you add the exponents
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3 years ago
PLZZZZ help as soon as possible
sammy [17]

Answer:

JK = 11

Step-by-step explanation:

JK ≅ JM

where

JK = 3x - 16

JM = 2x - 7

3x - 16 = 2x - 7

group like terms

3x - 2x = 16 - 7

x = 9

<u>plugin x=9 into JK</u>

JK = 3x - 16

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JK = 11

6 0
3 years ago
Read 2 more answers
Please help me i don’t understand this question at all.
Maru [420]

9514 1404 393

Answer:

  D.

Step-by-step explanation:

The wording "when x is an appropriate value" is irrelevant to this question. That phrase should be ignored. (You may want to report this to your teacher.)

When you look at the answer choices, you see that all of them are negative except the last one (D). When you look at the problem fraction, you see that it is positive.

The only reasonable choice is D.

__

Your calculator can check this for you.

  √12/(√3 +3) ≈ 3.4641/(1.7321 +3)

  = 3.4641/4.7321 ≈ 0.7321 = -1 +√3

__

If you want to "rationalize the denominator", then multiply numerator and denominator by the conjugate of the denominator. The conjugate is formed by switching the sign between terms.

  \displaystyle\frac{\sqrt{12}}{\sqrt{3}+3}=\frac{\sqrt{12}}{(\sqrt{3}+3)}\cdot\frac{(\sqrt{3}-3)}{(\sqrt{3}-3)}=\frac{\sqrt{36}-3\sqrt{12}}{3-9}\\\\=\frac{6-3\cdot2\sqrt{3}}{-6}=\boxed{-1+\sqrt{3}}

_____

<em>Additional comment</em>

We "rationalize the denominator" in this way to take advantage of the relation ...

  (a -b)(a +b) = a² -b²

Using this gets rid of the irrational root in the denominator, hence "rationalizes" the denominator.

We could also have multiplied by (3 -√3)/(3 -√3). This would have made the denominator positive, instead of negative. However, I chose to use (√3 -3) so you could see that all we did was change the sign from (√3 +3).

3 0
2 years ago
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