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eduard
3 years ago
5

Please help me asap​

Mathematics
2 answers:
8_murik_8 [283]3 years ago
7 0
1.) >
2.) is the fourth choice
3.) is 120
timurjin [86]3 years ago
5 0
1.>
2.4with exponent of 6 aka (D)
3.120
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Help needed....
LenKa [72]
The answer is C) y = 4x - 3
B and D can be eliminated by substituting 0 for x in each...the result is not -3
Now you have a 50/50 chance of getting the question correct
5 0
4 years ago
Please could you help me with this. 132 people took a driving test. 80 people predicted they would pass. 64 people
GenaCL600 [577]
You have to do 64 x 3 and that equals 192 the answer is 192 if that was what you was asking! Respond if not!
3 0
2 years ago
I need the answer please
Veseljchak [2.6K]
The trick to solving this problem is to know and remember that the sum of all the interior angles of a triangle is always 180 degrees.

Thus, (6x+1) + (5x-17) + (9x-24) = 180.

20x = -40, so x = -2 (answer)
7 0
4 years ago
Statements :
3241004551 [841]

Answer:

"e^x is irrational for every nonzero integer x"

Step-by-step explanation:

The original statement is

"e^x is rational for some nonzero integer x."

The negation is technically:

"It is NOT true that e^x is rational for some nonzero integer x."

So it's expressing that it's false that e^x can be rational for some nonzero integer x.

This just means that e^x is always irrational when x is a nonzero integer.

Which can be worded as

"e^x is irrational for every nonzero integer x"

4 0
3 years ago
Need step-by-step. Please answer correctly. This determines what grade I get. I will mark brainliest to the first person who ans
tatyana61 [14]

Answer:

  3√3

Step-by-step explanation:

For the problem shown here, your answer 3√3 is correct.

When there is a radical by itself in the denominator, you multiply numerator and denominator by a radical that results in the product being rational. For a square root, that will usually be the same square root:

  \dfrac{9}{\sqrt{3}}=\dfrac{9}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}}=\dfrac{9\sqrt{3}}{3}=\boxed{3\sqrt{3}}

__

If the problem has a sum in the denominator involving a square root, then you multiply numerator and denominator by the conjugate of that sum (the sum with the sign changed). This uses the special product "difference of squares" to eliminate the radical term.

<u>Example</u>:

  \dfrac{9}{2-\sqrt{3}}=\dfrac{9}{2-\sqrt{3}}\cdot\dfrac{2+\sqrt{3}}{2+\sqrt{3}}=\dfrac{9(2+\sqrt{3})}{2^2-(\sqrt{3})^2}=\dfrac{18+9\sqrt{3}}{4-3}\\\\=\boxed{18+9\sqrt{3}}

__

It is easy to demonstrate that none of the offered choices for this problem has the same value as 9/√3.

9/√3 ≈ 5.196. Offered choices have values of about 4.798, 1.732, 6.681, 23.196 -- none even close.

Please discuss this question with your teacher.

3 0
3 years ago
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