What is the probability of getting a number less than 4 and then roiling a 4?

Mathematics

1 answer:

Aleks [24]2 years ago

4 0

well there is -4 which is the oppsite of positive 4 if that what you ment

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Find the angles marked with the letters.

OLga [1]

Answer:

bigger b is 55°, the other b is 35°, c is 29°

Step-by-step explanation:

angles opposite each other on a point are the same so the bigger b is 55

55+90=145

180-145=35 so b is 35

alternate angles are the same so 35+116=151

180-151=29 so c=29

5 0

2 years ago

Is the sequence arithmetic, geometric, or neither

mestny [16]

<h3>Answer:</h3>

The first one is arithmetic, the second one is geometric, the third one is neither, the fourth one is geometric, the last one is arithmetic.

<h3>Step-by-step explanation:</h3>

The first question goes up by a rate of -100 each time, so its arithmetic.

The second one goes up by a rate of *5 each time, so its geometric.

The third one is inconsistent so it is neither.

The fourth one goes by a factor of *-6 each time, so its geometric.

The last one goes up by a rate of -10 each time, so its arithmetic.

4 0

3 years ago

If x2+1/x2=18, find the value of x3−1/x3

Elden [556K]

Answer:

There are 2 answers : 76 or -76

Step-by-step explanation:

$x^2+\dfrac{1}{x^2} =18\\\\(x-\dfrac{1}{x} )^2=x^2+\dfrac{1}{x^2} -2=18-2=16\\\\x-\dfrac{1}{x}=4\ or\ x-\dfrac{1}{x}-4\\\\(x-\dfrac{1}{x} )^3=x^3-\dfrac{1}{x^3} -3x^2*\dfrac{1}{x}+3x*\dfrac{1}{x^2}\\\\=x^3-\dfrac{1}{x^3}-3(x-\dfrac{1}{x})\\\\x^3-\dfrac{1}{x^3}=(x-\dfrac{1}{x} )^3+3(x-\dfrac{1}{x})\\\\if\ x-\dfrac{1}{x}=4\ then \ x^3-\dfrac{1}{x^3}=4^3+3*4=64-12=76\\\\if\ x-\dfrac{1}{x}=-4\ then \ x^3-\dfrac{1}{x^3}=(-4)^3+3*(-4)=-64-12=-76\\\\$