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Dafna1 [17]
3 years ago
8

1.

Mathematics
1 answer:
Art [367]3 years ago
4 0
1. The answer would be "-15" since integers are composed of whole numbers and negatives.

2. The statement "<span>Every real number is a rational number." is false, since real numbers are composed of both rational and irrational numbers.

3. The number "</span><span>8.52624 . . ." because this is the only non-terminating number, which makes it the only irrational number on the list.

4. "</span><span>Irrational numbers cannot be classified as rational numbers." is the only correct statement. No irrational numbers can be rational numbers, and the opposite is also true.

5. Only the statement "</span><span>Every irrational number is a real number.</span>" is true.
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Veronika [31]
<span>https://www.youtube.com/watch?v=GjYL0-BlejI So, to answer your question yes.</span>
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Which of the following is the value of 3x’ – x+2 when x=-4
Aleks [24]

Answer:

-6

Step-by-step explanation:

3x-x+2

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What number must you add to complete the square?
Travka [436]
B=9 (x+3)^2= x^2+6x+9
7 0
3 years ago
3,842.58 rounded to the nearest tenth
Nataliya [291]
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8 0
3 years ago
Find the standard form of the equation of the hyperbola satisfying the given conditions: X intercept +/- 6; foci at (-10,0) and
uysha [10]

Answer:

\frac{x^{2}}{36} - \frac{y^{2}}{64}=1

Step-by-step explanation:

Given an hyperbola with the following conditions:

  • Foci at (-10,0) and (10,0)
  • x-intercept +/- 6;

The following holds:

  • The center is midway between the foci, so the center must be at (h, k) = (0, 0).
  • The foci are 10 units to either side of the center, so c = 10 and c^2 = 100
  • The center lies on the origin, so the two x-intercepts must then also be the hyperbola's vertices.

Since the intercepts are 6 units to either side of the center, then a = 6 and a^2 = 36.

Then, a^2+b^2=c^2\\b^2=100-36=64

Therefore, substituting a^2 = 36. and b^2=64 into the standard form

\frac{x^{2}}{a^2} - \frac{y^{2}}{b^2}=1\\We \: have:\\ \dfrac{x^{2}}{36} - \dfrac{y^{2}}{64}=1

4 0
3 years ago
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