MATHEMATICS HIGH SCHOOL

If n ≠ 0, which of the following must be greater than n?I 2n
II n²
III 2 - n

Answers

Answer 1
Answer: tbh, none of them
ex. 2n
you can have n=-1
2(-1)
-2<-1
ex. n^2
n=1
1^2
=1
1=1
ex. 2-n
n=2
2-2
=0
0<2

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Find the slope of the line that passes through the points (-2,5) and (-3,-1)
Find the measure of the indicated angle to the nearest degree.
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Which of equation represents a function? A. x2 + y2 = 9 B. {(4, 2), (4, –2), (9, 3), (9, –3)} C. x = 4. 2x + y = 5

Answers

The answer to your questions is B. {(4, 2), (4, –2), (9, 3), (9, –3)} because you don't have a repeat of a number the X side which is your output. I hope that this is the answer that you were looking for and it has helped you.

Which of the following is not a solution for the variable y in the following inequality? 4y - 3 > 37 a. 16
b. 14
c. 12
d. 10

Answers

The answer it d.10 because 4(10)-3 would be equal to 37, so it would be 37=37.

How many real-number solutions does the equation have? 1. –7x² + 6x + 3 = 0
a.one solution
b.two solutions
c.no solutions
d.infinitely many solutions

2.–8x² – 8x – 2 = 0
a.one solution
b.two solutions
c.no solutions
d. infinitely many solutions

Answers

If the discriminant is negative no real solution exists.
If the discriminant is equal to 0 only one real solution exists.
If the discriminant is positive 2 real solutions exist.
The discriminant:
D = b² - 4 a c
1 ) - 7 x² + 6 x + 3 = 0
D = 6² - 4 · ( - 7 ) · 3 = 36 + 84 = 120  > 0
Answer: b) two solutions
2 ) - 8 x² - 8 x - 2 = 0
D = ( - 8 )² - 4 · ( - 8 ) · ( - 2 ) = 64 - 64 = 0
Answer: a ) one solution

One gallon of fuel mixture contains antifreeze in the ratio of 5 parts fuel to one part antifreeze. To this is added half a gallon of mixture which is one third antifreeze and two thirds fuel. What is the ratio of fuel to antifreeze in the final mixture?

Answers

1 gal (initial) = (5/6) parts fuel & (1/6) parts antifreeze= total of 6 parts
0.5 gal (add'l) = (2/3) parts fuel & (1/3) parts antifreeze = total of 3 parts

adding the two
Fuel:          1gal*(5/6 parts) + 0.5gal*(2/3 parts) = 7/6 parts in 1.5 total gallons
Antifreeze:  1gal*(1/6 parts) + 0.5gal*(1/3 parts) = 2/6 parts in 1.5 total gallons

thus with a common denominator provides a ratio of 7:2 (fuel:antifreeze)

Gavin buys 7 rulers for £3.50 how much would 9 cost

Answers

3.50 : 7 = 0.5
0.5 * 9 = 4.5
so the answer is £4.50

How many roots do the following equations have? -12x^2 - 25x+5 +x^3=0

Answers

Answer:

There are 3 roots of the given equation.

Step-by-step explanation:

Given the equation      

-12x^2-25x+5+x^3=0

we have to tell the number of roots of the given equation.

As the number of roots for an equation is equal to degree.

The degree of a polynomial is the highest power of its monomials  with non-zero coefficients.

Hence, number of roots is the highest power in the equation.

Now, the equation is -12x^2-25x+5+x^3=0

The highest power i.e degree of equation is 3.

hence, there are 3 roots of the given equation.
-12x^2 - 25x + 5 + x^(3) = 0
x^(3) - 12x^(2) - 25x + 5 = 0
x = \sqrt[3]{((-b^(3))/(27a^(3)) + (bc)/(6a^(2)) - (d)/(2a)) + \sqrt{((-b^(3))/(27a^(3)) + (bc)/(6a^(2)) - (d)/(2a))^(2) + ((c)/(3a) - (b^(2))/(9a^(2)))^(3)}} + \sqrt[3]{((-b^(3))/(27a^(3)) + (bc)/(6a^(2)) - (d)/(2a)) - \sqrt{((-b^(3))/(27a^(3)) + (bc)/(6a^(2)) - (d)/(2a))^(2) + ((c)/(3a) - (b^(2))/(9a^(2)))^(3)}} - (b)/(3a)
x = \sqrt[3]{((-(-12)^(3))/(27(1)^(3)) + ((-12)(-25))/(6(1)^(2)) - (5)/(2(1))) + \sqrt{((-(-12)^(3))/(27(1)^(3)) + ((-12)(-25))/(6(1)^(2)) - (5)/(2(1)))^(2) + ((-25)/(3(1)) - (-(-25)^(2))/(9(1)^(2)))^(3)}} + \sqrt[3]{((-(-12)^(3))/(27(1)^(3))}} + ((-12)(-25))/(6(1)^(2)) - (5)/(2(1))) - \sqrt{((-(-12)^(3))/(27(1)^(3)) + ((-12)(-25))/(6(1)^(2)) - (5)/(2(1)))^(2) + ((-25)/(3(1)) - (-(-25)^(2))/(9(1)^(2)))^(3) - (-12)/(3(1))}}
x = \sqrt[3]{((-(-1728))/(27(1)) + (300)/(6(1)) - (5)/(2)) + \sqrt{((-(-1728))/(27(1)^(3)) + (300)/(6(1)) - (5)/(2))^(2) + ((-25)/(3(1)) - (144)/(9(1)))^(3)}}} + \sqrt[3]{((-(-1728))/(27(1)) + (300)/(6(1)) - (5)/(2)) - \sqrt{((-(-1728))/(27(1)^(3)) + (300)/(6(1)) - (5)/(2))^(2) + ((-25)/(3(1)) - (144)/(9(1)))^(3)}}} - (-12)/(3)
x = \sqrt[3]{((1728)/(27) + (300)/(6) - 2(1)/(2)) + \sqrt{((1728)/(27) + (300)/(6) - 2(1)/(2))^(2) + ((-25)/(3) - (144)/(9))^(3)}} + \sqrt[3]{((1728)/(27) + (300)/(6) - 2(1)/(2)) - \sqrt{((1728)/(27) + (300)/(6) - 2(1)/(2))^(2) + ((-25)/(3) - (144)/(9))^(3)}} - 4
x = \sqrt[3]{(64 + 50 - 2(1)/(2)) + \sqrt{(64 + 50 - 2(1)/(2))^(2) + (-8(1)/(3) - 16)^(3)}} + sqrt[3]{(64 + 50 - 2(1)/(2)) - \sqrt{(64 + 50 - 2(1)/(2))^(2) + (-8(1)/(3) - 16)^(3)}} - 4
x = \sqrt[3]{(114 - 2(1)/(2)) + \sqrt{(114 - 2(1)/(2))^(2) + (-24(1)/(3))^(3)}} + \sqrt[3]{(114 - 2(1)/(2)) - \sqrt{(114 - 2(1)/(2))^(2) + (-24(1)/(3))^(3)}} - 4
x = \sqrt[3]{(112(1)/(2)) + \sqrt{(112(1)/(2))^(2) - (24(1)/(3))^(3)}} - \sqrt[3]{(112(1)/(2)) + \sqrt{(112(1)/(2))^(2) - (24(1)/(3))^(3)}} - 4
x = \sqrt[3]{112(1)/(2) + √(12656.25 - 14408.037)} + \sqrt[3]{112(1)/(2) + √(12656.25 - 14408.037)} - 4
x = \sqrt[3]{112(1)/(2) + √(-1751.787)} + \sqrt[3]{112(1)/(2) - √(-1751.787)} - 4
x = \sqrt[3]{112(1)/(2) + 41.855i} + \sqrt[3]{112(1)/(2) - 41.855i} - 4
x = -4 + \sqrt[3]{112(1)/(2) + 41.855i} + \sqrt[3]{112(1)/(2) - 41.855i}
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