what positive number satisfies the condition that twice the number minus three times the reciprocal of the number is equal to 1?

1 answer:

8 0

Solve:

"twice the number minus three times the reciprocal of the number is equal to 1."
3(1)
Let the number be n. Then 2n - ------- = 1
n

Mult all 3 terms by n to elim. the fractions:

2n^2 - 3 = n. Rearranging this, we get 2n^2 - n - 3 = 0.

We need to find the roots (zeros or solutions) of this quadratic equation.

Here a=2, b= -1 and c= -3. Let's find the discriminant b^2-4ac first:

disc. = (-1)^2 - 4(2)(-3) = 1 + 24 = 25.

That's good, because 25 is a perfect square.
-(-1) plus or minus 5 1 plus or minus 5
Then x = ------------------------------ = --------------------------
2(2) 4

x could be 6/4 = 3/2, or -5/4.

You must check both answers in the original equation. If the equation is true for one or the other or for both, then you have found one or more solutions.

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G(x) = -2x^3 – 15x^2 + 36x

shusha [124]

Consider the function $G(x) = -2x^3 - 15x^2 + 36x.$ First, factor it:

$G(x) = -2x^3 - 15x^2 + 36x=-x(2x^2+15x-36)=\\ \\=-x\cdot 2\cdot \left(x-\dfrac{-15-\sqrt{513}}{4}\right)\cdot \left(x-\dfrac{-15+\sqrt{513}}{4}\right).$

The x-intercepts are at points $\left(\dfrac{-15-\sqrt{513} }{4},0\right),\ (0,0),\ \left(\dfrac{-15+\sqrt{513} }{4},0\right).$

1. From the attached graph you can see that

function is positive for $x\in \left(-\infrty, \dfrac{-15-\sqrt{513} }{4}\right)\cup \left(0,\dfrac{-15+\sqrt{513} }{4}\right);$
function is negative for $x\in \left(\dfrac{-15-\sqrt{513} }{4},0\right)\cup \left(\dfrac{-15+\sqrt{513} }{4},\infty\right).$

2. Since

$G(-x) = -2(-x)^3 - 15(-x)^2 + 36(-x)=2x^3-15x^2-36x\neq G(x)\ \text{and }\neq -G(x)$ the function is neither even nor odd.

3. The domain is $x\in (-\infty,\infty),$ the range is $y\in (-\infty,\infty).$

8 0

3 years ago

Read 2 more answers

Obtuse angles measure greater than 90 degrees

Harman [31]

yes obtuse angles measure over 90 degrees and less 180 degrees

4 0

3 years ago

50 Points!

Advocard [28]

Well, 50 points total given, 25 points per user and 13 bonus for brainliest

anyway

1.
add them equations, the y's will cancel
x+2y=4
3x-2y=4 +
4x+0y=8

4x=8
divide by 4 both sides
x=2
sub back
x+2y=4
2+2y=4
minus 2
2y=2
divide 2
y=1
x=2
y=1
(x,y)
(2,1) is solution

2.
the solution is where they intersect
multiply 2nd equation by 2 and add to first
4x-14y=6
-4x+14y=-6 +
0x+0y=0
0=0
infinite solutions

that is because they are actually the same line
the solutions are (x,y) such that they satisfy -2x+7y=-3 or 4x-14y=6 (same equaiton)
infinite solutions

3.
multiply first equation by 2 and add to first
4x+2y=-6
1x-2y=-4 +
5x+0y=-10

5x=-10
divide by 5 both sides
x=-2
sub bac
x-2y=-4
-2-2y=-4
add 2
-2y=-2
divide by -2
y=1

x=-2
y=1
(x,y)
(-2,1)

4.
coincident means they are the same line
so

we see that we have to multiply 4 by 2 to get 8
multiply top equation by 2
8x+10y=16
8x+By=C
B=10 and C=16

5.
a. false, either 0, 1, or infinity solutions
change the word 'two' to 'one' or 'zero' or 'infinite', or change 'can' into 'can't'

b.false
'sometimes' to 'always'

c. true

d. false, change 'sometimes' to 'always'

EC
total cost=150
TC=childC+adultC
TC=3c+5a
150=3c+5a

40 tickets, c+a
40=c+a

the equations are
150=3c+5a and
40=c+a

eliminate
multiply 2nd equaton by -3 and ad to first one

150=3c+5a
-120=-3c-3a +
30=0c+2a

30=2a
divide by 2
15=a

sub back
40=c+a
40=c+15
minus 15
25=c

25 children tickets and 15 adult tickets were sold

ANSWERS:

1.
(2,1) is solution

2.
infinite solutions

3.
(-2,1)

4.
B=10 and C=16

5.
a. false,
change the word 'two' to 'one' or 'zero' or 'infinite', or change 'can' into 'can't'

b.false
'sometimes' to 'always'

c. true

d. false, change 'sometimes' to 'always'

EC
the equations are
150=3c+5a and
40=c+a
25 children tickets and 15 adult tickets were sold

7 0

3 years ago

Read 2 more answers

You earn $8.50 per hour. Then you receive a raise to $9.35 per hour. Find the percent increase. Then find your pay per hour if y

netineya [11]

Subtract $8.50 by $9.35

4 0

3 years ago

sophie was 28 years old when daughter Nichole was born. In how many years will Sophie be two years less than still times Nichole

Kamila [148]

It will take 6 years for that to happen.

You can find this by setting Nichole's age as x and Sophie's age as x + 28 since she will always be 28 years older.

Now we are looking for a time that Sophie's age is equal to 2 less than 6 times Nichole's age. So we can set up the below equation.

6(x) - 2 = x + 28
6x - 2 = x + 28
5x - 2 = 28
5x = 30
x = 6

This let's us know that it needs to be when Nichole is 6, which is 6 years in the future.

6 0

3 years ago