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aev [14]
3 years ago
10

Using the quadratic formula to solve 5x = 6x2 – 3, what are the values of x? StartFraction 5 plus-or-minus 3 StartRoot 11 EndRoo

t Over 12 EndFraction StartFraction 5 plus-or-minus StartRoot 97 EndRoot Over 12 EndFraction StartFraction 5 plus-or-minus StartRoot 47 EndRoot Over 12 EndFraction StartFraction negative 5 plus-or-minus StartRoot 97 EndRoot Over 12 EndFraction
Mathematics
2 answers:
Licemer1 [7]3 years ago
5 0

Answer:

Using the quadratic formula to solve 5x = 6x^2 – 3, what are the values of x?

5x = 6x^2 – 3

Subtract 5x from both sides:

0 = 6x^2 – 5x – 3

a = 6, b = -5, c = -3

x = (-b ± √(b^2 - 4ac))/(2a)

x = (-(-5) ± √((-5)^2 - 4(6)(-3)))/(2(6))

x = (5 ± √(25 + 72))/12

x = (5 ± √(97))/12

Step-by-step explanation:

iris [78.8K]3 years ago
3 0

Answer:

your answer is C

Step-by-step explanation:

thank me later;)

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A hypothetical population consists of eight individuals ages 13, 14, 17, 20, 21, 22, 24, & 30 years.
Alika [10]

Complete Question

A hypothetical population consists of eight individuals ages 13 14 17 20 21 22 24 30 years.  

A: what is the probability that a person in this population is a teenager?  

B: what is the probability of selecting a participant who is at least 20 years old?

We have that  probability that a person in this population is a teenager and probability of selecting a participant who is at least 20 years old is

From the question we are told

A hypothetical population consists of eight individuals ages 13, 14, 17, 20, 21, 22, 24, & 30 years.

  • P(T)=0.38
  • P(T')=0.63

a)

Generally the equation for the  probability that a person in this population is a teenager   is mathematically given as

P(T)=\frac{no of teens}{n}\\\\P(T)=\frac{3}{8}

P(T)=0.38

b)

Generally the equation for the probability of selecting a participant who is at least 20 years old  is mathematically given as

P(T)=\frac{ participants\ who\ is\ at\ least\ 20\ years old}{n}\\\\P(T)=\frac{5}{8}

P(T')=0.63

For more information on this visit

brainly.com/question/11234923?referrer=searchResults

6 0
2 years ago
A hummingbird can make up to 150 flaps in 4 seconds. How long will it take for it to flap its wings 1000 times?
andreev551 [17]
1000/150 = 6 2/3 * 4 = 26 2/3 seconds
6 0
3 years ago
Can someone PLEASE help me. No one will answer the question completely smh
Vesna [10]

Ok got it. If you have any questions just ask in the chat below :)

8 0
3 years ago
Consider the function f(x)=xln(x). Let Tn be the nth degree Taylor approximation of f(2) about x=1. Find: T1, T2, T3. find |R3|
Fynjy0 [20]

Answer:

R3 <= 0.083

Step-by-step explanation:

f(x)=xlnx,

The derivatives are as follows:

f'(x)=1+lnx,

f"(x)=1/x,

f"'(x)=-1/x²

f^(4)(x)=2/x³

Simialrly;

f(1) = 0,

f'(1) = 1,

f"(1) = 1,

f"'(1) = -1,

f^(4)(1) = 2

As such;

T1 = f(1) + f'(1)(x-1)

T1 = 0+1(x-1)

T1 = x - 1

T2 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2

T2 = 0+1(x-1)+1(x-1)^2

T2 = x-1+(x²-2x+1)/2

T2 = x²/2 - 1/2

T3 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2+f"'(1)/6(x-1)^3

T3 = 0+1(x-1)+1/2(x-1)^2-1/6(x-1)^3

T3 = 1/6 (-x^3 + 6 x^2 - 3 x - 2)

Thus, T1(2) = 2 - 1

T1(2) = 1

T2 (2) = 2²/2 - 1/2

T2 (2) = 3/2

T2 (2) = 1.5

T3(2) = 1/6 (-2^3 + 6 *2^2 - 3 *2 - 2)

T3(2) = 4/3

T3(2) = 1.333

Since;

f(2) = 2 × ln(2)

f(2) = 2×0.693147 =

f(2) = 1.386294

Since;

f(2) >T3; it is significant to posit that T3 is an underestimate of f(2).

Then; we have, R3 <= | f^(4)(c)/(4!)(x-1)^4 |,

Since;

f^(4)(x)=2/x^3, we have, |f^(4)(c)| <= 2

Finally;

R3 <= |2/(4!)(2-1)^4|

R3 <= | 2 / 24× 1 |

R3 <= 1/12

R3 <= 0.083

5 0
3 years ago
On a coordinate plane, 2 lines intersect around (0.4, 2.8). The graph on the left represents a system of equations. How can the
kompoz [17]

Answer:

How can the solution be located?

find the intersection

Which integers is the x-coordinate between?

0 and 1

Which integers is the y-coordinate between?

2 and 3

What is the solution approximated to the tenths place?

(0.4,2.8)

Step-by-step explanation:

3 0
3 years ago
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