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2 years ago

Solve for x^4 - 17x^2 + 16 = 0 Select all possible solitions

Mathematics

1 answer:

miskamm [114]2 years ago

5 0

Step-by-step explanation:

4-17x^2+16=0 s

Calculate multiply

20-17x^2=0

Reoder the term

-17x^2+20=0

Calculate the discriminat

D=0^2-4x(-17)x20

Simplify

D=1360

The quadric equation has 2 real solutions

So your Solution is :

2 real solutions

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A rectangle board is 1.2 meters long and 0.8 meters wide. What is the area of the board in millimeters

ivanzaharov [21]

Answer:

1200*800 = 960,000 mm²

Step-by-step explanation:

1 meter is 1000 mm and 0.8 meter is 800 mm

A=lw

5 0

2 years ago

2. Sandy made several investments. She bought 1000 shares of a company’s stock for $8.60/share, she bought a bond with a face va

vampirchik [111]

The stock price per share was $8.60
Number of shares bought 1000
Total price for the shares:
(Cost per share)*(Number of shares)
=8.60*1000
=$8600

The stock price after 1 year $9.15
Total number of shares is 1000
Current price=(current share price)*(number of shares)
9.15*1000
=$9150
current value=(Current price)-(buying price)
=9150-8600
=$550

Net Profit=(Current value)-(Expenses)
=550-14
=$536

6 0

2 years ago

Identify the functions that are continuous on the set of real numbers and arrange them in ascending order of their limits as x t

Studentka2010 [4]

Answer:

g(x)<j(x)<k(x)<f(x)<m(x)<h(x)

Step-by-step explanation:

1. $f(x)=\frac{x^2+x-20}{x^2+4}$

The denominator of f is defined for all real values of x

Therefore, the function is continuous on the set of real numbers

$\lim_{x\rightarrow 5}\frac{x^2+x-20}{x^2+4}=\frac{25+5-20}{25+4}=\frac{10}{29}=0.345$

3. $h(x)=\frac{3x-5}{x^2-5x+7}$

$x^2-5x+7=0$

It cannot be factorize .

Therefore, it has no real values for which it is not defined .

Hence, function h is defined for all real values.

$\lim_{x\rightarrow 5}\frac{3x-5}{x^2-5x+7}=\frac{15-5}{25-25+7}=\frac{10}{7}=1.43$

2. $g(x)=\frac{x-17}{x^2+75}$

The denominator of g is defined for all real values of x.

Therefore, the function g is continuous on the set of real numbers

$\lim_{x\rightarrow 5}\frac{x-17}{x^2+75}=\frac{5-17}{25+75}=\frac{-12}{100}=-0.12$

4. $i(x)=\frac{x^2-9}{x-9}$

x-9=0

x=9

The function i is not defined for x=9

Therefore, the function i is not continuous on the set of real numbers.

5. $j(x)=\frac{4x^2-7x-65}{x^2+10}$

The denominator of j is defined for all real values of x.

Therefore, the function j is continuous on the set of real numbers.

$\lim_{x\rightarrow 5}\frac{4x^2-7x-65}{x^2+10}=\frac{100-35-65}{25+10}=0$

6. $k(x)=\frac{x+1}{x^2+x+29}$

$x^2+x+29=0$

It cannot be factorize .

Therefore, it has no real values for which it is not defined .

Hence, function k is defined for all real values.

$\lim_{x\rightarrow 5}\frac{x+1}{x^2+x+29}=\frac{5+1}{25+5+29}=\frac{6}{59}=0.102$

7. $l(x)=\frac{5x-1}{x^2-9x+8}$

$x^2-9x+8=0$

$x^2-8x-x+8=0$

$x(x-8)-1(x-8)=0$

$(x-8)(x-1)=0$

$x=8,1$

The function is not defined for x=8 and x=1

Hence, function l is not defined for all real values.

8. $m(x)=\frac{x^2+5x-24}{x^2+11}$

The denominator of m is defined for all real values of x.

Therefore, the function m is continuous on the set of real numbers.

$\lim_{x\rightarrow 5}\frac{x^2+5x-24}{x^2+11}=\frac{25+25-24}{25+11}=\frac{26}{36}=\frac{13}{18}=0.722$

g(x)<j(x)<k(x)<f(x)<m(x)<h(x)

6 0

3 years ago

What is 23 over 6 as a mixed fraction?

Ulleksa [173]

23/6

Now how many times does 6 go in 23 ???
The answer is 3 times

So, Remaining
= 23 - (6×3)
= 23- 18
= 5

So, the mixed fraction would be
= 3 5/6

3 0

3 years ago

Read 2 more answers

Can someone help me find the relative maximum and minimum? please i really need help

bearhunter [10]

9514 1404 393

Answer:

relative maximum: -4
relative (and absolute) minimum: -5

Step-by-step explanation:

The curve has a relative maximum where values on either side are lower. This looks like a peak in the curve. There is one of those on the y-axis at y = -4.

The relative maximum is -4.

A relative minimum is a low point, where the curve is higher on either side. There are two of these, located symmetrically about the y-axis. The minimum appears to be about y = -5. (They might be at x = ± 1, but it is hard to tell.)

The relative minima are -5.

A minimum or maximum is absolute if no part of the curve is lower or higher. Here, the minima are absolute, while the maximum is only relative. (The left and right branches of the curve go higher than y=-4.)

_____

Identifying the points on the curve should be the easy part. Deciding what the coordinates are can be harder when the graph is like this one.

4 0

2 years ago