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3 years ago

Which is a solution of x2 – x – StartFraction 3 Over 4 EndFraction = 0?

Mathematics

2 answers:

V125BC [204]3 years ago

8 0

Answer:

3/2

Step-by-step explanation:

Anna [14]3 years ago

4 0

Answer:

x = - $\frac{1}{2}$ , x = $\frac{3}{2}$

Step-by-step explanation:

Given

x² - x - $\frac{3}{4}$ = 0 ( add $\frac{3}{4}$ to both sides )

x² - x = $\frac{3}{4}$

Solve using the method of completing the square

add ( half the coefficient of the x- term )² to both sides

x² + 2( - $\frac{1}{2}$ )x + $\frac{1}{4}$ = $\frac{3}{4}$ + $\frac{1}{4}$

(x - $\frac{1}{2}$ )² = 1 ( take the square root of both sides )

x - $\frac{1}{2}$ = ± 1 ( add $\frac{1}{2}$ to both sides )

x = $\frac{1}{2}$ ± 1

Thus

x = $\frac{1}{2}$ - 1 = - $\frac{1}{2}$

x = $\frac{1}{2}$ + 1 = $\frac{3}{2}$

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The top and bottom margins of a poster are each 15 cm and the side margins are each 10 cm. If the area of printed material on th

12345 [234]

Answer:

the dimension of the poster = 90 cm length and 60 cm width i.e 90 cm by 60 cm.

Step-by-step explanation:

From the given question.

Let p be the length of the of the printed material

Let q be the width of the of the printed material

Therefore pq = 2400 cm ²

q = $\dfrac{2400 \ cm^2}{p}$

To find the dimensions of the poster; we have:

the length of the poster to be p+30 and the width to be $\dfrac{2400 \ cm^2}{p} + 20$

The area of the printed material can now be: $A = (p+30)(\dfrac{2400 }{p} + 20)$

= $2400 +20 p +\dfrac{72000}{p}+600$

Let differentiate with respect to p; we have

$\dfrac{dA}{dp}= 20 - \dfrac{72000}{p^3}$

Also;

$\dfrac{d^2A}{dp^2}= \dfrac{144000}{p^3}$

For the smallest area $\dfrac{dA}{dp }=0$

$20 - \dfrac{72000}{p^2}=0$

$p^2 = \dfrac{72000}{20}$

p² = 3600

p =√3600

p = 60

Since p = 60 ; replace p = 60 in the expression q = $\dfrac{2400 \ cm^2}{p}$ to solve for q;

q = $\dfrac{2400 \ cm^2}{p}$

q = $\dfrac{2400 \ cm^2}{60}$

q = 40

Thus; the printed material has the length of 60 cm and the width of 40cm

the length of the poster = p+30 = 60 +30 = 90 cm

the width of the poster = $\dfrac{2400 \ cm^2}{p} + 20$ = $\dfrac{2400 \ cm^2}{60} + 20$ = 40 + 20 = 60

Hence; the dimension of the poster = 90 cm length and 60 cm width i.e 90 cm by 60 cm.

4 0

3 years ago

What is the product?

alexdok [17]

Answer: C

Explanation:i got it right on my test

4 0

2 years ago

Franky's age is 5 years less that 1/2 his mothers age. If Franky is 13 years old,write an equation to determine his mothers age

Alik [6]

Answer:

1/2x-5=13

Step-by-step explanation:

Let x= Frankie’s mother’s age

Half his mother’s age= 1/2x

5 less than that=1/2x-5

We know that Frankie is 13 so that equation is equal to Frankie’s age, which is 13.

Hope this helps.

4 0

3 years ago

Equation for ( 6, 13) ( 7900, 5 )

tia_tia [17]

For this case we have that by definition, the equation of the line in the slope-intersection form is given by:

$y=mx+b$

Where:

m: It is the slope of the line

b: It is the cut-off point with the y axis

We have the following points through which the line passes:

$(x_ {1}, y_ {1}) :( 6,13)\\(x_ {2}, y_ {2}): (7900,5)$

We find the slope of the line:

$m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}} = \frac {5-13} {7900-6} = \frac {-8} {7894} = - \frac {4} {3947}$

Thus, the equation of the line is of the form:

$y = - \frac {4} {3947} x + b$

We substitute one of the points and find b:

$13 = - \frac {4} {3947} (6) + b\\13 = - \frac {24} {3947} + b\\13+ \frac {24} {3947} = b\\b = \frac {51335} {3947}$

Finally, the equation is:

$y = - \frac {4} {3947} x + \frac {51335} {3947}$

Answer:

$y = - \frac {4} {3947} x + \frac {51335} {3947}$

5 0

3 years ago

There are 500 rabbits in Lancaster on February 1st. If the amount of rabbits triples every month. Write a function that represen

Anuta_ua [19.1K]

Answer:

10,93,500.

Step-by-step explanation:

On first February total number of rabbits were 500.

Now it is given that after each month rabbit population will become triple of initial.

So, after one month rabbits will become 3×500.

After two months they will become $3^{2}$ ×500.

Thus , after m months total number of rabbits will become $3^{m}$ ×500.

Now,

On August 1 , 7 months will get passed from February 1 so putting m = 7 in the equation we get ,

Total number of rabbits = $3^{7}$ ×500 = 10,93,500.

8 0

3 years ago