All categories

3 years ago

What is the missing constant term in the perfect square that starts with x^2+\dfrac{1}{2}xx

Mathematics

1 answer:

grandymaker [24]3 years ago

7 0

Answer:

1/16

Step-by-step explanation:

Given the expression;

$x^2+\dfrac{1}{2}x$

To find a constant that will make it a perfect square, we will use the square if the half of coefficient of x

Coefficient of x = 1/2

half of Coefficient of x = 1/2*(1/2)

Half of Coefficient of x = 1/4

square of the half of coefficient of x = (1/4)^2

square if the half of coefficient of x = 1/16

hence the constant that will make it a perfect square is 1/16

You might be interested in

On a spelling test, Marcy got 15% of the questions incorrect. If there were 40 questions on the spelling test, than how many que

marissa [1.9K]

Answer:

Step-by-step explanation:

Marcy got 15% of the 40 wrong

.15 *40

If she got 6 wrong, then she got the rest right

40-6 = 34

She got 34 right

4 0

3 years ago

Solve the following equation for x.

Crazy boy [7]

X=(-4) or X= 4
add 112 to both sides
divide both sides by 7
then take the square root of 16

5 0

3 years ago

Read 2 more answers

Cindy shoots a bottle rocket straight up into the air such that its height h, in feet, at time t, in seconds, is given by the fu

dezoksy [38]

Answer:

t = [0.493,4.2]

Step-by-step explanation:

The height of the rocket straight up into the air is given by :

$h(t)=-16t^2+85t+5$

We need to find the interval the rocket be at least 43 feet above the ground.

$43\ge -16t^2+85t+5\\\\t\ge 0.493\ and\ t\le 4.82\ s$

So, the required interval during which the rocket is at least 43 feet above the ground is [0.493,4.2].

6 0

3 years ago

. A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs. Suppose that three bulbs ar

yaroslaw [1]

Answer:

a) 59.34%

b) 44.82%

c) 26.37%

d) 4.19%

Step-by-step explanation:

(a)

There are in total 4+5+6 = 15 bulbs. If we want to select 3 randomly there are K ways of doing this, where K is the combination of 15 elements taken 3 at a time

$K=\binom{15}{3}=\frac{15!}{3!(15-3)!}=\frac{15!}{3!12!}=\frac{15.14.13}{6}=455$

As there are 9 non 75-W bulbs, by the fundamental rule of counting, there are 6*5*9 = 270 ways of selecting 3 bulbs with exactly two 75-W bulbs.

So, the probability of selecting exactly 2 bulbs of 75 W is

$\frac{270}{455}=0.5934=59.34\%$

(b)

The probability of selecting three 40-W bulbs is

$\frac{4*3*2}{455}=0.0527=5.27\%$

The probability of selecting three 60-W bulbs is

$\frac{5*4*3}{455}=0.1318=13.18\%$

The probability of selecting three 75-W bulbs is

$\frac{6*5*4}{455}=0.2637=26.37\%$

Since the events are disjoint, the probability of taking 3 bulbs of the same kind is the sum 0.0527+0.1318+0.2637 = 0.4482 = 44.82%

(c)

There are 6*5*4 ways of selecting one bulb of each type, so the probability of selecting 3 bulbs of each type is

$\frac{6*5*4}{455}=0.2637=26.37\%$

(d)

The probability that it is necessary to examine at least six bulbs until a 75-W bulb is found, supposing there is no replacement, is the same as the probability of taking 5 bulbs one after another without replacement and none of them is 75-W.

As there are 15 bulbs and 9 of them are not 75-W, the probability a non 75-W bulb is $\frac{9}{15}=0.6$