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OverLord2011 [107]

2 years ago

12

Which of the numbers listed below are solutions to the equation? Check all that apply. x2 = 81 A. 18 B. 40.5 C. -9 D. 162 E. 9 F

. 6561

2 answers:

Harman [31]2 years ago

5 0

If 81 equals to x² (so the unknown number is 81 when x is squared), the to find x you do the square root of 81.

x = ±√81

x = <span>±9

Answers are either C and E</span>

Murljashka [212]2 years ago

3 0

X^2 = 81...take the square root of both sides, eliminating the ^2
x = (+-) sqrt 81
x = (+-) 9

answers are : -9 and 9

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In a factory, two thirds of the floor are is taken up by the production line.Out of the remaining floor area,three fifths is tak

KATRIN_1 [288]

Answer:the floor area of the production line is 20000 m^2

Step-by-step explanation:

Let x represent the total area of the factory.

In the factory, two thirds of the floor are is taken up by the production line. This means that the area of the production line is 2/3 × x = 2x/3 m^2

The remaining floor area would be

x - 2x/3 = x/3 m^2

Out of the remaining floor area,three fifths is taken up by the office space. This means that the area taken up by the office space would be 3/5 × x/3 = x/5 m^2

The remaining area would be

x/3 - x/5 = 2x/15

The rest is warehouse space. The warehouse space occupies 2000m2. This means that

2x/15 = 2000

2x = 15 × 2000 = 30000

x = 30000/2 = 15000m^2

The floor area of the production line would be

(2 × 30000)/3 = 20000 m^2

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2 years ago

A department store sells t-shirts for $16.00. Every month that a t-shirt doesn’t sell, the store reduces the selling price by 25

Reil [10]

Answer:

Step-by-step explanation:

A = 25% of 16 = $=\frac{25}{100}*8=\frac{1}{4}*8=2$

Selling price after reduction = 16 - 2 = $14

B= 25% of 14

$=\frac{25}{100}*14=\frac{1}{4}*14=\frac{7}{2}$

= 3.5

Selling price = 14 - 3.5 = $10.5

C = 25% of 10.5

$=\frac{25}{100}*10.5=\frac{10.5}{4}$

= 2.625 = $ 2.63

Selling price = 10.5 - 2.63 = $ 7.87

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2 years ago

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Makayla has $8 to buy tickets at the school fair. Each ticket costs @1.5. Which inequality best represents how many tickets she

cricket20 [7]

Answer:

$n\le 5$

Step-by-step explanation:

Let n be the number of tickets Makaya has to buy.

If the cost of one ticket is $1.5, then the cost of n tickets is $1.5n.

Makayla has $8 to buy tickets at the school fair, thus

$1.5n\le 8\\ \\n\le \dfrac{8}{1.5}\\ \\n\le \dfrac{80}{15}\\ \\n\le \dfrac{16}{3}\\ \\n\le 5\dfrac{2}{3}$

The maximum number of tickets Makaya can buy is 5, so

$n\le 5$

5 0

2 years ago

Dividends Per Share Seventy-Two Inc., a developer of radiology equipment, has stock outstanding as follows: 60,000 shares of cum

anygoal [31]

Answer and Step-by-step explanation:

The computation of dividends per share on each class of stock for each of the four years is shown below:-

Particulars 1st year 2nd-year 3rd-year 4th year

Preferred dividend

paid a $34,000 $38,000 $36,000 $36,000

Number of preferred

stock b 60,000 60,000 60,000 60,000

Dividend per share

(a ÷ b) $0.57 $0.63 $0.60 $0.60

Dividend paid to common

stockholders c $0 $38,000 $44,000 $64,000

Number of common stock

shares d 410,000 410,000 410,000 410,000

Dividend per share

on common stock $0 $0.093 $0.11 $0.16

(c ÷ d)

Working note:

Preferred dividend = Number of preferred stock shares × Par value per share × Percentage of dividend

= 60,000 × $20 × 3%

= $36,000

Preferred stock

For 1st year

= $34,000

For 2nd-year

Dividend in year 2+ Dividend balance in year 1

= $36,000 + ($36,000 - $34,000)

= $38,000

For 3rd-year

= $36,000

For 4th year

= $36,000

Common stock dividend

Particulars 1 year 2 year 3 year 4 year

Total dividend paid $34,000 $76,000 $80,000 $100,000

Less:

Preferred stock

dividend $34,000 $38,000 $36,000 $36,000

Dividend paid to common

stockholders $0 $38,000 $44,000 $64,000

8 0

2 years ago

The center of a circle is at the origin on a coordinate grid. The vertex of a parabola that opens upward is at (0, 9). If the ci

zhannawk [14.2K]

Answer:

"The maximum number of solutions is one."

Step-by-step explanation:

Hopefully the drawing helps visualize the problem.

The circle has a radius of 9 because the vertex is 9 units above the center of the circle.

The circle the parabola intersect only once and cannot intercept more than once.

The solution is "The maximum number of solutions is one."

Let's see if we can find an algebraic way:

The equation for the circle given as we know from the problem without further analysis is so far $x^2+y^2=r^2$ .

The equation for the parabola without further analysis is $y=ax^2+9$ .

We are going to plug $ax^2+9$ into $x^2+y^2=r^2$ for $y$ .

$x^2+y^2=r^2$

$x^2+(ax^2+9)^2=r^2$

To expand $(ax^2+9)^2$ , I'm going to use the following formula:

$(u+v)^2=u^2+2uv+v^2$ .

$(ax^2+9)^2=a^2x^4+18ax^2+81$ .

$x^2+y^2=r^2$

$x^2+(ax^2+9)^2=r^2$

$x^2+a^2x^4+18ax^2+81=r^2$

So this is a quadratic in terms of $x^2$

Let's put everything to one side.

Subtract $r^2$ on both sides.

$x^2+a^2x^4+18ax^2+81-r^2=0$

Reorder in standard form in terms of x:

$a^2x^4+(18a+1)x^2+(81-r^2)=0$

The discriminant of the left hand side will tell us how many solutions we will have to the equation in terms of $x^2$ .

The discriminant is $B^2-4AC$ .

If you compare our equation to $Au^2+Bu+C$ , you should determine $A=a^2$

$B=(18a+1)$

$C=(81-r^2)$

The discriminant is

$B^2-4AC$

$(18a+1)^2-4(a^2)(81-r^2)$

Multiply the (18a+1)^2 out using the formula I mentioned earlier which was:

$(u+v)^2=u^2+2uv+v^2$

$(324a^2+36a+1)-4a^2(81-r^2)$

Distribute the 4a^2 to the terms in the ( ) next to it:

$324a^2+36a+1-324a^2+4a^2r^2$

$36a+1+4a^2r^2$

We know that $a>0$ because the parabola is open up.

We know that $r>0$ because in order it to be a circle a radius has to exist.

So our discriminat is positive which means we have two solutions for $x^2$ .

But how many do we have for just $x$ .

We have to go further to see.

So the quadratic formula is:

$\frac{-B \pm \sqrt{B^2-4AC}}{2A}$

We already have $B^2-4AC}$

$\frac{-(18a+1) \pm \sqrt{36a+1+4a^2r^2}}{2a^2}$

This is t he solution for $x^2$ .

To find $x$ we must square root both sides.

$x=\pm \sqrt{\frac{-(18a+1) \pm \sqrt{36a+1+4a^2r^2}}{2a^2}}$

So there is only that one real solution (it actually includes 2 because of the plus or minus outside) here for x since the other one is square root of a negative number.

That is,

$x=\pm \sqrt{\frac{-(18a+1) \pm \sqrt{36a+1+4a^2r^2}}{2a^2}}$

means you have:

$x=\pm \sqrt{\frac{-(18a+1)+\sqrt{36a+1+4a^2r^2}}{2a^2}}$

or

$x=\pm \sqrt{\frac{-(18a+1)-\sqrt{36a+1+4a^2r^2}}{2a^2}}$ .

The second one is definitely includes a negative result in the square root.

18a+1 is positive since a is positive so -(18a+1) is negative

2a^2 is positive (a is not 0).

So you have (negative number-positive number)/positive which is a negative since the top is negative and you are dividing by a positive.

We have confirmed are max of one solution algebraically. (It is definitely not 3 solutions.)

If r=9, then there is one solution.

If r>9, then there is two solutions as this shows:

$x=\pm \sqrt{\frac{-(18a+1)+\sqrt{36a+1+4a^2r^2}}{2a^2}}$

r=9 since our circle intersects the parabola at (0,9).

Also if (0,9) is intersection, then

$0^2+9^2=r^2$ which implies r=9.

Plugging in 9 for r we get:

$x=\pm \sqrt{\frac{-(18a+1)+\sqrt{36a+1+4a^2(9)^2}}{2a^2}}$

$x=\pm \sqrt{\frac{-(18a+1)+\sqrt{36a+1+324a^2}}{2a^2}}$

$x=\pm \sqrt{\frac{-(18a+1)+\sqrt{(18a+1)^2}}{2a^2}}$

$x=\pm \sqrt{\frac{-(18a+1)+18a+1}{2a^2}}$

$x=\pm \sqrt{\frac{0}{2a^2}}$

$x=\pm 0$

$x=0$

The equations intersect at x=0. Plugging into $y=ax^2+9$ we do get $y=a(0)^2+9=9$ .

After this confirmation it would be interesting to see what happens with assume algebraically the solution should be (0,9).

This means we should have got x=0.

$0=\frac{-(18a+1)+\sqrt{36a+1+4a^2r^2}}{2a^2}$

A fraction is only 0 when it's top is 0.

$0=-(18a+1)+\sqrt{36a+1+4a^2r^2}$

Add 18a+1 on both sides:

$18a+1=\sqrt{36a+1+4a^2r^2$

Square both sides:

$324a^2+36a+1=36a+1+4a^2r^2$

Subtract 36a and 1 on both sides:

$324a^2=4a^2r^2$

Divide both sides by $4a^2$ :

$81=r^2$

Square root both sides:

$9=r$

The radius is 9 as we stated earlier.

Let's go through the radius choices.

If the radius of the circle with center (0,0) is less than 9 then the circle wouldn't intersect the parabola. So It definitely couldn't be the last two choices.

7 0

2 years ago

Read 2 more answers