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scoundrel [369]

2 years ago

15

One solution each is given for four quadratic equations. Assuming that each quadratic equation has two solutions, what is the se

cond solution
for each equation?
I = - 4 + 51
I = 4 - 5
I = 5 + 4i
I = -5 - 4
I = 4 + 51
H
II
5 - 4
II
- 51
Reset
Next

1 answer:

Romashka-Z-Leto [24]2 years ago

5 0

Answer:

5+4i=4i+5

5i-4=-4+5i

-5i-4=-4-5i

Step-by-step explanation:

5+4i=4i+5

5i-4=-4+5i

-5i-4=-4-5i

Find the two that are the same with the terms swiched around and you get the answer.

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7k-2(3k+1)-9 simplifyed

tamaranim1 [39]

Answer:

k= -11

Step-by-step explanation:

Let's simplify step-by-step.

7k−2(3k+1)−9

Distribute:

=7k+(−2)(3k)+(−2)(1)+−9

=7k+−6k+−2+−9

Combine Like Terms:

=7k+−6k+−2+−9

=(7k+−6k)+(−2+−9)

=k + −11

6 0

3 years ago

Water flows through a pipe at a rate of 12 cups every 5 hours. Express this rate of flow in gallons per week.

horrorfan [7]

Answer:

Well 403.2 gallons per week

Step-by-step explanation:

So 24*7

168

168/5

33 3/5

33 3/5* 12

403.2

7 0

3 years ago

The members of the city cultural center have decided to put on a play once a night for a week. Their auditorium holds

fgiga [73]

A) At $3,300 for 600 seats, the average price per seat is 3300/600 = $5.50. The mix of tickets that results in that average can be found using an X diagram as shown below. The numbers on the right are the differences along the diagonals. When they are multiplied by 2, they add to 600. This shows that the required sales for revenue of $3,300 are
200 adult tickets
400 student tickets

b) When 3 student tickets are sold for each adult ticket, the average seat price is
(3*$4.50 +7.50)/4 = $5.25
Then the shortfall in revenue is ...
$3,300 -480*$5.25 = $780

Hope I was able to help!<3 mark brainly I would appreciate it!:)

8 0

2 years ago

All boxes with a square base, an open top, and a volume of 60 ft cubed have a surface area given by S(x)equalsx squared plus

Karo-lina-s [1.5K]

Answer:

The absolute minimum of the surface area function on the interval $(0,\infty)$ is $S(2\sqrt[3]{15})=12\cdot \:15^{\frac{2}{3}} \:ft^2$

The dimensions of the box with minimum surface area are: the base edge $x=2\sqrt[3]{15}\:ft$ and the height $h=\sqrt[3]{15} \:ft$

Step-by-step explanation:

We are given the surface area of a box $S(x)=x^2+\frac{240}{x}$ where x is the length of the sides of the base.

Our goal is to find the absolute minimum of the the surface area function on the interval $(0,\infty)$ and the dimensions of the box with minimum surface area.

1. To find the absolute minimum you must find the derivative of the surface area ( $S'(x)$ ) and find the critical points of the derivative ( $S'(x)=0$ ).

$\frac{d}{dx} S(x)=\frac{d}{dx}(x^2+\frac{240}{x})\\\\\frac{d}{dx} S(x)=\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(\frac{240}{x}\right)\\\\S'(x)=2x-\frac{240}{x^2}$

Next,

$2x-\frac{240}{x^2}=0\\2xx^2-\frac{240}{x^2}x^2=0\cdot \:x^2\\2x^3-240=0\\x^3=120$

There is a undefined solution $x=0$ and a real solution $x=2\sqrt[3]{15}$ . These point divide the number line into two intervals $(0,2\sqrt[3]{15})$ and $(2\sqrt[3]{15}, \infty)$

Evaluate S'(x) at each interval to see if it's positive or negative on that interval.

$\begin{array}{cccc}Interval&x-value&S'(x)&Verdict\\(0,2\sqrt[3]{15}) &2&-56&decreasing\\(2\sqrt[3]{15}, \infty)&6&\frac{16}{3}&increasing \end{array}$

An extremum point would be a point where f(x) is defined and f'(x) changes signs.

We can see from the table that f(x) decreases before $x=2\sqrt[3]{15}$ , increases after it, and is defined at $x=2\sqrt[3]{15}$ . So f(x) has a relative minimum point at $x=2\sqrt[3]{15}$ .

To confirm that this is the point of an absolute minimum we need to find the second derivative of the surface area and show that is positive for $x=2\sqrt[3]{15}$ .

$\frac{d}{dx} S'(x)=\frac{d}{dx}(2x-\frac{240}{x^2})\\\\S''(x) =\frac{d}{dx}\left(2x\right)-\frac{d}{dx}\left(\frac{240}{x^2}\right)\\\\S''(x) =2+\frac{480}{x^3}$

and for $x=2\sqrt[3]{15}$ we get:

$2+\frac{480}{\left(2\sqrt[3]{15}\right)^3}\\\\\frac{480}{\left(2\sqrt[3]{15}\right)^3}=2^2\\\\2+4=6>0$

Therefore S(x) has a minimum at $x=2\sqrt[3]{15}$ which is:

$S(2\sqrt[3]{15})=(2\sqrt[3]{15})^2+\frac{240}{2\sqrt[3]{15}} \\\\2^2\cdot \:15^{\frac{2}{3}}+2^3\cdot \:15^{\frac{2}{3}}\\\\4\cdot \:15^{\frac{2}{3}}+8\cdot \:15^{\frac{2}{3}}\\\\S(2\sqrt[3]{15})=12\cdot \:15^{\frac{2}{3}} \:ft^2$

2. To find the third dimension of the box with minimum surface area:

We know that the volume is 60 $ft^3$ and the volume of a box with a square base is $V=x^2h$ , we solve for h

$h=\frac{V}{x^2}$

Substituting V = 60 $ft^3$ and $x=2\sqrt[3]{15}$

$h=\frac{60}{(2\sqrt[3]{15})^2}\\\\h=\frac{60}{2^2\cdot \:15^{\frac{2}{3}}}\\\\h=\sqrt[3]{15} \:ft$

The dimension are the base edge $x=2\sqrt[3]{15}\:ft$ and the height $h=\sqrt[3]{15} \:ft$

6 0

3 years ago

Help ME! I’m literly struggling

miv72 [106K]

Answer:

See explanation below

Step-by-step explanation:

BD - diagonal Added Construction

m∠CBD = m∠ADB Alternate Interior Angles Theorem

BD ≅ DB Reflexive Property

m∠A = m∠C Opposite ∠'s Congruent Theorem

ΔABD ≅ ΔCDB AAS or SAS

BC ≅ DA CPCTC

AC - diagonal Added Construction

m∠BCA = m∠CAD Alternate Interior Angles Theorem

AC ≅ CA Reflexive Property

m∠B = m∠D Opposite ∠'s Congruent Theorem

ΔABC ≅ ΔCDA AAS or SAS

AB ≅ CD CPCTC

6 0

2 years ago