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3 years ago

At which point do the two equations

Mathematics

1 answer:

Lena [83]3 years ago

4 0

The lines intersect at two places.

They intersect at approximately (-2,8) and (2,3)

Looking at the choices they would be at (1.8,3.2) and (-2.8,7.8)

The answer is D. Both A and B.

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Solving Square Roots Worksheet (x - k)^2 : Part 1

iVinArrow [24]

Answer:

1. x = +/- 2 $\sqrt{2}$ - 7

2. x = $3$ +/- $2i\sqrt{3}$

3. n = $2$ +/- $i\sqrt{2}$

Step-by-step explanation:

1. Divide both sides by 2: (x + 7)^2 = 8

Square root both sides: x + 7 = +/- 2 $\sqrt{2}$

Subtract 7 from both sides: x = +/- 2 $\sqrt{2}$ - 7

2. Square root both sides: x - 3 = $\sqrt{-12}$

Since there is a negative inside the radical, we need to have an imaginary number: $i=\sqrt{-1}$ . So, $\sqrt{-12} =i\sqrt{12} =2i\sqrt{3}$

Add 3 to both sides: x = $3$ +/- $2i\sqrt{3}$

3. Divide by -5 from both sides: (n - 2)^2 = -2

Square root both sides: n - 2 = $\sqrt{-2}$

Again, we have to use i: $n-2=\sqrt{-2} =i\sqrt{2}$

Add 2 to both sides: n = $2$ +/- $i\sqrt{2}$

Hope this helps!

8 0

3 years ago

Read 2 more answers

The average ticket price for a concert at the opera house was $50. The average attendance was 4000. When the ticket price was ra

kotegsom [21]

Answer

given,

opera house ticket = $50

attendance = 4000 persons

now,

opera house ticket = $52

attendance = 3800 person

assuming these are the points on the demand curve

(x, p) = (4000,50) and (x,p) = (3800,52)

using point slope formula

$p-50 = \dfrac{50-52}{4000-3800}(x - 4000)$

$p-50 = \dfrac{-2}{200}(x - 4000)$

$p-50 = \dfrac{-x}{100}+ 40$

$p = \dfrac{-x}{100}+ 90$

R(x) = x . p

$R(x) = x (\dfrac{-x}{100}+ 90)$

$R(x) = \dfrac{-x^2}{100}+ 90x)$

$\dfrac{d}{dx}(R(x)) = \dfrac{d}{dx}(\dfrac{-x^2}{100}+ 90x))$

$\dfrac{d}{dx}(R(x)) = (\dfrac{-2x}{100})+90)$

at $\dfrac{d}{dx}(R(x)) = 0$

$\dfrac{-2x}{100}= -90$

x = 4500

$\dfrac{d^2}{d^2x}(R(x)) = -ve$

hence at x =4500 the revenue is maximum

for maximum revenue ticket price will be

$p = \dfrac{-4500}{100}+ 90$

p = $45

6 0

3 years ago

A footbridge is 5 feet wide and is built with wood planks that are 6 inches wide. How many planks wide is the path?

Ede4ka [16]

Here we have to know the conversion between feet and inches.

1 Feet = 12 inches (This is given and is to be known)

So the 5 feet wide footbridge when converted to inches would be about 5*12 = 60 inches

The width of the footbridge in inches = 60 inches

Each wooden plank is 6 inches,

So the number of wooden planks that are required to build the foot bridge of 60 inches = 60/6 = 10 planks

Hence, the path path is 10 planks wide

7 0

3 years ago

Terrence says pi is a rational number since it is the ratio of a circle’s circumference to its diameter. Why is Terrence incorre

QveST [7]

Terrence is not correct.

Terrence is not correct because pi is an irrational number, and it goes on forever with no actual sequence.

4 0

3 years ago

The maximum safe load of a bridge is 1500kg to the nearest 10kg. An average soldier is 75kg to the nearest kilogram. Work out an

soldier1979 [14.2K]

<h2>i cant answer it without text sorry. i just cant see the question :(</h2>

5 0

3 years ago