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3 years ago

How to graph y=12 and x=0?

Mathematics

1 answer:

Effectus [21]3 years ago

7 0

Just look at the graph

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Brandon drew a scale drawing of a hotel. a room in the hotel is 18 inches wide in the drawing. the actual room is 12 feet wide,

Contact [7]

3inches=1 1/2 feet I believe but I'm not to sure about it you will have to the math this is just my answer to you guys sorry if its wrong at least I tried

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3 years ago

Consider the quadratic equation x 2 = 4x - 5. How many solutions does the equation hase?

lara [203]

Answer:

The equation has 2 non real solutions.

Step-by-step explanation:

Given:

$x ^2 = 4x - 5$

To Find:

The solutions of the equation = ?

Solution:

Lets find the solution using the quadratic equation formula

$x ^2 = 4x - 5$

$x^2-4x +5 = 0$

$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Here

a = 1

b =-4

c = 5

Now Substituting the values,

$x = \frac{-(-4) \pm \sqrt{(-4)^2-4(1)(5)}}{2(1)}$

$x = \frac{4\pm \sqrt{16-20}}{2}$

$x = \frac{4\pm \sqrt{-4}}{2}$

$x = \frac{4\pm \sqrt{4}\times \sqrt{-1}}{2}$

$x = \frac{4\pm 2 \sqrt{-1}}{2}$

$x = \frac{4\pm 2i}{2}$

$x= \frac{4+2i}{2}$ $x= \frac{4-2i}{2}$

x= 2 + i x= 2-i

5 0

3 years ago

Select all the equations where b=11 solution

Vadim26 [7]

Answer:

b + 18 = 7

77 = 7b

9 = b - 2

7 0

3 years ago

Find the imaginary part of\[(\cos12^\circ+i\sin12^\circ+\cos48^\circ+i\sin48^\circ)^6.\]

iren [92.7K]

Answer:

The imaginary part is 0

Step-by-step explanation:

The number given is:

$x=(\cos(12)+i\sin(12)+ \cos(48)+ i\sin(48))^6$

First, we can expand this power using the binomial theorem:

$(a+b)^k=\sum_{j=0}^{k}\binom{k}{j}a^{k-j}b^{j}$

After that, we can apply De Moivre's theorem to expand each summand: $(\cos(a)+i\sin(a))^k=\cos(ka)+i\sin(ka)$

The final step is to find the common factor of i in the last expansion. Now:

$x^6=((\cos(12)+i\sin(12))+(\cos(48)+ i\sin(48)))^6$

$=\binom{6}{0}(\cos(12)+i\sin(12))^6(\cos(48)+ i\sin(48))^0+\binom{6}{1}(\cos(12)+i\sin(12))^5(\cos(48)+ i\sin(48))^1+\binom{6}{2}(\cos(12)+i\sin(12))^4(\cos(48)+ i\sin(48))^2+\binom{6}{3}(\cos(12)+i\sin(12))^3(\cos(48)+ i\sin(48))^3+\binom{6}{4}(\cos(12)+i\sin(12))^2(\cos(48)+ i\sin(48))^4+\binom{6}{5}(\cos(12)+i\sin(12))^1(\cos(48)+ i\sin(48))^5+\binom{6}{6}(\cos(12)+i\sin(12))^0(\cos(48)+ i\sin(48))^6$

$=(\cos(72)+i\sin(72))+6(\cos(60)+i\sin(60))(\cos(48)+ i\sin(48))+15(\cos(48)+i\sin(48))(\cos(96)+ i\sin(96))+20(\cos(36)+i\sin(36))(\cos(144)+ i\sin(144))+15(\cos(24)+i\sin(24))(\cos(192)+ i\sin(192))+6(\cos(12)+i\sin(12))(\cos(240)+ i\sin(240))+(\cos(288)+ i\sin(288))$

The last part is to multiply these factors and extract the imaginary part. This computation gives:

$Re x^6=\cos 72+6cos 60\cos 48-6\sin 60\sin 48+15\cos 96\cos 48-15\sin 96\sin 48+20\cos 36\cos 144-20\sin 36\sin 144+15\cos 24\cos 192-15\sin 24\sin 192+6\cos 12\cos 240-6\sin 12\sin 240+\cos 288$

$Im x^6=\sin 72+6cos 60\sin 48+6\sin 60\cos 48+15\cos 96\sin 48+15\sin 96\cos 48+20\cos 36\sin 144+20\sin 36\cos 144+15\cos 24\sin 192+15\sin 24\cos 192+6\cos 12\sin 240+6\sin 12\cos 240+\sin 288$

(It is not necessary to do a lengthy computation: the summands of the imaginary part are the products sin(a)cos(b) and cos(a)sin(b) as they involve exactly one i factor)

A calculator simplifies the imaginary part Im(x⁶) to 0

4 0

3 years ago

Estimation the 90 to the nearest tenth

alina1380 [7]

If you round 90 to the nearest tenth it is 90

8 0

3 years ago