Two more than three times a number is 18.

Answer:

-6

Step-by-step explanation:

x + 12 = 6

 - 12      - 12

so

12-12=0

6-12=-6

put into equation

x + 0 = -6

x = -6

Answer:

Ekta and Preyal

Step-by-step explanation:

Answer:4 in each bad 2 left over

Step-by-step explanation:

Answer:

4 in each bag and 2 left over

Step-by-step explanation:

divide 14 by 3

3 goes into 14, 4 times

14 - 12 = 2

4 in each bag and then 2 left over

Answer: 2(−5x^2 +9)(5x^2 +9)

Step-by-step explanation:

Given: 162-50x^4

Factor out 2:  2(81-25x^4)

Rewrite:  2(9^2-(5x^2)^2)

Use difference of squares:  2(9-5x^2)(9+5x^2)

Reorder (if you need to) 2(-5x^2+9)(5x^2+9)

Answer:

6(27 - 8x^4)

Step-by-step explanation:

162 - 50x^4 /2

2(81 - 24x^4) /3

6(27 - 8x^4)

Answer:

zero

Step-by-step explanation:

[tex]\overrightarrow{u} = \overrightarrow{a} - \overrightarrow{b}\\\\\overrightarrow{v}= \overrightarrow{a} + \overrightarrow{b}\\\\\overrightarrow{u} . \overrightarrow{v} = a^2 - b^2 \\\\\overrightarrow{u} . \overrightarrow{v} = 2^2 - 2^2 = 0[/tex]

So, vector u and vector v are perpendicular to each other.

Answer:

See Below (Boxed Solutions).

Step-by-step explanation:

We are given the two complex numbers:

[tex]\displaystyle z = \sqrt{3} - i\text{ and } w = 6\left(\cos \frac{5\pi}{12} + i\sin \frac{5\pi}{12}\right)[/tex]

First, convert z to polar form. Recall that polar form of a complex number is:

[tex]z=r\left(\cos \theta + i\sin\theta\right)[/tex]

We will first find its modulus r, which is given by:

[tex]\displaystyle r = |z| = \sqrt{a^2+b^2}[/tex]

In this case, a = √3 and b = -1. Thus, the modulus is:

[tex]r = \sqrt{(\sqrt{3})^2 + (-1)^2} = 2[/tex]

Next, find the argument θ in [0, 2π). Recall that:

[tex]\displaystyle \tan \theta = \frac{b}{a}[/tex]

Therefore:

[tex]\displaystyle \theta = \arctan\frac{(-1)}{\sqrt{3}}[/tex]

Evaluate:

[tex]\displaystyle \theta = -\frac{\pi}{6}[/tex]

Since z must be in QIV, using reference angles, the argument will be:

[tex]\displaystyle \theta = \frac{11\pi}{6}[/tex]

Therefore, z in polar form is:

[tex]\displaystyle z=2\left(\cos \frac{11\pi}{6} + i \sin \frac{11\pi}{6}\right)[/tex]

Part A)

Recall that when multiplying two complex numbers z and w:

[tex]zw=r_1\cdot r_2 \left(\cos (\theta _1 + \theta _2) + i\sin(\theta_1 + \theta_2)\right)[/tex]

Therefore:

[tex]\displaystyle zw = (2)(6)\left(\cos\left(\frac{11\pi}{6} + \frac{5\pi}{12}\right) + i\sin\left(\frac{11\pi}{6} + \frac{5\pi}{12}\right)\right)[/tex]

Simplify. Hence, our polar form is:

[tex]\displaystyle\boxed{zw = 12\left(\cos\frac{9\pi}{4} + i\sin \frac{9\pi}{4}\right)}[/tex]

To find the complex form, evaluate:

[tex]\displaystyle zw = 12\cos \frac{9\pi}{4} + i\left(12\sin \frac{9\pi}{4}\right) =\boxed{ 6\sqrt{2} + 6i\sqrt{2}}[/tex]

Part B)

Recall that when raising a complex number to an exponent n:

[tex]\displaystyle z^n = r^n\left(\cos (n\cdot \theta) + i\sin (n\cdot \theta)\right)[/tex]

Therefore:

[tex]\displaystyle z^{10} = r^{10} \left(\cos (10\theta) + i\sin (10\theta)\right)[/tex]

Substitute:

[tex]\displaystyle z^{10} = (2)^{10} \left(\cos \left(10\left(\frac{11\pi}{6}\right)\right) + i\sin \left(10\left(\frac{11\pi}{6}\right)\right)\right)[/tex]

Simplify:

Simplify using coterminal angles. Thus, the polar form is:

[tex]\displaystyle \boxed{z^{10} = 1024\left(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}\right)}[/tex]

And the complex form is:

[tex]\displaystyle z^{10} = 1024\cos \frac{\pi}{3} + i\left(1024\sin \frac{\pi}{3}\right) = \boxed{512+512i\sqrt{3}}[/tex]

Part C)

Recall that:

[tex]\displaystyle \frac{z}{w} = \frac{r_1}{r_2} \left(\cos (\theta_1-\theta_2)+i\sin(\theta_1-\theta_2)\right)[/tex]

Therefore:

[tex]\displaystyle \frac{z}{w} = \frac{(2)}{(6)}\left(\cos \left(\frac{11\pi}{6} - \frac{5\pi}{12}\right) + i \sin \left(\frac{11\pi}{6} - \frac{5\pi}{12}\right)\right)[/tex]

Simplify. Hence, our polar form is:

[tex]\displaystyle\boxed{ \frac{z}{w} = \frac{1}{3} \left(\cos \frac{17\pi}{12} + i \sin \frac{17\pi}{12}\right)}[/tex]

And the complex form is:

[tex]\displaystyle \begin{aligned} \frac{z}{w} &= \frac{1}{3} \cos\frac{5\pi}{12} + i \left(\frac{1}{3} \sin \frac{5\pi}{12}\right)\right)\\ \\ &=\frac{1}{3}\left(\frac{\sqrt{2}-\sqrt{6}}{4}\right) + i\left(\frac{1}{3}\left(- \frac{\sqrt{6} + \sqrt{2}}{4}\right)\right) \\ \\ &= \boxed{\frac{\sqrt{2} - \sqrt{6}}{12} -\frac{\sqrt{6}+\sqrt{2}}{12}i}\end{aligned}[/tex]

Part D)

Let a be a cube root of z. Then by definition:

[tex]\displaystyle a^3 = z = 2\left(\cos \frac{11\pi}{6} + i\sin \frac{11\pi}{6}\right)[/tex]

From the property in Part B, we know that:

[tex]\displaystyle a^3 = r^3\left(\cos (3\theta) + i\sin(3\theta)\right)[/tex]

Therefore:

[tex]\displaystyle r^3\left(\cos (3\theta) + i\sin (3\theta)\right) = 2\left(\cos \frac{11\pi}{6} + i\sin \frac{11\pi}{6}\right)[/tex]

If two complex numbers are equal, their modulus and arguments must be equivalent. Thus:

[tex]\displaystyle r^3 = 2\text{ and } 3\theta = \frac{11\pi}{6}[/tex]

The first equation can be easily solved:

[tex]r=\sqrt[3]{2}[/tex]

For the second equation, 3θ must equal 11π/6 and any other rotation. In other words:

[tex]\displaystyle 3\theta = \frac{11\pi}{6} + 2\pi n\text{ where } n\in \mathbb{Z}[/tex]

Solve for the argument:

[tex]\displaystyle \theta = \frac{11\pi}{18} + \frac{2n\pi}{3} \text{ where } n \in \mathbb{Z}[/tex]

There are three distinct solutions within [0, 2π):

[tex]\displaystyle \theta = \frac{11\pi}{18} , \frac{23\pi}{18}\text{ and } \frac{35\pi}{18}[/tex]

Hence, the three roots are:

[tex]\displaystyle a_1 = \sqrt[3]{2} \left(\cos\frac{11\pi}{18}+ \sin \frac{11\pi}{18}\right) \\ \\ \\ a_2 = \sqrt[3]{2} \left(\cos \frac{23\pi}{18} + i\sin\frac{23\pi}{18}\right) \\ \\ \\ a_3 = \sqrt[3]{2} \left(\cos \frac{35\pi}{18} + i\sin \frac{35\pi}{18}\right)[/tex]

Or, approximately:

[tex]\displaystyle\boxed{ a _ 1\approx -0.4309 + 1.1839i,} \\ \\ \boxed{a_2 \approx -0.8099-0.9652i,} \\ \\ \boxed{a_3\approx 1.2408-0.2188i}[/tex]

Answer:

classified info jk juss use a mf calculater

Step-by-step explanation:

Answer:  90

Step-by-step explanation:

[tex]\displaystyle\ \Large \boldsymbol{} (\sqrt{0,04}-\sqrt{(-1,2)^2}+\sqrt{121 } ) \cdot \sqrt{81} = \\\\\\(0,2-1,2+11)\cdot 9=(11-1)\cdot 9=90[/tex]

Answer:

228573

Step-by-step explanation:

a = 219 (first term)

an = 2018 (last term)

Sn->Sum of n terms

Sn=n/2(a + an)         [Where n is no. of terms] -> eq 1

To find number of terms,

an = a + (n-1)d     [d->Common Difference] -> eq 2

d= 226-219 = 7

=> d=7

Substituting in eq 2,

2018 = 219 + (n-1)(7)

1799 = (n-1)(7)

1799 = 7n-7

1799 = 7(n-1)

1799/7 = n-1

257 = n-1

n=258

Substituting values in eq 1,

Sn = 258/2(219+2018)

    = 129(2237)

    = 228573

Answer:

She borrowed 4000 from bank.

Step-by-step explanation:

Let 'y' be the amount borrowed from bank. Then 10000-y is the amount borrowed from her mother-in-law.

Let x= interest amount gained by bank . Then 1080- x = interest gained by mother-in-law

I1= interest rate by bank= 9%

I2= interest rate by mother-in-law=12%

Time(T) = 1 year

Now, By Simple Interest formula:

x=PTR/100

Or, x=(y*1*9)/100

Or,100x=9y

or,9y-100x=0...........................Equation (i)

Again 1080-x= ((10000-y)*1*12)/100

Or, 108000-100x=120000-12y

Or, 12y-100x=12000.................Equation(ii)

Solving equation (i) and (ii), we get

y= 4000, which the amount borrowed from bank.

Answer:

abc is a triangle so ,

a is ( 9,6 )

b is ( 9,3 )

and c is ( 3,3 )

Step-by-step explanation:

V=4/3πr^3

V=4/3π(3.85)^3

V=4/3π(57.066625)

V=4/3 (179.280089865)

V=239.04011982

V=239 ft^3

Answer:

9/8 or 1.125

Step-by-step explanation:

We want to find the area of a rectangular postage stamp

The area of a rectangle can be found by multiplying the length by the width

Given length: 3/2

Given width: 3/4

Area = 3/2 * 3/4 = 9/8 or 1.125

The area of a 2D form is the amount of space within its perimeter. The area of the stamp in square inches is 1 1/8 inches².

The area of a 2D form is the amount of space within its perimeter. It is measured in square units such as cm², m², and so on. To find the area of a square formula or another quadrilateral, multiply its length by its width.

Given that a rectangular postage stamp has a length of 3/2 inches and a width of 3/4 inch. Therefore, the area of the stamp in square inches is,

Area of the stamp = Length × Width

                              = 3/2 inches × 3/4 inches

                              = 9/8 inches²

                              = 1 1/8 inches²

Hence, the area of the stamp in square inches is 1 1/8 inches².

Learn more about the Area here:

https://brainly.com/question/1631786

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[tex]\\ \rm\Rrightarrow -30=5(x+1)[/tex]

[tex]\\ \rm\Rrightarrow -30=5x+5[/tex]

[tex]\\ \rm\Rrightarrow 5x=-30-5[/tex]

[tex]\\ \rm\Rrightarrow 5x=-35[/tex]

[tex]\\ \rm\Rrightarrow x=\dfrac{-35}{-5}[/tex]

[tex]\\ \rm\Rrightarrow x=7[/tex]

Answer:

x = -7

Step-by-step explanation:

-30 = 5 (x -1 )

5 ( x + 1 ) =-30

5 (x + 1 ) = - 30

     5            5

x + 1 = -6

x + 1 -1 = -6 -1

x = - 7

Answer:

divide, 2x+9

Step-by-step explanation:

got it right

It follows from the definition of the derivative and basic properties of arithmetic. Let f(x) and g(x) be functions. Their derivatives, if the following limits exist, are

[tex]\displaystyle f'(x) = \lim_{h\to0}\frac{f(x+h)-f(x)}h\text{ and }g'(x)\lim_{h\to0}\frac{g(x+h)-g(x)}h[/tex]

The derivative of f(x) + g(x) is then

[tex]\displaystyle \big(f(x)+g(x)\big)' = \lim_{h\to0}\big(f(x)+g(x)\big) \\\\ \big(f(x)+g(x)\big)' = \lim_{h\to0}\frac{\big(f(x+h)+g(x+h)\big)-\big(f(x)+g(x)\big)}h \\\\ \big(f(x)+g(x)\big)' = \lim_{h\to0}\frac{\big(f(x+h)-f(x)\big)+\big(g(x+h)-g(x)\big)}h \\\\ \big(f(x)+g(x)\big)' = \lim_{h\to0}\frac{f(x+h)-f(x)}h+\lim_{h\to0}\frac{g(x+h)-g(x)}h \\\\ \big(f(x)+g(x)\big)' = f'(x) + g'(x)[/tex]

Answer:

1) 0.7104 = 71%

2) 0.6615 = 66%

3) 0.8944 = 89%

Step-by-step explanation:

1)

Z(low)=-2.333 0.009815329

Z(upper)=0.583 0.720165536

2)

Z(low)=0.333 0.63055866

Z(upper)=8322.908 1

3)

Z(low)=-10.417 0

Z(upper)=1.25 0.894350226

Answer: 7/8

Cos2x has 3 formulas, Sinx is given in the question, we should use the formula with sinus. I guess that's the solution.

Answer:

Do you need it in percentage, a graph or just normal annual calculation?

Answer:

0.1009

Step-by-step explanation:

To convert percentage into decimal, you need to divide the percentage by 100

10.09/100

= 0.1009

Answer:

1320

Step-by-step explanation:

Use the formula for sum of series, s(a) = n/2(2a + (n-1)d)

The terms increase by 6, so d is 6

a is the first term, -56

n is the terms you want to find, 33

Plug in the numbers, 33/2 (2(-56)+(32)6)

Simplify into 33(80)/2 and you get 1320

Pay per shelf = $3.25

No of shelfs per hour = 5

Total hours per day = 8

Total days to find pay of = 7

= 3.25×5×8×7

= 910

Therefore she is paid $910 after 1 week.

Must click thanks and mark brainliest

Answer:

Perpendicular bisector

Step-by-step explanation:

The construction is of a perpendicular bisector, because the line in bisected and the angle formed by the bisector is a right angle.

Answer:

sqrt(3) /3

Step-by-step explanation:

1 / sqrt(3)

Multiply the top and bottom by sqrt(3)

1/ sqrt(3) * sqrt(3)/ sqrt(3)

sqrt(3) /  sqrt(3)*sqrt(3)

sqrt(3) /3

Answer:

[tex] = { \sf{ \frac{1}{ \sqrt{3} } }} \\ \\ { \sf{ = \frac{1}{ \sqrt{3} } . \frac{ \sqrt{3} }{ \sqrt{3} } }} \\ \\ = { \sf{ \frac{ \sqrt{3} }{ {( \sqrt{3}) }^{2} } = \frac{ \sqrt{3} }{3} }} [/tex]

9514 1404 393

Answer:

  (d)  Right, scalene

Step-by-step explanation:

The little square in the upper left corner tells you that is a right angle. Any triangle with a right angle is a right triangle. This one is scalene, because the sides are all different lengths.

__

Additional comment

An obtuse triangle cannot be equilateral, and vice versa.

An equilateral triangle has all sides the same length, and all angles the same measure: 60°. It is an acute triangle.

Answer:

C BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.

Step-by-step explanation:

In a certain class, a teacher distributed a few candies and a few bars among the students such that each student got an equal number of candies and an equal number of bars and no candies or bars remained undistributed. How many students were there in the class?

(1) The teacher distributed 180 candies and 40 bars.

(2) The total number of items received by each student was less than 20.

A Statement (1) ALONE is sufficient, but statement (2) ALONE is not sufficient to answer the question asked.

B Statement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient to answer the question asked.

C BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.

D EACH statement ALONE is sufficient to answer the question asked.

E Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.

180 candies and 40 bars

The highest common factor of 180 candies and 40 bars = 20

Answer:

günah(3x) + 2

Step-by-step explanation:

Gösterilen sinüzoidal fonksiyonun denklemini yazınız? A) y = cos x + 2 B) y = cos(3x) + 2 C) y = günah x + 2 D) y =

Answer:

y = sin(3x) + 2

Answer:

1 1/4

Step-by-step explanation:

2 3/4 - 1 1/2

3 3/4 - 1 2/4

1 1/4

Answer:

37.8 %

Step-by-step explanation:

Let CP = 100

[tex]SP =\frac{100+profit}{100}*CP\\\\=\frac{120}{100}*100[/tex]

SP = 120

New CP:

CP decreased by 10%

 [tex]Decreased \ amount=\frac{10}{100}*CP\\\\=\frac{10}{100}*100[/tex]

 = 10

New CP = 100- 10 = 90

New SP:

SP increased by 10%

Increase amount = [tex]\frac{10}{100}*old \ SP[/tex]

                            [tex]= \frac{10}{100}*120\\\\= 12[/tex]

New SP = 120 + 12 = 132

Profit = new SP - new CP

         = 132 - 90 = 42

Profit percentage = [tex]\frac{Profit}{CP}*100[/tex]

                             [tex]= \frac{42}{90}*100\\[/tex]

                             = 46.67%

Step-by-step explanation:

Here your ans..

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