What do you think your favorite way of solving an equation will be and why?

(graphing, substitution, elimination)

[tex]\\ \sf\longmapsto 2^3+6^5[/tex]

[tex]\\ \sf\longmapsto 2^3+(2\times 3)^5[/tex]

[tex]\\ \sf\longmapsto 8+2^5\times 3^5[/tex]

[tex]\\ \sf\longmapsto 8+32\times 243[/tex]

[tex]\\ \sf\longmapsto 40+7776[/tex]

[tex]\\ \sf\longmapsto 7784[/tex]

Answer:

2*2*2= 8

6*6*6*6*6= 7,776

7,776+8=

7,784

Answer:

Perimeter square = 8 sqrt(pi)

Step-by-step explanation:

The perimeter of a square is 4*s

The area of a circle is Area = pi * r^2

The circumference of a circle is C = 2*pi * r

C = 4 pi

4pi = 2*pi * r

r = 2

So the area of the circle is pi * r^2 = pi * 2^2 = 4pi

The square has the same area

Area = 4*pi

Square = 4*pi

s^2 = 4*pi

s = sqrt(4*pi)

s = 2*sqrt(pi)

The perimeter = 4 * 2 * sqrt(pi)

The perimeter = 8 * sqrt(pi)

Answer:

first odd integer=x

second odd integer=x+2

third odd integer=x+4

x+x+2+x+4=2097

x+x+x+2+4=2097

3x+6=2097

3x=2097-6

3x=2091

3x/3=2091/3

x=697

therefore, x=697

x+2=697+2=699

x+4=697+4=701

Step-by-step explanation:

A natural number is a positive whole number.

A whole number is a positive number with no fractions or decimals.

A interger is a whole number negative or positive.

A rational number is a number that terminates or continue with repeating digits.

A irrational number is a number that doesn't terminate or continue with repeating digits.

1. Rational Number

2. Natural,Whole,Interger,Rational

3. Whole,Rational,Interger

4. Rational

5.Irrational

6.Rational

7.Natural,Whole,Interger,Rational

8.Interger,Rational

9.Irrational

Answer:

D

Step-by-step explanation:

Sector of Area=(theta/360)*pi*r^2

Sector of area=(135/360)*pi*19^2=(1083)/8 pi

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:

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]

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:

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:

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

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:

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

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:

abc is a triangle so ,

a is ( 9,6 )

b is ( 9,3 )

and c is ( 3,3 )

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:

Ekta and Preyal

Step-by-step explanation:

[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

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]

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:

1 1/4

Step-by-step explanation:

2 3/4 - 1 1/2

3 3/4 - 1 2/4

1 1/4

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.

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:

classified info jk juss use a mf calculater

Step-by-step explanation:

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: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:

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:

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

#SPJ2

Answer:

divide, 2x+9

Step-by-step explanation:

got it right

Answer:

33ft^3

Step-by-step explanation:

radius is half the diameter, half of 2=1 and 1^2=1

3(1)(11)=33

Answer: V = 33 ft³

Step-by-step explanation:

π = 3

r = (1/2)d = (1/2) (2) = 1 ft

h = 11 ft

Given Formula

V = π r² h

Substitute values into the formula

V = (3) (1)² (11)

Simplify exponents

V = (3) (1) (11)

Simplify by multiplication

V = 33 ft³

Hope this helps!! :)

Please let me know if you have any questions