What is an equation of the line that passes through the point (-1, 4) and is perpendicular to the line x-6у = 12?

Answer:

20

Step-by-step explanation:

This is because If we are looking for arena then we need to multiply the base by the height if we do this then our answer will be 40, but because we are looking at a triangle we will have to devide that by 2. Which will give us the final answer of 20

Answer:

area = 20 inches

Step-by-step explanation:

Area of a triangle = [tex]\frac{1}{2}[/tex] × Base × Height

= [tex]\frac{1}{2}[/tex] × 8 ×5

= [tex]\frac{1}{2}[/tex] × 40

= [tex]\frac{40}{2}[/tex]

= 20

area = 20 inches

Answer:

B. 7 to the right of the 9

9+(-7)

=9-7=2

so, the 7 is on the right side of 9

jet1 = 549 mi/h

jet2 = 620 mi/h

x= jet1

x + 71 = jet2

6 hours

6 * x

6 * x + 71

6x + 6x + 426 = 7014

12x = 7014 - 426

12x = 6588

x = 6588/12

x= 549

x + 71 = 620

wyzant

Answer:

6 in.

Step-by-step explanation:

To find the area of the rectangle, substitute the given values into the formula A = lw. Then, substitute 36 inches squared into the formula A = s2. Find the square root of both sides of the equation to find s.

Answer:

irrational

Step-by-step explanation:

√5 is a non terminating and non recurring number therefore it can't be written as quotient of two numbers so it is irrational.

Note:if you need to ask any question please let me know.

Given z = 3 + i, right away we can find

(a) square

z ² = (3 + i )² = 3² + 6i + i ² = 9 + 6i - 1 = 8 + 6i

(b) modulus

|z| = √(3² + 1²) = √(9 + 1) = √10

(d) polar form

First find the argument:

arg(z) = arctan(1/3)

Then

z = |z| exp(i arg(z))

z = √10 exp(i arctan(1/3))

or

z = √10 (cos(arctan(1/3)) + i sin(arctan(1/3))

(c) square root

Any complex number has 2 square roots. Using the polar form from part (d), we have

√z = √(√10) exp(i arctan(1/3) / 2)

and

√z = √(√10) exp(i (arctan(1/3) + 2π) / 2)

Then in standard rectangular form, we have

[tex]\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right)\right)[/tex]

and

[tex]\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right)\right)[/tex]

We can simplify this further. We know that z lies in the first quadrant, so

0 < arg(z) = arctan(1/3) < π/2

which means

0 < 1/2 arctan(1/3) < π/4

Then both cos(1/2 arctan(1/3)) and sin(1/2 arctan(1/3)) are positive. Using the half-angle identity, we then have

[tex]\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}[/tex]

[tex]\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}[/tex]

and since cos(x + π) = -cos(x) and sin(x + π) = -sin(x),

[tex]\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}[/tex]

[tex]\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}[/tex]

Now, arctan(1/3) is an angle y such that tan(y) = 1/3. In a right triangle satisfying this relation, we would see that cos(y) = 3/√10 and sin(y) = 1/√10. Then

[tex]\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10+3\sqrt{10}}{20}}[/tex]

[tex]\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10-3\sqrt{10}}{20}}[/tex]

[tex]\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}[/tex]

[tex]\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}[/tex]

So the two square roots of z are

[tex]\boxed{\sqrt z = \sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}[/tex]

and

[tex]\boxed{\sqrt z = -\sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}[/tex]

Answer:

[tex]\displaystyle \text{a. }8+6i\\\\\text{b. }\sqrt{10}\\\\\text{c. }\\\sqrt{\sqrt{\frac{5}{2}}+\frac{3}{2}}+i\sqrt{\frac{\sqrt{10}-3}{2}},\\-\sqrt{\sqrt{\frac{5}{2}}+\frac{3}{2}}-i\sqrt{\frac{\sqrt{10}-3}{2}}\\\\\\\text{d. }\\\text{Exact: }z=\sqrt{10}\left(\cos\left(\arctan\left(\frac{1}{3}\right)\right), i\sin\left(\arctan\left(\frac{1}{3}\right)\right)\right),\\\text{Approximated: }z=3.16(\cos(18.4^{\circ}),i\sin(18.4^{\circ}))[/tex]

Step-by-step explanation:

Recall that [tex]i=\sqrt{-1}[/tex]

Part A:

We are just squaring a binomial, so the FOIL method works great. Also, recall that [tex](a+b)^2=a^2+2ab+b^2[/tex].

[tex]z^2=(3+i)^2,\\z^2=3^2+2(3i)+i^2,\\z^2=9+6i-1,\\z^2=\boxed{8+6i}[/tex]

Part B:

The magnitude, or modulus, of some complex number [tex]a+bi[/tex] is given by [tex]\sqrt{a^2+b^2}[/tex].

In [tex]3+i[/tex], assign values:

[tex]|z|=\sqrt{3^2+1^2},\\|z|=\sqrt{9+1},\\|z|=\sqrt{10}[/tex]

Part C:

In Part A, notice that when we square a complex number in the form [tex]a+bi[/tex], our answer is still a complex number in the form

We have:

[tex](c+di)^2=a+bi[/tex]

Expanding, we get:

[tex]c^2+2cdi+(di)^2=a+bi,\\c^2+2cdi+d^2(-1)=a+bi,\\c^2-d^2+2cdi=a+bi[/tex]

This is still in the exact same form as [tex]a+bi[/tex] where:

Thus, we have the following system of equations:

[tex]\begin{cases}c^2-d^2=3,\\2cd=1\end{cases}[/tex]

Divide the second equation by [tex]2d[/tex] to isolate [tex]c[/tex]:

[tex]2cd=1,\\\frac{2cd}{2d}=\frac{1}{2d},\\c=\frac{1}{2d}[/tex]

Substitute this into the first equation:

[tex]\left(\frac{1}{2d}\right)^2-d^2=3,\\\frac{1}{4d^2}-d^2=3,\\1-4d^4=12d^2,\\-4d^4-12d^2+1=0[/tex]

This is a quadratic disguise, let [tex]u=d^2[/tex] and solve like a normal quadratic.

Solving yields:

[tex]d=\pm i \sqrt{\frac{3+\sqrt{10}}{2}},\\d=\pm \sqrt{\frac{{\sqrt{10}-3}}{2}}[/tex]

We stipulate [tex]d\in \mathbb{R}[/tex] and therefore [tex]d=\pm i \sqrt{\frac{3+\sqrt{10}}{2}}[/tex] is extraneous.

Thus, we have the following cases:

[tex]\begin{cases}c^2-\left(\sqrt{\frac{\sqrt{10}-3}{2}}\right)^2=3\\c^2-\left(-\sqrt{\frac{\sqrt{10}-3}{2}}\right)^2=3\end{cases}\\[/tex]

Notice that [tex]\left(\sqrt{\frac{\sqrt{10}-3}{2}}\right)^2=\left(-\sqrt{\frac{\sqrt{10}-3}{2}}\right)^2[/tex]. However, since [tex]2cd=1[/tex], two solutions will be extraneous and we will have only two roots.

Solving, we have:

[tex]\begin{cases}c^2-\left(\sqrt{\frac{\sqrt{10}-3}{2}}\right)^2=3 \\c^2-\left(-\sqrt{\frac{\sqrt{10}-3}{2}}\right)^2=3\end{cases}\\\\c^2-\sqrt{\frac{5}{2}}+\frac{3}{2}=3,\\c=\pm \sqrt{\sqrt{\frac{5}{2}}+\frac{3}{2}[/tex]

Given the conditions [tex]c\in \mathbb{R}, d\in \mathbb{R}, 2cd=1[/tex], the solutions to this system of equations are:

[tex]\left(\sqrt{\sqrt{\frac{5}{2}}+\frac{3}{2}}, \sqrt{\frac{\sqrt{10}-3}{2}}\right),\\\left(-\sqrt{\sqrt{\frac{5}{2}}+\frac{3}{2}},- \frac{\sqrt{10}-3}{2}}\right)[/tex]

Therefore, the square roots of [tex]z=3+i[/tex] are:

[tex]\sqrt{z}=\boxed{\sqrt{\sqrt{\frac{5}{2}}+\frac{3}{2}}+i\sqrt{\frac{\sqrt{10}-3}{2}} },\\\sqrt{z}=\boxed{-\sqrt{\sqrt{\frac{5}{2}}+\frac{3}{2}}-i\sqrt{\frac{\sqrt{10}-3}{2}}}[/tex]

Part D:

The polar form of some complex number [tex]a+bi[/tex] is given by [tex]z=r(\cos \theta+\sin \theta)i[/tex], where [tex]r[/tex] is the modulus of the complex number (as we found in Part B), and [tex]\theta=\arctan(\frac{b}{a})[/tex] (derive from right triangle in a complex plane).

We already found the value of the modulus/magnitude in Part B to be [tex]r=\sqrt{10}[/tex].

The angular polar coordinate [tex]\theta[/tex] is given by [tex]\theta=\arctan(\frac{b}{a})[/tex] and thus is:

[tex]\theta=\arctan(\frac{1}{3}),\\\theta=18.43494882\approx 18.4^{\circ}[/tex]

Therefore, the polar form of [tex]z[/tex] is:

[tex]\displaystyle \text{Exact: }z=\sqrt{10}\left(\cos\left(\arctan\left(\frac{1}{3}\right)\right), i\sin\left(\arctan\left(\frac{1}{3}\right)\right)\right),\\\text{Approximated: }z=3.16(\cos(18.4^{\circ}),i\sin(18.4^{\circ}))[/tex]

Answer:

225

Step-by-step explanation:

x + y + z = 1305

x = 360

y = 360×2 = 720

z = 1305 - 360 - 720 = 225

Answer:

225

Step-by-step explanation:

since the first number is 360 and the second number is twice the first number meaning it's (360×2) which is 720,you have to add the two and subtract by 1305 to find the value of the third number(which I have represented by x)

360+720+x=1305

1080+x=1305

x=1305-1080

=225

therefore the third number is 225,you can prove it by adding the three to see whether you will get 1305.

I hope this helps

Answer:

Here is the answer to your question. Stay blessed. Nice to help you.

Answer:

True

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Algebra I

Algebra II

Step-by-step explanation:

Step 1: Define

Identify

Point (5, 6)

Point (-3, 8)

Step 2: Find distance d

Simply plug in the 2 coordinates into the distance formula to find distance d

3²: It is an example of numeric expression

is a mathematical sentence that encompasses, power, root, multiplication, division, addition and subtraction.

In this question - the following example 3² was given. This example can be classified as a numerical expression, because a power is a multiplication of equal factors.

So, this example is a numeric expression.

Answer:

a^6b^4

Step-by-step explanation:

(a³b²)²

a^6b^4

9514 1404 393

Answer:

  a⁶b⁴

Step-by-step explanation:

The relevant rules of exponents are ...

  (ab)^c = (a^c)(b^c)

  (a^b)^c = a^(bc)

__

These let us simplify the expression as follows:

  [tex](a^3b^2)^2=(a^3)^2(b^2)^2=a^{3\cdot2}b^{2\cdot2}=\boxed{a^6b^4}[/tex]

_____

Additional comment

It might be helpful to remember that an exponent signifies repeated multiplication.

  a·a·a = a³ . . . . . . 'a' is a factor 3 times, so the exponent is 3.

Similarly, ...

  (a³)² = (a³)·(a³) = (a·a·a)·(a·a·a) = a⁶

Answer:

15/22

Step-by-step explanation:

3/11 ÷ 2/5

3/11 x 5/2

3/11 x 5/2 = 15/22

3/11 divided by 2/5 is equal to 15/22.

Here, we have,

To divide fractions, we can multiply the first fraction by the reciprocal of the second fraction.

The reciprocal of a fraction is obtained by flipping the numerator and denominator.

Let's calculate 3/11 divided by 2/5:

(3/11) ÷ (2/5)

To divide, we multiply by the reciprocal:

(3/11) * (5/2)

Now, multiply the numerators together and the denominators together:

(3 * 5) / (11 * 2) = 15/22

Therefore, 3/11 divided by 2/5 is equal to 15/22.

To learn more on division click:

brainly.com/question/21416852

#SPJ6

Answer:

4  2/5

Step-by-step explanation:

4 is 4

0.40 is equal to 4/10 which is equal to 2/5

hoped this helped

Answer:

22/5

Step-by-step explanation:

Answer:

$680

Step-by-step explanation:

original price=$800

discount = 15%

discount in dollars= $800 *15%

discount=$120

cost=$800-$120

cost=$680

Answer:

V = 192π cm³ ≈ 603 cm³

Step-by-step explanation:

V = πR²h

V = π(8/2)²(12)

V = 192π cm³ ≈ 603 cm³

J = 3   Hooray, but why?

Answer:

Step-by-step explanation:

Radius r = 4 cm

Height h = 12 cm

Volume = πr²h = 192π cm³ ≈ 576 cm³

Answer:

6 * 100 = 600

14 * 10  = 140

12 * 1    = 12

------------------------

752

Answer:

x=4.25

------------------------------------------

8+8=4x-1

16=4x-1

4x=16+1

4x=17

x= 17/4

I have no idea if i am correct just a guesstimate

Have a good day

Answer:

Y = (-2,-4)

Step-by-step explanation:

The ratio would be 1:5

so

1/5(6 - -4) + -4 , 1/5(-2 -3) +3

Endpoints of the segment XZ are [tex]X(x_1,y_1)[/tex] and [tex]Z(x_2,y_2)[/tex].

If a point Y(h, k) divides it in the ratio of m : n.

Then the coordinates of the point Y will be,

[tex](\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n})[/tex]

If the endpoints are [tex]X(-4,3)[/tex] and [tex]Z(6,-2)[/tex]and  point [tex]Y(h.,k)[/tex] divides the this segment in the ratio of [tex]1:5[/tex].

(h, k) = [tex](\frac{1(6)+(5)(-4)}{1+5},\frac{1(-2)+5(3)}{1+5})[/tex]

        = [tex](-\frac{14}{6},\frac{13}{6})[/tex]

        = [tex](-\frac{7}{3},\frac{13}{6})[/tex]

Lean more,

https://brainly.com/question/24407442

Answer:

8i+3j

Step-by-step explanation:

let point P2(0,3)

point P1(-8,0)

vector P1P2= Position vector of P2- position vector of P1

vector= (0,3)-(-8,0)

vector= (8,3)

vector=8i+3j

Answer: New price =  rs. 1710

Step-by-step explanation:

Given information

Original price (market price) = rs.1800

Discount rate = 5%

Given expression deducted from the question

New price = Original price × (1 - discount rate)

Substitute values into the expression

New price = 1800 × (1 - 5%)

Simplify parentheses

New price = 1800 × 0.95

Simplify by multiplication

New price =[tex]\boxed{rs.1710}[/tex]

Hope this helps!! :)

Please let me know if you have any questions

the correct answer would be 54 degrees. you would subtract 90 and 54 from 360, leaving 108. you would have to divide by 2 since we have 1 and 2 angles, that leaves 54. hope this helps!

Answer:

last option 180 degrees

Step-by-step explanation:

let ∠1 and angle ∠2 be 2x as both are of same angles

angle sum proprty = 360

90 + 54+ ∠1 +∠2 = 360

144 + 2x = 360

2x = 360 - 144

2x = 216

x = 216 ÷ 2

x = 108

∠2 = 108 degrees

Step-by-step explanation:

z = 5y/8

x = 3y/8

2x + 3y - z = 50

6y/8 + 3y - 5y/8 = 50

6y/8 + 24y/8 - 5y/8 = 50

25y/8 = 50

25y = 400

y = 16

x = 3y/8 = 3×16/8 = 3×2 = 6

z = 5y/8 = 5×16/8 = 5×2 = 10

Step-by-step explanation:

one to one relationship

this is the answer

Answer:

Explaination down below

Step-by-step explanation:

Before you start this question, make sure you know this rule. When multiplying fractions, Multiply the numerator by the other numerator and multiply the denominator by the other one.

[tex]\frac{1}{5}[/tex] x [tex]\frac{1}{2}[/tex]. 1*1=1, so one is the numerator. 2*5= 10 so 10 is the denominator.

The final answer is [tex]\frac{1}{10}[/tex].

You could also look at it another way. What is one half of one-fifth?

That is another way to look at it.

If this helped, please mark me as brainliest. Thank you! ;)

Answer:

[tex]{ \sf{ = - (\frac{y}{x}) {}^{ - 1} }}[/tex]

Answer:

I can't understand what you have written and what to find

Answer:

96

Step-by-step explanation:

The mean number of 5 nights is 90. It means that the quantity of guests who came these nights is 90*5=450

The mean number of the six nights is 91. So the quantity of guests is 6*91=546

546-450= 96- the quantity of guests that came on the extra night