Chi needs to simplify the expression below. (1.25 -0.4)-7+4x3 Which operation should she perform first?

I need an answer quickly ​

Answer:

3rd option, 2√10

Step-by-step explanation:

x²=4×10

or, x=√(4×10)

or, x=2√10

Answer: C, 2√10

Step-by-step explanation:

Answer:

1(6+13mn+7m^2n^2)

Step-by-step explanation:

The coefficients and variables in this equation have a common factor of 1, therefore the equation will go back to it's original state if you use 1 as the common factor.

I hope this helps and sorry if am wrong

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Explanation:

We can prove this by contradiction.

Let's say

and

A+B = C

We'll show that a contradiction happens based on this.

If A is rational, then A = p/q where p,q are two integers. The q cannot be zero.

If C is rational, then C = r/s for some other integers. We can't have s be zero.

Note the following

A+B = C

B = C - A

B = r/s - p/q

B = qr/qs - ps/qs

B = (qr - ps)/qs

B = (some integer)/(some other integer)

This shows B is rational. But this is where the contradiction happens: We stated earlier that B was irrational. A number cannot be both rational and irrational at the same time. The very definition "irrational" literally means "not rational".

In short, I've shown that if A+B = C such that A,C are rational, then B must be rational as well.

The template is

rational + rational = rational

Therefore, we've shown that if A is rational and B is irrational, then C cannot possibly be rational. C is irrational.

Another template is

rational + irrational = irrational

Answer:

A) (√3-√2)/(√3+√2)

Step-by-step explanation:

i think so

The following are the statement justifications and explanatory reasons ;

(a) AD + DB = AB; Is justified by segment addition postulate

Reason

Segment addition postulate states that let A and B represent two points,

Then a third D can be on the line joining A and B, if and only if the

relationship between the points is according to the equation, AD + DB = AB

only

(b) m∠1 + m∠2 = m∠CDB; Is justified by angle addition postulate

Reason

Angle addition postulate states that given that a point B lies between an angle m∠AOC, then we have;

m∠AOC = m∠AOB + m∠BOC

(c) ∠2 ≅ ∠6; Is justified by vertically opposite angles theorem

Reason

Vertical angles formed by the crossing of two lines are always equal

(d) If D is the midpoint of [tex]\overline{AB}[/tex], then AD = [tex]\dfrac{1}{2}[/tex]×AB; Definition of midpoint

Reason

The midpoint of a line is the point that is of equal distance from both ends of the of the line. It is the point halfway between the endpoints

(e) If [tex]\underset{DF}{\rightarrow}[/tex] bisects ∠CDB then ∠1 ≅ ∠2; Definition of angle bisector

Reason

An angle bisector is a ray or line that divides an angle into two angles that are congruent

(f) m∠ADF + m∠FDB = 180°; Linear pair postulate

Reason

The linear pair postulate states that where we have two angles that form a linear pear (their addition forms a straight line), then the two angles are supplementary  (their sum is 180°)

(g) ∠ADF and ∠4 are supplements, then m∠ADF + m∠4 = 180°; Definition of supplementary angle

Reason

Supplementary angle are defined as angles that when added together, have a sum of 180°

(h) If ∠4 is complementary to ∠5, and ∠6 is complementary to ∠5, then ∠4 ≅ ∠6; Transitive property

Reason

Transitive property states that given three real numbers, a, b, and c, if a = b and c = b, then a = c

Learn more about angle properties postulates, and theorems here;

https://brainly.com/question/14437065

Answer:

So the sqrt(46) is between 6 and 7

Step-by-step explanation:

sqrt(46)

5*5 = 25

6*6=36

7*7=49

So the sqrt(46) is between 6 and 7

Answer:

? ≈ 37°

Step-by-step explanation:

Using the cosine ratio in the right triangle

cos? = [tex]\frac{adjacent}{hypotenuse}[/tex] = [tex]\frac{4}{5}[/tex] , then

? = [tex]cos^{-1}[/tex] ([tex]\frac{4}{5}[/tex] ) ≈ 37° ( to the nearest degree )

LOL

Answer:

D. The over 30s have a larger range and interquartile range than the under 30s

Step-by-step explanation:

In a data set, the range is the difference between the maximum and minimum. So, the range for under 30s is 20, while the range for over 30s is 24. Additionally, the interquartile range is the difference between Q3 and Q1. For a boxplot, Q3 is the line where the box ends and Q1 is the line where the box begins. Therefore, the IQR for the under 30s is 8, and the IQR for over 30s is 11. So, D must be correct.

Hi there!  

»»————-　★　————-««

I believe your answer is:  

[tex]\frac{1}{6}[/tex]

»»————-　★　————-««  

Here’s why:  

⸻⸻⸻⸻

[tex]\boxed{\text{Box Probability}}\\\\\rightarrow \frac{\text{# of boxed flawed}}{\text{# of boxes checked}} \\\\\rightarrow \frac{3}{18}\\\\\rightarrow \frac{3/3}{18/3}\\\\\rightarrow\boxed{\frac{1}{6}}[/tex]

⸻⸻⸻⸻

»»————-　★　————-««  

Hope this helps you. I apologize if it’s incorrect.  

Answer:

(-3)

Step-by-step explanation:

follow me if you want

Answer:

Step-by-step explanation:

[tex]\frac{sinxcos^3x-cos xsin^3x}{cos^42x-sin^42x} \\=\frac{sin x cos x(cos^2x-sin ^2 x)}{(cis^2 2x+sin^2 2x)(cos^2 2x-sin ^22x)} \\=\frac{2sin x cos x cos 2x}{2(1)(cos 4x)} \\=\frac{sin 2x cos 2x}{2 cos 4x} \\=\frac{2 sin 2x cos 2x}{4 cos 4x} \\=\frac{sin 4x}{4 cos 4x} \\=\frac{1}{4} tan 4x[/tex]

Answer:

hope it help you

Step-by-step explanation:

mark me brailiest answer

Answer:

149 inches squared

Step-by-step explanation:

top rectangle: 25 * 7 = 175

second rectangle: 8 * (25 - 17) = 8^2 = 64

triangle in bottom right: 1/2 * (13 - 8) * (15 - 11) = 10

175 + 64 + 10 = 149 sq in

hopefully got this right!

Answer: x= 11√3= 19.0525 = 19.1

Step-by-step explanation:

Let the reference angle be 30

so

cos 30 = b/h

√3/2 = x/22

or, 22√3 = 2x

or. x = (22√3)/2

so, x = 11√3

Answer:

x = 19.1 cm

Step-by-step explanation:

→ Find the name of the side you are not given

Opposite

→ Find a formula without opposite in it

Cos = Adjacent ÷ Hypotenuse

→ Rearrange to make adjacent the subject

Adjacent = Cos × Hypotenuse

→ Substitute in the values

Adjacent = Cos ( 30 ) × 22

→ Simplify

Adjacent = 19.1

Basically count/add up the total amount of degrees that are include in the angle <FHD.

-- (central angles)

So, 35 + 65 = 100 degrees

Answer:

216

Step-by-step explanation:

Volume of a rectangular prism = area of base * depth

Area of base: 72

Depth: 3

Volume = 72 * 3 = 216

Answer:

252

Step-by-step explanation:

To be divisible by 3, it's digits have to add to a number that is a multiple of 3.

To be divisible by 4 its last 2 digits have to be divisible by 3.

So let's start with 1x1 which won't work because 1x1 is odd. so let's go to 2x2 and see what happens.

212 that's divisible by 4 but not 3

222 divisible by 3 but not 4

232 divisible by 4 but not 3

242 not divisible by either one.

252 I think this might be your answer

The digits add up to 9 which is a multiple of 3 and the last 2 digits are divisible by 4

Answer:

range

Step-by-step explanation:

Answer:

B. range.

Step-by-step explanation:

others are:

» Standard variation.

» Interquatile range.

» Quatiles, deciles and percentiles.

» variance.

[tex]{ \underline{ \blue{ \sf{christ \: † \: alone}}}}[/tex]

Answer:

8n-2 = 6n+9

2n-2 = 9

2n = 11

n = 5.5

So C is correct

Let me know if this helps!

Answer:

y = 10

Step-by-step explanation:

To find the y coordinate of the midpoint, take the y coordinates of the endpoints and average

(0+y)/2 = 5

Multiply each  die by 2

0+y = 10

y = 10

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:

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:

divide, 2x+9

Step-by-step explanation:

got it right

Answer:

9 years older

Step-by-step explanation:

The ratio of their ages is 8 : 5 = 8x : 5x ( x is a multiplier )

In 30 years their ages will be 8x + 30 and 5x + 30 and the ratio 6 : 5 , so

[tex]\frac{8x+30}{5x+30}[/tex] = [tex]\frac{6}{5}[/tex] ( cross- multiply )

5(8x + 30) = 6(5x + 30) ← distribute parenthesis on both sides

40x + 150 = 30x + 180 ( subtract 30x from both sides )

10x + 150 = 180 ( subtract 150 from both sides )

10x = 30 ( divide both sides by 10 )

x = 3

Then

Michell is 8x = 8 × 3 = 24 years old

Connor is 5x = 5 × 3 = 15 years old

Mitchell is 24 - 15 = 9 years older than Connor

9514 1404 393

Answer:

  equals can be substituted anytime anywhere

Step-by-step explanation:

cos(60°) = 1/2, so wherever one appears, the other can be substituted. This is allowed by the substitution property of equality.

__

If you don't substitute at some point, you find the answer to be ...

  x = 10/cos(60°)

Most of us are interested in a numerical value for x, so we prefer that cos(60°) be replaced by a numerical value.

Answer:

7t + 6 + 9v

Step-by-step explanation:

7t + 6 + 3v + 6v (since 3v and 6v are like terms you will add them both.)

7t + 6 + 9v

Hope this helps, thank you :) !!

Answer:

7t+6+9v

Step-by-step explanation:

7t+6+3v+6v

7t has no opponent it is =7t

6 is on it own =6

3v+6v=9v,reason is 3v has an opponent which is 6v so addition of 3v and 6v is =9v

so ur ans. is =7t+6+9v

a) 26 x (-48) + (-48) x (-36) = ( –1248) + ( + 1728) = – 1248+ 1728 = 480

b) 625 x (-35) + (-625) x 65 = ( –21875) + ( –40625) = – 21875 –40625 = –62500

I hope I helped you^_^