James and his family spend a morning picking peaches. They fill several bags. This line plot shows the weight of each bag. How many more pounds does one of the heaviest bags weigh than one of the lightest bags? Enter your answer as a fraction in simplest form by filling in the boxes.

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:

B

Step-by-step explanation:

option B is not similar.

the ratio of each side isn't same

Answer:

8n-2 = 6n+9

2n-2 = 9

2n = 11

n = 5.5

So C is correct

Let me know if this helps!

Answer:

305.11

Step-by-step explanation:

Just use a calculator. A centimeter is 2.5 inches. Divide 775 by that.

Answer:

Area = 5 × 8

= 40 square units

Answer:

40^2

Step by Step Solution:

I counted the difference between the length and the width, which was 5 and 8, then using the formula for area, lw=a^2, I did 5(8)=40^2.  Some people leave out the squared part of the area, but 40^2 would be the most correct option if they do not square any of the answers, just put 40 that'll probably be accepted too.

Answer:

x(ax-1)-y(y+1)

Step-by-step explanation:

you have to group the like terms

ax^2-x-y^2-y

x(ax-1)-y(y+1)

I hope this helps

Answer:

x=12

Step-by-step explanation:

LM = RS

9x+27 = 135

Subtract 27 from each side

9x+27-27 =135-27

9x=108

Divide each side by 9

9x/9 = 108/9

x = 12

the boxes are equivalent so the one with a single dash is equal to the other with a single dash.

the one with 2 dashes is equal to the other with 2 dashes so on and so forth

SR=LM

LM=9x+27

RS=135

9x+27=135

so I solve it in my own weird way but you can solve it differently. 135-27=108

108/9=12

so your answer is 12

The surface area of the rectangular pyramid is the sum of the area of its individual surface. The surface area of the pyramid is 384 square inches.

Each surface of the rectangular pyramid has the shape of a triangle.

So that the area of each surface of the pyramid = [tex]\frac{1}{2}[/tex] x base x height.

From the given question;

The area of one of the two lateral faces = [tex]\frac{1}{2}[/tex] x b x h

                                                                 = [tex]\frac{1}{2}[/tex] x 10 x 15

The area of one of the two lateral faces = 75 square inches

Thus,

the area of the first two given lateral faces = 2 x 75

                                                            = 150 square inches

The area of the first two given lateral faces = 150 square inches

Also,

The area of one of the other two lateral faces = [tex]\frac{1}{2}[/tex] x b x h

                                                                             = [tex]\frac{1}{2}[/tex] x 18 x 13

The area of one of the other two lateral faces =  117 square inches

So that;

the area of the first two other lateral faces = 2 x 117

                                                            = 234 square inches

The area of the first two other lateral faces = 234 square inches

Thus,

the surface area of the pyramid = 150 + 234

                                                              = 384 square inches

Therefore, the surface area of the rectangular pyramid is 384 square inches.

For more clarifications: https://brainly.com/question/23564399

Answer:

Step-by-step explanation:

The surface area of the rectangular pyramid is the sum of the area of its individual surface. The surface area of the pyramid is 384 square inches.

Each surface of the rectangular pyramid has the shape of a triangle.

So that the area of each surface of the pyramid =  x base x height.

From the given question;

The area of one of the two lateral faces =  x b x h

                                                                =  x 10 x 15

The area of one of the two lateral faces = 75 square inches

Thus,

the area of the first two given lateral faces = 2 x 75

                                                           = 150 square inches

The area of the first two given lateral faces = 150 square inches

Also,

The area of one of the other two lateral faces =  x b x h

                                                                            =  x 18 x 13

The area of one of the other two lateral faces =  117 square inches

So that;

the area of the first two other lateral faces = 2 x 117

                                                           = 234 square inches

The area of the first two other lateral faces = 234 square inches

The area of the rectangle is B x H = 18 x 10 = 180 square inches

Thus,

the surface area of the pyramid = 150 + 234+180

                                                             = 564 square inches

Therefore, the surface area of the rectangular pyramid is 564 square inches.

Step-by-step explanation:

-9x+15=3(2-x)

6-6x

2. collect like terms

-9x+15= 6-6x

15-6 = 6x+9x

11= 15x

3. divide both sides by the coefficient of X which is 15

x= 11/15

Answer:

x = 6.2

Step-by-step explanation:

Since this is a right triangle, we can use trig functions

tan theta = opp / adj

tan 32 = x/ 10

10 tan 32 = x

x=6.24869

Rounding to the nearest tenth

x = 6.2

Answer:

x=6.2 (Rounded to the nearest tenth)

Step-by-step explanation:

This problem gives you an angle(32°), and ask for the dimension of the opposite side to that angle(x), along with another dimension the adjacent side(10).

Since you have the opposite and adjacent sides, you can use tangent. opposite (x) over adjacent (10). Tan(32) =[tex]\frac{x}{10}[/tex]. You want (x) so multiply tan(32) by 10. Then round to the nearest tenth.

Remember to put calculator in degree mode!

tan (32) = 0.6248693519 multiply by ten 6.248693519. Round to nearest 10th 6.2.

Hope 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:

your answer is 30

Step-by-step explanation:

I hope this help

[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:

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:

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

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:

abc is a triangle so ,

a is ( 9,6 )

b is ( 9,3 )

and c is ( 3,3 )

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:

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]