The centre circle has a diameter of 3 cm and the difference in the radii of each larger circle is 4 cm. What is the circumference of the yellow circle? Use π=3.14

Answer:

AB= 5.582

Step-by-step explanation:

Centeral angle /360° = AB length/2 pi r

[tex] \frac{80}{360} = \frac{ab}{4} \\ ab = 5.582[/tex]

Answer:

5.6

Step-by-step explanation:

the length of arc AB =

80/360 × 2× 3.14×4

= 2/9 × 3.14 × 8

= 5.58 => 5.6

Answer:

according to me the ans is 3.

Answer:

Mean=2.53

median=2

mode=2

range=3

Step-by-step explanation:

1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 4

MEAN

Add up all data values to get the sum

Count the number of values in your data set

Divide the sum by the count

38/15=2.53

MEDIAN

Arrange data values from lowest to the highest value

The median is the data value in the middle of the set

If there are 2 data values in the middle the median is the mean of those 2 values.

MODE

Mode is the value or values in the data set that occur most frequently.

RANGE

18-15=3

Answer:

12√3

Step-by-step explanation:

sin 60° = 18/h

h = 18/sin 60°

h = 12√3

Answer:

14.0625 = [tex]\frac{225}{16}[/tex]

6.41666666666... = [tex]\frac{77}{12}[/tex]

Hope that this helps!

Answer:

1,500

Step-by-step explanation:

a + 5d = 39 (1)

a + 18d = 78 (2)

Subtract (1) from (2) to eliminate a

18d - 5d = 78 - 39

13d = 39

d = 39/13

d = 3

Substitute d = 3 into (1)

a + 5d = 39 (1)

a + 5(3) = 39

a + 15 = 39

a = 39 - 15

a = 24

Sum of the first 25 terms

Sn = n/2[2a + (n – 1)d]

S25 = 25/2{2*24 + (25-1)3}

= 12.5{48 + (24)3}

= 12.5{48 + 72)

= 600 + 900

= 1,500

S25 = 1,500

Answer:

Mean: 2.125

median: 2

range: 4

Answer:

Step-by-step explanation:

[tex] \frac{4f^{2} }{3} \div \frac{1}{4f} [/tex]

[tex] \frac{4 {f}^{2}}{3} (4f)[/tex]

[tex]4 \frac{4 {f}^{2} }{3} f[/tex]

[tex] \frac{16 {f}^{2} }{3} f[/tex]

[tex] \frac{16 {f}^{3} }{3} [/tex]

Answer:

96

Step-by-step explanation:

Total of 4 test at 90

90 * 4 = 360

Current total

88 * 3 = 264

Score needed

360 - 264 = 96

Answer:

96

Step-by-step explanation:

this is how i solved it:

88 x 3 = 264 ( the sum of the three test score )

now i just gotta look for a number to add to 264 that will give me 90 (the wanted mean score) if i divide the sum by 4 (the four test scores).

so the equation would be:

(264 + x) / 4 = 90

264 + x = 360

x= 96

Answer:

yeah what you need help with

Answer:

The answer to the problem is 15.

Answer:

Step-by-step explanation: hope u like that cus i sike that C?

Answer:

B looks like a 180 degree angle cause its js a straight line and isnt curved or bent. Brainliest plz?

Step-by-step explanation:

Answer:

2, probably

Step-by-step explanation:

Answer:

180 D

Step-by-step explanation:

Answer:

a. (-4,8)

Step-by-step explanation:

the two lines intersect at this point

Answer:

10x+ y

Step-by-step explanation:

The unit's digit is y and the ten's digit is x.

The ten's digit has a zero placed beside it .

So multiply x by 10  giving 10 x and then add the unit's digit .

This wil give 10x+ y

The number is 10 x + y

This can be elaborated through the use of numbers . Suppose we have unit's digit as 6 and the ten's digit as 5.

Multiply 10 by 5 and add 6

5*10 +6= 50+6= 56

Step-by-step explanation:

which class are you

just to confirm

Answer:

slope=y2-y1/x2-x1

=5-5/-1-3

=0/-4

=0

Step-by-step explanation:

Answer:

slope = 0

Step-by-step explanation:

Calculate the slope m using the slope formula

m = [tex]\frac{y_{2}-y_{1} }{x_{2}-x_{1} }[/tex]

with (x₁, y₁ ) = (3, 5) and (x₂, y₂ ) = (- 1, 5)

m = [tex]\frac{5-5}{-1-3}[/tex] = [tex]\frac{0}{-4}[/tex] = 0

Answer:

8

Step-by-step explanation:

6g = 48

/6 /6

divide 6 by both sides

g = 8

hope this helped!

Answer:

360 cubic inches (remember your units!)

Step-by-step explanation:

the formula for volume of a rectangular prism is length times width times height times depth so you have to do 8 times 5 times 9 witch is 360 cubic inches

Answer:

[tex]\displaystyle \frac{dy}{dx} = \frac{-cos(x) - ysec(xy)tan(xy)}{-sin(y) + xsec(xy)tan(xy)}[/tex]

General Formulas and Concepts:

Pre-Algebra

Distributive Property

Algebra I

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

Trig Differentiation

Derivative Rule [Chain Rule]:                                                                                       [tex]\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)[/tex]

Implicit Differentiation

Step-by-step explanation:

Step 1: Define

Identify

sin(x) + cos(y) + sec(xy) = 251

Step 2: Differentiate

Topic: AP Calculus AB/BC (Calculus I/II)

Unit: Implicit Differentiation

Book: College Calculus 10e

Step-by-step explanation:

x = 7 cos 210 = 7×(-½√3) = -3.5√3

y = 7 sin 210 = 7×(-½) = -3.5

point C (-3.5√3 , -3.5)

Answer:

x = 4.4

Step-by-step explanation:

Flat cost = $57.5/month

Cost of 1GB = $4

But Aubrey wants to keep her bill at $75.1/month.

Let 'x' be the number of GBs she can use while staying within her budget.

So, the equation will be → 4x + 57.5 = 75.1

Now, solve the equation :-

[tex]=> 4x + 57.5 - 57.5 = 75.1 - 57.5[/tex]

[tex]=> 4x = 17.6[/tex]

[tex]=> \frac{4x}{4} = \frac{17.6}{4}[/tex]

[tex]=> x = 4.4[/tex]

Answer:

48°

Step-by-step explanation:

The angle CRS looks like a "L" shape, meaning that both lines are perpindicular to each other, resulting in a right angle (which is 90°)

90° + 42° = 132°

180° - 132° = 48°

Answer:

<RCS = 48 degrees

Step-by-step explanation:

I'm pretty sure that is a right triangle

180-90-42=48

Answer:

n geometry, the notion of a connection makes precise the idea of transporting data[further explanation needed] along a curve or family of curves in a parallel and consistent manner. There are various kinds of connections in modern geometry, depending on what sort of data one wants to transport. For instance, an affine connection, the most elementary type of connection, gives a means for parallel transport of tangent vectors on a manifold from one point to another along a curve. An affine connection is typically given in the form of a covariant derivative, which gives a means for taking directional derivatives of vector fields, measuring the deviation of a vector field from being parallel in a given direction.

Connections are of central importance in modern geometry in large part because they allow a comparison between the local geometry at one point and the local geometry at another point. Differential geometry embraces several variations on the connection theme, which fall into two major groups: the infinitesimal and the local theory. The local theory concerns itself primarily with notions of parallel transport and holonomy. The infinitesimal theory concerns itself with the differentiation of geometric data. Thus a covariant derivative is a way of specifying a derivative of a vector field along another vector field on a manifold. A Cartan connection is a way of formulating some aspects of connection theory using differential forms and Lie groups. An Ehresmann connection is a connection in a fibre bundle or a principal bundle by specifying the allowed directions of motion of the field. A Koszul connection is a connection which defines directional derivative for sections of a vector bundle more general than the tangent bundle.

Connections also lead to convenient formulations of geometric invariants, such as the curvature (see also curvature tensor and curvature form), and torsion tensor.

Step-by-step explanation:

Answer:

n geometry, the notion of a connection makes precise the idea of transporting data[further explanation needed] along a curve or family of curves in a parallel and consistent manner. There are various kinds of connections in modern geometry, depending on what sort of data one wants to transport. For instance, an affine connection, the most elementary type of connection, gives a means for parallel transport of tangent vectors on a manifold from one point to another along a curve. An affine connection is typically given in the form of a covariant derivative, which gives a means for taking directional derivatives of vector fields, measuring the deviation of a vector field from being parallel in a given direction.

Connections are of central importance in modern geometry in large part because they allow a comparison between the local geometry at one point and the local geometry at another point. Differential geometry embraces several variations on the connection theme, which fall into two major groups: the infinitesimal and the local theory. The local theory concerns itself primarily with notions of parallel transport and holonomy. The infinitesimal theory concerns itself with the differentiation of geometric data. Thus a covariant derivative is a way of specifying a derivative of a vector field along another vector field on a manifold. A Cartan connection is a way of formulating some aspects of connection theory using differential forms and Lie groups. An Ehresmann connection is a connection in a fibre bundle or a principal bundle by specifying the allowed directions of motion of the field. A Koszul connection is a connection which defines directional derivative for sections of a vector bundle more general than the tangent bundle.

Connections also lead to convenient formulations of geometric invariants, such as the curvature (see also curvature tensor and curvature form), and torsion tensor.

Step-by-step explanation:

Answer:

[tex]-2 -\sqrt{3}[/tex]

Step-by-step explanation:

First consider numerator

[tex]sin \frac{\frac{7\pi}{6}}{2} = sin \frac{7\pi}{12}= sin (\frac{\pi}{4} + \frac{\pi}{3})[/tex]

Using the formula : sin (A + B) = sin A cos B + cos A sin B

[tex]sin \frac{\pi}{4} = \frac{\sqrt{2} }{2}, \ cos \frac{\pi}{4} = \frac{\sqrt{2} }{2}\\\\sin \frac{\pi}{3} = \frac{\sqrt{2} }{2}, \ cos \frac{\pi}{3} = \frac{1}{2}[/tex]

[tex]sin(\frac{\pi}{4} + \frac{\pi}{3}) = sin \frac{\pi}{4} \cdot cos \frac{\pi}{3} + cos \frac{\pi}{4} \cdot sin \frac{\pi}{3}[/tex]

                [tex]=\frac{\sqrt{2} }{2} \cdot \frac{1}{2} + \frac{\sqrt{2} }{2} \cdot \frac{\sqrt{3} }{2} \\\\= \frac{\sqrt{2} }{4} + \frac{\sqrt{6} }{4}\\\\=\frac{\sqrt{2} +\sqrt{6} }{4}[/tex]

Second consider denominator

[tex]cos \frac{\frac{7\pi}{6}}{2} = cos \frac{7\pi}{12}= cos (\frac{\pi}{4} + \frac{\pi}{3})[/tex]

Using the formula : cos (A + B) = cos A cos B - sin A sin B

[tex]sin \frac{\pi}{4} = \frac{\sqrt{2} }{2}, \ cos \frac{\pi}{4} = \frac{\sqrt{2} }{2}\\\\sin \frac{\pi}{3} = \frac{\sqrt{2} }{2}, \ cos \frac{\pi}{3} = \frac{1}{2}[/tex]

[tex]cos(\frac{\pi}{4} + \frac{\pi}{3}) = cos \frac{\pi}{4} \cdot cos \frac{\pi}{3} -sin \frac{\pi}{4} \cdot sin \frac{\pi}{3}[/tex]

[tex]=\frac{\sqrt{2} }{2} \cdot \frac{1}{2} - \frac{\sqrt{2} }{2} \cdot \frac{\sqrt{3} }{2}\\\\=\frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4}\\\\= \frac{\sqrt{2} -\sqrt{6} }{4}[/tex]

Therefore,

           [tex]tan \frac{7\pi}{12} = \frac{sin \frac{7\pi}{12}}{cos\frac{7\pi}{12}}[/tex]

                     [tex]= \frac{\frac{\sqrt{2} +\sqrt{6} }{4}} {\frac{\sqrt{2} -\sqrt{6} }{4} }\\\\=\frac{\sqrt{2} +\sqrt{6} }{4} \times \frac{4 }{\sqrt{2} -\sqrt{6}}\\\\=\frac{\sqrt{2} +\sqrt{6} }{\sqrt{2} -\sqrt{6}}[/tex]

Either we can stop here or Rationalize the denominator:

[tex]\frac{\sqrt{2} +\sqrt{6} }{\sqrt{2} -\sqrt{6}} \times \frac{\sqrt{2} +\sqrt{6} }{\sqrt{2} +\sqrt{6}} = \frac{(\sqrt{2} +\sqrt{6})^{2} }{(\sqrt{2})^2 -(\sqrt{6})^2} = \frac{2 + 6 +2\sqrt{12} }{2-6} = \frac{8+2\sqrt{12} }{-4} = \frac{8+ 4\sqrt{3} }{-4} = -2-\sqrt{3}[/tex]

Answer:

number 3 sir

Step-by-step explanation: