What is the radius of a circle if it’s center is (1,2) and one point on the circle is (1,6)?

Answer:

Square root 75

Step-by-step explanation:

You can just enter it in the calculator to see which is equal to 5 square root 3

The solution to the question is:

c is 6 = [tex]\sqrt{a^{2} + b^{2} -2abcosC }[/tex]

b is 5 = [tex]\sqrt{a^{2} + c^{2} -2accosB }[/tex]

cosB is 2 = [tex]\frac{a^{2} + c^{2} - b^{2} }{2ac}[/tex]

a is 4 = [tex]\sqrt{b^{2} + c^{2} -2bccosA }[/tex]

cosA is 3 = [tex]\frac{b^{2} + c^{2} -a^{2} }{2bc}[/tex]

cosC is 1 = [tex]\frac{b^{2} + a^{2} - c^{2} }{2ab}[/tex]

it is used to relate the three sides of a triangle with the angle facing one of its sides.

The square of the side facing the included angle is equal to the some of the squares of the other sides and the product of twice the other two sides and the cosine of the included angle.

Analysis:

If c is the side facing the included angle C, then

[tex]c^{2}[/tex] = [tex]a^{2}[/tex] + [tex]b^{2}[/tex] -2ab cos C-----------------1

then c =  [tex]\sqrt{a^{2} + b^{2} -2abcosC }[/tex]

if b is the side facing the included angle B, then

[tex]b^{2}[/tex] = [tex]a^{2}[/tex] + [tex]c^{2}[/tex] -2accosB-----------------2

b =  [tex]\sqrt{a^{2} + c^{2} -2accosB }[/tex]

from equation 2, make cosB the subject of equation

2ac cosB =  [tex]a^{2}[/tex] +  [tex]c^{2}[/tex] - [tex]b^{2}[/tex]

cosB =  [tex]\frac{a^{2} + c^{2} - b^{2} }{2ac}[/tex]

if a is the side facing the included angle A, then

[tex]a^{2}[/tex] = [tex]b^{2}[/tex] + [tex]c^{2}[/tex] -2bccosA--------------------3

a =  [tex]\sqrt{b^{2} + c^{2} -2bccosA }[/tex]

from equation 3, making cosA subject of the equation

2bcosA =  [tex]b^{2}[/tex] +  [tex]c^{2}[/tex]  - [tex]a^{2}[/tex]

cosA =  [tex]\frac{b^{2} + c^{2} -a^{2} }{2bc}[/tex]

from equation 1, making cos C the subject

2abcosC =  [tex]b^{2}[/tex] + [tex]a^{2}[/tex] -  [tex]c^{2}[/tex]

cos C =  [tex]\frac{b^{2} + a^{2} - c^{2} }{2ab}[/tex]

In conclusion,

c is 6 = [tex]\sqrt{a^{2} + b^{2} -2abcosC }[/tex]

b is 5 = [tex]\sqrt{a^{2} + c^{2} -2accosB }[/tex]

cosB is 2 = [tex]\frac{a^{2} + c^{2} - b^{2} }{2ac}[/tex]

a is 4 = [tex]\sqrt{b^{2} + c^{2} -2bccosA }[/tex]

cosA is 3 = [tex]\frac{b^{2} + c^{2} -a^{2} }{2bc}[/tex]

cosC is 1 = [tex]\frac{b^{2} + a^{2} - c^{2} }{2ab}[/tex]

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The water consumption based on the information for three Months will be 22267m³.

The total water consumption for the three months will be:

= 7219 + 7390 + 7658

= 22267m³

In conclusion, the information simply want you to add the values of the water given.

Therefore, the water consumption based on the information for three Months will be 22267m³.

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Answer: OD, 1/5

Step-by-step explanation:

Well what I did was take DE and seen how many time it could fit into BC. BC would take up a total of 5 DE's. So since we already have one which is DE then we would have 1/5.

HOPE THIS HELPS! ^_^

[tex]\\ \rm\Rrightarrow \dfrac{IJ}{IL}=\dfrac{MN}{MP}[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{24}{12}=\dfrac{MN}{3.8}[/tex]

[tex]\\ \rm\Rrightarrow 2=\dfrac{MN}{3.8}[/tex]

[tex]\\ \rm\Rrightarrow MN=3.8(2)[/tex]

[tex]\\ \rm\Rrightarrow MN=7.6[/tex]

Answer:

Yes they are.

Step-by-step explanation:

By simply moving 2x from the LHS to the RHS we'll get it's equivalent to be 3x.

i.e 5x - 2x = 3x

Answer:

H = 5

Step-by-step explanation:

rectangular prism:

V = L x W x H

400 cm3 = 16 x 5 x H

400 cm3 = 80 x H

400/80 = 5

h = 5 cm

95% sure this is correct! :)

Answer:

5 cm

Step-by-step explanation:

Formula used :

Volume of the rectangular prism = Length × Width × Height

=========================================================

Given :

⇒ Volume = 400 cm³

⇒ Length = 16 cm

⇒ Width = 5 cm

===========================================================

Solving :

⇒ 400 cm³ = 16 cm × 5 cm × Height

⇒ Height = 400 cm³ / 80 cm²

⇒ Height = 5 cm

[tex]\huge\huge\bold{\green a \blue n \pink s \purple w \orange e \red r}[/tex]

volume of rectangular prism

Formula to calculate:

[tex]\large\bold{\green V\green = \green l \green * \green w \green * \green h}[/tex]

[tex]\large\bold{V= 7.5*4*2}[/tex]

[tex]\huge\boxed{\red V\red = \red 6 \red 0 \: {\red i \red n}^{\red 3}}[/tex]

length= 7.5 in

width= 2 in

height= 4 in

The volume of the rectangular prism or cuboid.

Formula: volume= length × width × height

so, V= 7.5 × 2 × 4

[tex]\bold{V= 60 \: {\: in}^{3}}[/tex]

hope this helps!

The binomial expression (p + q ) n (p+q)n to calculate the binomial distribution given will give a value of 10.

From the information given, the expression (p + q ) n (p+q)n is given to calculate a binomial distribution with n = 5 and p = 0.3.

p = 0.3

q = 1 - 0.3 = 0.7

n = 5

Therefore, expression ( p + q ) n (p+q)n will b:

= (0.3 + 0.7)× 5 + (0.3 + 0.7)× 5

= 5 + 5

= 10

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Answer:

1

Step-by-step explanation:

216 is term 1
36 is term 2
6 is term 3
1 is term 4

Total electrons

Answer:

Given:

To calculate how many electrons have the same mass as one dust particle, divide the mass of a dust particle by the mass of an electron:

[tex]\begin{aligned}\implies \dfrac{\textsf{Mass of a dust particle}}{\textsf{Mass of an electron}} & = \sf \dfrac{7.5 \times 10^{-10}}{9.1 \times 10^{-31}}\\\\& = \sf \dfrac{7.5}{9.1} \times \dfrac{10^{-10}}{10^{-31}}\\\\& = \sf 0.8241... \times 10^{(-10-(-31))}\\\\& = \sf 0.8241... \times 10^{21}\\\\& = \sf 8.2 \times 10^{20}\end{aligned}[/tex]

Therefore, approximately [tex]\sf 8.2 \times 10^{20}[/tex] electrons have the same mass as one dust particle.

Hi student, let me help you out! :)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

We are asked to find the G.C.F. of 8x-6.

[tex]\triangle~\fbox{\bf{KEY:}}[/tex]

So what is the G.C.F. of 8x-6? We can list all of their common factors and then select the greatest one, like so:

[tex]\longmapsto\bf{Factors\;of\;8x:}[/tex] 1, 2, 4, 8, x

[tex]\longmapsto\bf{Factors\;of\;-6:}[/tex] -1, 1, 2, -3, 3, -6, 6

So the GCF is:

2

Now, we can also factor it out by placing it outside the parentheses ():

2(4x-3)

See, we factored it out by dividing both terms of the expression by the G.C.F.

[tex]\ddot\bigstar[/tex] Remember this...

[tex]\fbox{\bf{We\;factor\;out\;the\;G.C.F.\;by\;dividing\;all\;the\;terms\;of\;the\;expression\;by\;it}}[/tex]

Hope this helps you out! :D

Ask in comments if any queries arise.

~Just a smiley person helping fellow students :)

Answer: 2, you have to use math apps man it will save your life

Answer:

x = 3
x = -6

Step-by-step explanation:

Hello!

Let's first set the equation to 0.

To factor, first think about two numbers that multiply to -18, but add up to 3. The two numbers are 6 and -3.

Now, expand 3x to 6x and -3x. Then, factor by grouping.

Now, set each factor to 0 and solve for x

x = 3, x = -6

Answer:

[tex]12\dfrac 12\%[/tex]

Step by step explanation:

[tex]\text{Let it be}~ x\%,\\\\~~~~~72 \cdot x \% = 9\\\\\\\implies 72\cdot \dfrac{x}{100} = 9\\\\\\\implies x = \dfrac{9 \times 100}{72}\\\\\\\implies x = \dfrac{100}{8}\\\\\\\implies x = \dfrac{25}2\\\\\\\implies x = 12 \dfrac 12\\\\\\\text{Hence 9 is}~ 12\dfrac 12\% ~ \text{of}~ 72[/tex]

The length of the segment HI in the figure is 32.9

To do this, we make use of the following secant-tangent equation:

HI² = KI * JI

From the figure, we have:

KI = 21 + 24 = 45

JI = 24

So, we have:

HI² = 45 * 24

Evaluate the product

HI² = 1080

Take the square root of both sides

HI = 32.9

Hence, the length of the segment HI is 32.9

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Answer:

7.5

Step-by-step explanation:

10 is to 8 as    x is to 6

10/8  = x/6

6 (10/8) = x = 7.5

Answer:

Step-by-step explanation:

The two triangles are similar because the corresponding angle in each is equal to the angle corresponding angles in the other.

Put in simpler language. ΔAEB ~ ΔADC by parallel lines properties. If that's the case 10/(10 + 8) = x/(x + 6)

Equation

AE/AD = AB/AC                  Substitute values

Solution

10/(10 + 8) = x/(x + 6)           Cross Multiply

10*(x + 6) = x* (10 + 8)          Combine

10*(x + 6) = 18x                     Remove the Brackets

10x + 60 = 18x                      Subtract 10x from both sides

8x = 60                                 Divide by 8.

x = 60/8

x = 7.5

Answer: A

Answer:   18.7346314730578

This value is approximate.

=================================================

Explanation:

Use the law of sines to find side b

sin(A)/a = sin(B)/b

sin(112)/37 = sin(28)/b

b*sin(112) = 37*sin(28)

b = 37*sin(28)/sin(112)

b = 18.7346314730578

Round that value however you need to.

The portion on your screenshot says "round to the nearest", but it doesn't say to the nearest what. Is it nearest tenth? Hundredth? Something else? That part is cut off.  So I'll just write down all of the decimal digits my calculator is showing, and let you do the final rounding step.

The total perimeter of the chicken coop if the length of one side is 3√5 yards is; 15√5 yards.yards

We are told that Isabelle had a five sided chicken coop in her yard and that the length was equal in all the sides.

Now, we need to find the total perimeter of the chicken coop.

The length of one side of the coop from online sources is 3√5 yards.

Since the coop is 5 sided, it means that it is a pentagon and we know that;

Perimeter of a pentagon = 5a

Where a is length of one side of the pentagon. Thus;

Perimeter = 5 x 3√5

Perimeter = 15 √5 yards

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Using proportions, it is found that the number of books that was left over was of 505.

A proportion is a fraction of a total amount, and the measures are related using a rule of three.

The number of books received by Marie is given by:

M = 772 - 345 = 427.

Melanie received one-sixth of Marie, hence:

Me = 427/6 = 71.

The total amount is 25 times Melanie's amount, hence:

T = 25 x 71 = 1775.

The amount left over is:

L = 1775 - (772 + 427 + 71) = 505

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The ratio of nonfiction books to fiction books checked out is 1 : 3

The given parameters are:

Total = 20

Fictions = 15

This means that:

Non Fictions = 20 - 15

Evaluate

Non Fictions = 5

The ratio is then represented as:

Ratio = Non Fiction : Fiction

This gives

Ratio = 5 : 15

Simplify

Ratio = 1 : 3

Hence, the ratio of nonfiction books to fiction books checked out is 1 : 3

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Answer:

n¹⁵

Key:

Step-by-step explanation:

[tex]\frac{n^{10}}{n^{-5}}\\=n^{10-\left(-5\right)}\\n^{10} = n10\\^{-5} = -5\\n10 - -5 = n15\\= n^{15}[/tex]

The value of the composite function (p + q)(2) is 1

The function expression is given as:

(p + q)(2)

As a general rule

(p + q)(x) = p(x) + q(x)

Substitute 2 for x

(p + q)(2) = p(2) + q(2)

From the table, we have:

p(2) = 3 and q(2) =-2

So, we have:

(p + q)(2) = 3 - 2

Evaluate

(p + q)(2) = 1

Hence, the value of the composite function (p + q)(2) is 1

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Answer: (p+q)(2)=1

Step-by-step explanation:

Edge

Answer:

-3

Step-by-step explanation:

g(-3) = -2x+3x

-2(-3) +3(-3)

6+-9 = -3

give brainliest please

hope this helps :)

Answer:

just commenting so you can give the other guy brainliest !

Step-by-step explanation:

Answer:

Hi,

Step-by-step explanation:

[tex](fog)(x)=g(f(x))=g(3x+2)=(3x+2)^2+1=9x^2+6x+5[/tex]

Answer:

hi

Step-by-step explanation:

Answer:

8.0

Step-by-step explanation:

The triangle is a right triangle, so we are able to use trigonometric functions. Relative to the angle of 37°, we have the opposite side which is 6 and the adjacent side which is x. The trig function that uses the opposite and adjacent is tangent(SohCahToa). We can set up the following equation:

[tex]tan(37) = \frac{opposite}{adjacent}[/tex]

[tex]tan(37) = \frac{6}{x}[/tex]

tan(37) evaluates to about , so we can plug it in and solve for x:

[tex]0.7536 = \frac{6}{x} \\0.7536x = 6\\x = 7.962[/tex]

To the nearest tenth, x rounds to 8.0.

The largest interval centered about x = 0 is ([tex]-\frac{\pi}{2}, \frac{\pi}{2}[/tex]) for which the given initial-value problem has a unique solution: [tex]y'' + tan(x)y = e^x[/tex], y(0) = 1, y'(0) = 0.

An initial-value problem (IVP) is one kind of mathematical problem that contains obtaining a solution to a differential equation along with a set of initial conditions. It is commonly encountered in the field of differential equations.

Since it is mentioned in the theorem that in an initial-value problem y' + p(t)y = g(t), y([tex]t_o[/tex]) = [tex]y_o[/tex] has a unique solution if p(t) and g(t) are continuous functions on the interval (a, b) containing [tex]t_o[/tex].

Since [tex]e^x[/tex] is continuous everywhere and tanx is continuous on ([tex]-\frac{\pi}{2}, \frac{\pi}{2}[/tex]) containing 0.

Since R, a set of real numbers, and ([tex]-\frac{\pi}{2}, \frac{\pi}{2}[/tex]), both contain 0.

Thus the largest interval centered about x = 0 is ([tex]-\frac{\pi}{2}, \frac{\pi}{2}[/tex]).

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Replace x with -2 and solve:

A) -2 + -2 = -4 which is not less than or equal to -8

B) 2 + -2 = 0, this is greater than or equal to -8

C) -2 + -2 = -4, this is not less than -8

D) 2 + -2 = 0, this is not greater than 8

Answer: B is true