HELP !!!!
Circle A has center of (4, 5), and a radius of 3 and circle B has a center of (1, 7), and a radius of 9. What steps will help show that circle A is similar to circle B? (6 points)

Group of answer choices

Translate circle A using the rule (x + 3, y − 2).

Dilate circle A by a scale factor of 3.

Rotate circle A 90° about the center.

Reflect circle A over the x-axis.

Answer:

8 = x²

Step-by-step explanation:

Logarithmic form: logₓ(8) = 2

Exponential form: 8 = x²

Answer:

[tex]A.\ \tan(x) \to 2.\ \sin(x) \sec(x)[/tex]

[tex]B.\ \cos(x) \to 5. \sec(x) - \sec(x)\sin^2(x)[/tex]

[tex]C.\ \sec(x)csc(x) \to 3. \tan(x) + \cot(x)[/tex]

[tex]D. \frac{1 - (cos(x))^2}{cos(x)} \to 1. \sin(x) \tan(x)[/tex]

[tex]E.\ 2\sec(x) \to\ 4.\ \frac{\cos(x)}{1 - \sin(x)} +\frac{1-\sin(x)}{\cos(x)}[/tex]

Step-by-step explanation:

Given

[tex]A.\ \tan(x)[/tex]

[tex]B.\ \cos(x)[/tex]

[tex]C.\ \sec(x)csc(x)[/tex]

[tex]D.\ \frac{1 - (cos(x))^2}{cos(x)}[/tex]

[tex]E.\ 2\sec(x)[/tex]

Required

Match the above with the appropriate identity from

[tex]1.\ \sin(x) \tan(x)[/tex]

[tex]2.\ \sin(x) \sec(x)[/tex]

[tex]3.\ \tan(x) + \cot(x)[/tex]

[tex]4.\ \frac{cos(x)}{1 - sin(x)} + \frac{1 - \sin(x)}{cos(x)}[/tex]

[tex]5.\ \sec(x) - \sec(x)(\sin(x))^2[/tex]

Solving (A):

[tex]A.\ \tan(x)[/tex]

In trigonometry,

[tex]\frac{sin(x)}{\cos(x)} = \tan(x)[/tex]

So, we have:

[tex]\tan(x) = \frac{\sin(x)}{\cos(x)}[/tex]

Split

[tex]\tan(x) = \sin(x) * \frac{1}{\cos(x)}[/tex]

In trigonometry

[tex]\frac{1}{\cos(x)} =sec(x)[/tex]

So, we have:

[tex]\tan(x) = \sin(x) * \sec(x)[/tex]

[tex]\tan(x) = \sin(x) \sec(x)[/tex] --- proved

Solving (b):

[tex]B.\ \cos(x)[/tex]

Multiply by [tex]\frac{\cos(x)}{\cos(x)}[/tex] --- an equivalent of 1

So, we have:

[tex]\cos(x) = \cos(x) * \frac{\cos(x)}{\cos(x)}[/tex]

[tex]\cos(x) = \frac{\cos^2(x)}{\cos(x)}[/tex]

In trigonometry:

[tex]\cos^2(x) = 1 - \sin^2(x)[/tex]

So, we have:

[tex]\cos(x) = \frac{1 - \sin^2(x)}{\cos(x)}[/tex]

Split

[tex]\cos(x) = \frac{1}{\cos(x)} - \frac{\sin^2(x)}{\cos(x)}[/tex]

Rewrite as:

[tex]\cos(x) = \frac{1}{\cos(x)} - \frac{1}{\cos(x)}*\sin^2(x)[/tex]

Express [tex]\frac{1}{\cos(x)}\ as\ \sec(x)[/tex]

[tex]\cos(x) = \sec(x) - \sec(x) * \sin^2(x)[/tex]

[tex]\cos(x) = \sec(x) - \sec(x)\sin^2(x)[/tex] --- proved

Solving (C):

[tex]C.\ \sec(x)csc(x)[/tex]

In trigonometry

[tex]\sec(x)= \frac{1}{\cos(x)}[/tex]

and

[tex]\csc(x)= \frac{1}{\sin(x)}[/tex]

So, we have:

[tex]\sec(x)csc(x) = \frac{1}{\cos(x)}*\frac{1}{\sin(x)}[/tex]

Multiply by [tex]\frac{\cos(x)}{\cos(x)}[/tex] --- an equivalent of 1

[tex]\sec(x)csc(x) = \frac{1}{\cos(x)}*\frac{1}{\sin(x)} * \frac{\cos(x)}{\cos(x)}[/tex]

[tex]\sec(x)csc(x) = \frac{1}{\cos^2(x)}*\frac{\cos(x)}{\sin(x)}[/tex]

Express [tex]\frac{1}{\cos^2(x)}\ as\ \sec^2(x)[/tex] and [tex]\frac{\cos(x)}{\sin(x)}\ as\ \frac{1}{\tan(x)}[/tex]

[tex]\sec(x)csc(x) = \sec^2(x)*\frac{1}{\tan(x)}[/tex]

[tex]\sec(x)csc(x) = \frac{\sec^2(x)}{\tan(x)}[/tex]

In trigonometry:

[tex]tan^2(x) + 1 =\sec^2(x)[/tex]

So, we have:

[tex]\sec(x)csc(x) = \frac{\tan^2(x) + 1}{\tan(x)}[/tex]

Split

[tex]\sec(x)csc(x) = \frac{\tan^2(x)}{\tan(x)} + \frac{1}{\tan(x)}[/tex]

Simplify

[tex]\sec(x)csc(x) = \tan(x) + \cot(x)[/tex]  proved

Solving (D)

[tex]D.\ \frac{1 - (cos(x))^2}{cos(x)}[/tex]

Open bracket

[tex]\frac{1 - (cos(x))^2}{cos(x)} = \frac{1 - cos^2(x)}{cos(x)}[/tex]

[tex]1 - \cos^2(x) = \sin^2(x)[/tex]

So, we have:

[tex]\frac{1 - (cos(x))^2}{cos(x)} = \frac{sin^2(x)}{cos(x)}[/tex]

Split

[tex]\frac{1 - (cos(x))^2}{cos(x)} = \sin(x) * \frac{sin(x)}{cos(x)}[/tex]

[tex]\frac{sin(x)}{\cos(x)} = \tan(x)[/tex]

So, we have:

[tex]\frac{1 - (cos(x))^2}{cos(x)} = \sin(x) * \tan(x)[/tex]

[tex]\frac{1 - (cos(x))^2}{cos(x)} = \sin(x) \tan(x)[/tex] --- proved

Solving (E):

[tex]E.\ 2\sec(x)[/tex]

In trigonometry

[tex]\sec(x)= \frac{1}{\cos(x)}[/tex]

So, we have:

[tex]2\sec(x) = 2 * \frac{1}{\cos(x)}[/tex]

[tex]2\sec(x) = \frac{2}{\cos(x)}[/tex]

Multiply by [tex]\frac{1 - \sin(x)}{1 - \sin(x)}[/tex] --- an equivalent of 1

[tex]2\sec(x) = \frac{2}{\cos(x)} * \frac{1 - \sin(x)}{1 - \sin(x)}[/tex]

[tex]2\sec(x) = \frac{2(1 - \sin(x))}{(1 - \sin(x))\cos(x)}[/tex]

Open bracket

[tex]2\sec(x) = \frac{2 - 2\sin(x)}{(1 - \sin(x))\cos(x)}[/tex]

Express 2 as 1 + 1

[tex]2\sec(x) = \frac{1+1 - 2\sin(x)}{(1 - \sin(x))\cos(x)}[/tex]

Express 1 as [tex]\sin^2(x) + \cos^2(x)[/tex]

[tex]2\sec(x) = \frac{\sin^2(x) + \cos^2(x)+1 - 2\sin(x)}{(1 - \sin(x))\cos(x)}[/tex]

Rewrite as:

[tex]2\sec(x) = \frac{\cos^2(x)+1 - 2\sin(x)+\sin^2(x)}{(1 - \sin(x))\cos(x)}[/tex]

Expand

[tex]2\sec(x) = \frac{\cos^2(x)+1 - \sin(x)- \sin(x)+\sin^2(x)}{(1 - \sin(x))\cos(x)}[/tex]

Factorize

[tex]2\sec(x) = \frac{\cos^2(x)+1(1 - \sin(x))- \sin(x)(1-\sin(x))}{(1 - \sin(x))\cos(x)}[/tex]

Factor out 1 - sin(x)

[tex]2\sec(x) = \frac{\cos^2(x)+(1- \sin(x))(1-\sin(x))}{(1 - \sin(x))\cos(x)}[/tex]

Express as squares

[tex]2\sec(x) = \frac{\cos^2(x)+(1-\sin(x))^2}{(1 - \sin(x))\cos(x)}[/tex]

Split

[tex]2\sec(x) = \frac{\cos^2(x)}{(1 - \sin(x))\cos(x)} +\frac{(1-\sin(x))^2}{(1 - \sin(x))\cos(x)}[/tex]

Cancel out like factors

[tex]2\sec(x) = \frac{\cos(x)}{1 - \sin(x)} +\frac{1-\sin(x)}{\cos(x)}[/tex] --- proved

[tex](\stackrel{x_1}{-4}~,~\stackrel{y_1}{-4})\hspace{10em} \stackrel{slope}{m} ~=~ - \cfrac{3}{4} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-4)}=\stackrel{m}{- \cfrac{3}{4}}(x-\stackrel{x_1}{(-4)}) \implies y +4= -\cfrac{3}{4} (x +4) \\\\\\ y+4=-\cfrac{3}{4}x-3\implies {\Large \begin{array}{llll} y=-\cfrac{3}{4}x-7 \end{array}}[/tex]

Answer:

y = -3/4x - 7

Step-by-step explanation:

Given: slope = m = -3/4

Plug this value into the standard slope-intercept equation of y = mx + b.

y = -3/4x + b

To find b, we want to plug in a value that we know is on this line: in this case, point (-4, -4). Plug in the x and y values into the x and y of the standard equation.

-4 = -3/4(-4) + b

To find b, multiply the slope and the input of x(-4)

-4 = 3 + b

Now, subtract 3 from both sides to isolate b.

-7 = b

Plug this into your standard equation.

y = -3/4x - 7

This is your equation.

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Answer:

The length of the diagonal is 12.53

Step-by-step explanation:

Since we have a rectangle, the diagonal is the hypotenuse of a right triangle. So we can use Pythagorean Theorem.

a^2 + b^2 = c^2

We know the two legs are 6 and 11, so we can find the hypotenuse (which is the diagonal) see image.

See image.

Answer:

To solve this problem, we can set up the following equation:

Huilan's age + Thomas's age = 79

Let H be Huilan's age and T be Thomas's age. We can then rewrite the equation as:

H + T = 79

We are told that Huilan is 7 years older than Thomas, so we can write the following equation:

H = T + 7

Substituting the second equation into the first equation, we get:

(T + 7) + T = 79

Combining like terms, we get:

2T + 7 = 79

Subtracting 7 from both sides, we get:

2T = 72

Dividing both sides by 2, we get:

T = 36

Therefore, Thomas's age is 36 years.

Step-by-step explanation:

Answer:

let Thomas age be a

there fore, hullans age is a+7

(a+7)+a=79

2a=79-7

2a=72

a=36

Answer:

[tex]\int\ {\frac{x^4}{1 + x^{10}}} \, dx = \frac{1}{5}( \arctan(x^5)) + c[/tex]

Step-by-step explanation:

Given

[tex]\int\ {\frac{x^4}{1 + x^{10}}} \, dx[/tex]

Required

Integrate

We have:

[tex]\int\ {\frac{x^4}{1 + x^{10}}} \, dx[/tex]

Let

[tex]u = x^5[/tex]

Differentiate

[tex]\frac{du}{dx} = 5x^4[/tex]

Make dx the subject

[tex]dx = \frac{du}{5x^4}[/tex]

So, we have:

[tex]\int\ {\frac{x^4}{1 + x^{10}}} \, dx[/tex]

[tex]\int\ {\frac{x^4}{1 + x^{10}}} \, \frac{du}{5x^4}[/tex]

[tex]\frac{1}{5} \int\ {\frac{1}{1 + x^{10}}} \, du[/tex]

Express x^(10) as x^(5*2)

[tex]\frac{1}{5} \int\ {\frac{1}{1 + x^{5*2}}} \, du[/tex]

Rewrite as:

[tex]\frac{1}{5} \int\ {\frac{1}{1 + x^{5)^2}}} \, du[/tex]

Recall that: [tex]u = x^5[/tex]

[tex]\frac{1}{5} \int\ {\frac{1}{1 + u^2}}} \, du[/tex]

Integrate

[tex]\frac{1}{5} * \arctan(u) + c[/tex]

Substitute: [tex]u = x^5[/tex]

[tex]\frac{1}{5} * \arctan(x^5) + c[/tex]

Hence:

[tex]\int\ {\frac{x^4}{1 + x^{10}}} \, dx = \frac{1}{5}( \arctan(x^5)) + c[/tex]

now, we're assuming this is some rectangular area, and thus is simply the product of both quantities, let's change all mixed fractions to improper fractions firstly.

[tex]\stackrel{mixed}{12\frac{1}{2}}\implies \cfrac{12\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{25}{2}} ~\hfill \stackrel{mixed}{11\frac{3}{4}} \implies \cfrac{11\cdot 4+3}{4} \implies \stackrel{improper}{\cfrac{47}{4}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{25}{2}\cdot \cfrac{47}{4}\implies \cfrac{1175}{8}\implies 146\frac{7}{8}~ft^2[/tex]

Answer:

45 degree , bc the ends are all 90 degree so divide 90 by two since the line crosses right in between

Answer:

45°

That looks like an Isosceles triangle so two sides are equal

The angle near O would be 90° cause that looks like an right angle

180-90=90

To find each angle divide by 2

90÷2=45

Answer:

10 females, 15 males

Step-by-step explanation:

Let the number of females in the class be x. As the number of males is 5 more than the number of females, which is x, the number of males would be shown as x+5.

The number of females plus the number of males equal 25, so we get this equation:

x + (x+5) = 25

x + x + 5 = 25

2x + 5 = 25

2x = 20

x = 10.

So, there are 10 females, and since there are 5 more males than females, there are 10 + 5 = 15 males.

Answer: A. Quadrilateral

Step-by-step explanation:

The name for a polygon with 4 sides is a quadrilateral. Quadrilateral is a two-dimensional shape with four straight sides and four angles. The most common type of quadrilateral is the rectangle, which has four right angles. Other types of quadrilaterals include squares, parallelograms, trapezoids, and rhombuses.

Answer:

quadrilateral!!

Step-by-step explanation:

look at the beginning of the word, quad. Quad means 4 and is referring to a 4 sided figure.

Answer:

1/24

Step-by-step explanation:

Solve for y =4

48x=2

Answer:

B. 0.20 per century

Step-by-step explanation:

I calculated it logically

Answer:

1st question :55%

2nd ":122.22%

3rd ":81.82%

Answer:

1/8 pound per person

Step-by-step explanation:

Answer:

it would be 10392

Step-by-step explanation:

10392÷8=1299

Answer:

10392

Step-by-step explanation:

10391 ÷ 8 = 1298.875

Not divisible by 8

10392 ÷ 8 = 1299

Divisible by 8

The sample space when choosing one marble will be B s=(purple, pink, red, blue, green, orange, yellow}

A sample space is a set of potential results from a random experiment. The letter "S" is used to denote the sample space. Events are the subset of possible experiment results. Depending on the experiment, a sample space may contain a variety of outcomes.

A paper bag has seven colored marbles. The marbles are pink, red, green, blue, purple, yellow, and orange. The sample space when choosing one marble will be all the colors. The correct option is B.

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Answer:

-132

Step-by-step explanation:

Split the summation to make the starting value of  

k equal to 1.

Answer:S6 = - 132

Step-by-step explanation:

Step-by-step explanation:

as the description says, the higher the ratio the better.

9/10 = 0.9

12/15 = 4/5 = 0.8

so, Tasha has the better batting average, as she has the higher ratio.

9/10 = 27/30

12/15 = 24/30

27/30 is higher than 24/30.

Answer:

The shopper can but 12 jackets.

Step-by-step explanation:

80*.6 (100% - 40% = paying 60% or .6) = 48

576 / 48 = 12

Here we have a problem of percentage discounts, such that we need to find how much a given price decreases, and how many jackets we can buy considering the decreased price.

We will find that the number is 12.

Now let's see how we can get that result:

If something has an original price P, and we discount an x% of that price, the new price will be:

[tex]np = P\cdot(1 - x\%/100\%)[/tex]

Here we know that the regular price of a jacket is $80

And we have a discount of 40%

This means that the new price of these jackets will be:

[tex]np = \$80\cdot (1 - 40\%/100\%) = \$80\cdot (1 - 0.4) = \$80\cdot 0.6 = \$48[/tex]

Now we want to know how many of these jackets we can buy with $576

This will be equal to the quotient between the money we have and the new price of each jacket:

[tex]N = \$576/\$48 = 12[/tex]

This means that we can buy exactly 12 jackets at the sale.

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The design of a simulation that would allow Ian to determine how likely it would

be to guess all answers correctly on this quiz can be Illustrated through a coin.

Ian can design a simulation to determine the likelihood of guessing all answers correctly on the quiz by using a fair coin. He can flip the coin four times, with each flip representing a guess on the quiz. If the coin lands heads up, he marks it as a true answer, and if it lands tails up, he marks it as a false answer.

He can repeat this process a large number of times (e.g. 1000) and count the number of times that the sequence T, T, F, T is generated. The proportion of times that this sequence is generated out of the total number of simulations will be an estimate of the probability of guessing all answers correctly on the quiz by chance.

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Answer:

Hello There!!

Step-by-step explanation:

The answer is 15 as 3a=3×a so 3×5=15.

hope this helps,have a great day!!

~Pinky~

Answer:

[tex]y = 0.75x - 6[/tex]

Step-by-step explanation:

Given

Parallel to:

[tex]y = 0.75x[/tex]

Passes through

[tex](8,0)[/tex]

Required

The equation

The slope intercept form of an equation is:

[tex]y = mx + b[/tex]

Where:

[tex]m \to[/tex] slope

So, by comparison:

[tex]m = 0.75[/tex]

For a line parallel to [tex]y = 0.75x[/tex], it means they have the same slope of:

[tex]m = 0.75[/tex]

The equation of the line is then calculated using:

[tex]y = m(x - x_1) + y_1[/tex]

Where:

[tex]m = 0.75[/tex]

[tex](x_1,y_1) = (8,0)[/tex]

So, we have:

[tex]y = 0.75(x - 8) + 0[/tex]

Open bracket

[tex]y = 0.75x - 6 + 0[/tex]

[tex]y = 0.75x - 6[/tex]

Answer:

the answer is

B. 145° 12' 0"

Answer:

(1/6) of the students did not pass.

Ans: (1/6)30 = 5 of the students did not pass

To solve this question, we need to first figure out, in fractional form, the number of students who did not pass. The problem tells us that 5/6 of the students passed. Therefore, we know that 1/6 of the students did not pass, because 5/6 + 1/6 = 1. So, all we need to do is multiply 1/6 by the total number of students:

1/6 * 36 = 6

So, 6 students did not pass.

The given polynomial is a factor of P(x) because P(-5)=0.

The Remainder Theorem states that if a synthetic division of a polynomial by x = a results in a zero remainder, then x = an is a zero of the polynomial (thanks to the Remainder Theorem), and x an is also a factor of the polynomial (courtesy of the Factor Theorem).

The Factor Theorem states that x - c is only a factor of P(x) when P(c) = 0.

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Answer:

research it many my answer will be wrong

3

Step-by-step explanation:

-3 square is 9, adding 3 you will have 12. 10-6 is 4. so dividing 12 by 4 would give you 3.

Answer:

Step-by-step explanation:

distance = [tex]\sqrt{(-5/2+1/2)^{2} +(9/4+3/4)^{2} }=\sqrt{13}[/tex]=3.61

Answer:

x=17 degrees

Step-by-step explanation:

We know that a right angle is equal to 90 degrees. Therefore, two angles between a right angle have to equal 90. Therefore, 73+x=90. Isolate x and get x=90-73. x=17. Therefore, x=17 degrees.

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