Write an equation for a circle with a diameter that has endpoints at (–4, –7) and (–2, –5). Round to the nearest tenth if necessary. Question 9 options: (x + 3)2 + (y + 6)2 = 2 (x + 3)2 + (y + 6)2 = 8 (x – 3)2 + (y – 6)2 = 2 (x – 3)2 + (y – 6)2 = 8

Answer:

[tex]\large\boxed{\dfrac{6x^2-54x}{x-9}\cdot\dfrac{7x}{6x}=7x}[/tex]

Step-by-step explanation:

[tex]\dfrac{6x^2-54x}{x-9}=\dfrac{6x(x-9)}{x-9}=6x\qquad\text{canceled}\ (x-9)\\\\\dfrac{7x}{6x}=\dfrac{7}{6}\qquad\text{canceled}\ x\\\\\dfrac{6x^2-54x}{x-9}\cdot\dfrac{7x}{6x}=6x\cdot\dfrac{7}{6}=7x\qquad\text{canceled 6}[/tex]

Answer:

B. [tex]f(x)=x^4-x^3+2x^2-4x-8[/tex]

Step-by-step explanation:

If [tex]2i[/tex] is a root of f(x), then the complex conjugate [tex]-2i[/tex] is also a solution. If f(x) should have exactly 2 real roots, then by the Fundamental Theorem of Algebra, the minimum degree of f(x) is 4.

Hence the first and last options are eliminated.

By the Remainder Theorem, [tex]f(2i)=0[/tex].

Let us check for options B and C.

For option B.

[tex]f(x)=x^4-x^3+2x^2-4x-8[/tex]

[tex]\implies f(2i)=(2i)^4-(2i)^3+2(2i)^2-4(2i)-8[/tex]

[tex]\implies f(2i)=16i^4-8i^3+8i^2-8i-8[/tex]

[tex]\implies f(2i)=16+8i-8-8i-8=0[/tex]

For option C

[tex]f(x)=x^4-x^3-6x^2+4x+8[/tex]

[tex]\implies f(2i)=(2i)^4-(2i)^3-6(2i)^2+4(2i)+8[/tex]

[tex]\implies f(2i)=16i^4-8i^3-24i^2+8i+8[/tex]

[tex]\implies f(2i)=-16+8i+24+8i+8\ne0[/tex]

Therefore the correct choice is B.

Answer:

C = pi * d formula

C = 314 ft

Step-by-step explanation:

To find the distance around the theatre, we want to use circumference.

C= 2 * pi *r  where r is the radius

or

C = pi * d  where d is the diameter

C = pi *100

C = 100pi

Using 3.14 as an approximation for pi

C = 3.14*100

C = 314 ft

Answer:

c 5.1

Step-by-step explanation:

The diameter of the circle is 102.

The radius is half of the diameter

r =d/2

r =10.2 /2 =5.1

Answer:

The distance between (-3, 2) and (0,3) is √10.

Step-by-step explanation:

As we go from (-3,2)  to  (0,3), x increases by 3 and y increases by 1.

Think of a triangle with base 3 and height 1.  Use the Pythagorean Theorem to find the length of the hypotenuse, which represents the distance between the points (-3, 2) and (0, 3):

distance = √(3² + 1²) = √10

The distance between (-3, 2) and (0,3) is √10.

For this case we have that by definition, the distance between two points is given by:

[tex]d = \sqrt {(x_ {2} -x_ {1}) ^ 2+ (y_ {2} -y_ {1}) ^ 2}[/tex]

We have the following points:

[tex](x_ {1}, y_ {1}): (- 3,2)\\(x_ {2}, y_ {2}) :( 0,3)[/tex]

Substituting:

[tex]d = \sqrt {(0 - (- 3)) ^ 2+ (3-2) ^ 2}\\d = \sqrt {(3) ^ 2 + (1) ^ 2}\\d = \sqrt {9 + 1}\\d = \sqrt {10}[/tex]

Answer:

The distance between the points is [tex]\sqrt {10}[/tex]

Answer:

c. when all angles are right angles

Step-by-step explanation:

Answer:

With Two points is enough to describe this graph

[tex]\left[\begin{array}{ccc}\frac{4}{6} &0\\0&-4\\\end{array}\right][/tex]

Step-by-step explanation:

Interception with Y axis (y=0), thus we have the following equation:

[tex]0=6*x+4[/tex]

[tex]x=4/6[/tex]

The point for this is (4/6 , 0)

Interception with x axis (x=0), thus we have the following equation:

[tex]Y=4*0-4[/tex]

[tex]y=-4[/tex]

The point for this is (0, -4)

Answer:

a=1

Step-by-step explanation:

Let's find out!

So we have the point (x,y)=(a,3) is on the equation 5x+y=8.

Let's replace x with a and y with 3. This gives us:

5x+y=8

5a+3=8

Subtract 3 on both sides:

5a    =5

Divide both sides by 5:

a     =5/5

a     =1

So the point has to be (1,3)

Answer:

a=2*n*pi where n is an integer

Step-by-step explanation:

[tex]\frac{\cos^2(a)}{1-\tan(a)}+\frac{\sin^3(a)}{\sin(a)-\cos(a)}[/tex]

The denominators are different here so I'm going to try to make them the same.

I'm going to write everything in terms of sine and cosine.

That means I'm rewriting tan(a) as sin(a)/cos(a)

[tex]\frac{\cos^2(a)}{1-\frac{\sin(a)}{\cos(a)}}+\frac{\sin^3(a)}{\sin(a)-\cos(a)}[/tex]

I'm going to multiply top and bottom of the first fraction by cos(a) to clear the mini-fraction from the bigger fraction.

[tex]\frac{\cos^2(a)}{1-\frac{\sin(a)}{\cos(a)}} \cdot \frac{\cos(a)}}{\cos(a)}+\frac{\sin^3(a)}{\sin(a)-\cos(a)}[/tex]

Distributing and Simplifying:

[tex]\frac{\cos^3(a)}{\cos(a)-\sin(a)}+\frac{\sin^3(a)}{\sin(a)-\cos(a)}[/tex]

Now I see the bottom's aren't quite the same but they are almost... They are actually just the opposite. That is -(cos(a)-sin(a))=-cos(a)+sin(a)=sin(a)-cos(a).

Or -(sin(a)-cos(a))=-sin(a)+cos(a)=cos(a)-sin(a).

So to get the denominators to be the same I'm going to multiply either fraction by -1/-1... I'm going to do this to the second fraction.

[tex]\frac{\cos^3(a)}{\cos(a)-\sin(a)}+\frac{\sin^3(a)}{\sin(a)-\cos(a)} \cdot \frac{-1}{-1}[/tex]

[tex]\frac{\cos^3(a)}{\cos(a)-\sin(a)}+\frac{-\sin^3(a)}{\cos(a)-\sin(a)}[/tex]

The bottoms( the denominators) are the same now.  We can write this as one fraction, now.

[tex]\frac{\cos^3(a)-\sin^3(a)}{\cos(a)-\sin(a)}[/tex]

I don't know if you know but we can factor a difference of cubes.

The numerator is in the form of a^3-b^3.

The formula for factoring that is (a-b)(a^2+ab+b^2).

[tex]\frac{(\cos(a)-\sin(a))(\cos^2(a)+\cos(a)\sin(a)+\sin^2(a)}{\cos(a)-\sin(a)}[/tex]

There is a common factor of cos(a)-sin(a) on top and bottom you can "cancel it".  

So we now have

[tex]\cos^2(a)+\cos(a)\sin(a)+\sin^2(a)[/tex]

We can actually simplify this even more.

[tex]\cos^2(a)+\sin^2(a)=1[/tex] is a Pythagorean Identity.

So we rewrite [tex]\cos^2(a)+\cos(a)\sin(a)+\sin^2(a)[/tex]

as [tex]1+\cos(a)\sin(a)[/tex]

So that is what we get after simplifying left hand side.

So I guess we are trying to solve for a.

[tex]1+\cos(a)\sin(a)=\sin(a)+\cos(a)[/tex]

Subtract sin(a) and cos(a) on both sides.

[tex]\cos(a)\sin(a)-\sin(a)-\cos(a)+1=0[/tex]

This can be factored as

[tex](\sin(a)-1)(\cos(a)-1)=0[/tex]

So we just need to solve the following two equations:

[tex]\sin(a)-1=0 \text{ and } \cos(a)-1=0[/tex]

[tex]\sin(a)=1 \text{ and } cos(a)=1 \text{ I just added one on both sides}[/tex]

Now we just need to think of the y-coordinates on the unit circle that are 1

and the x-coordinates being 1 also (not at the same time of course).

List thinking of the y-coordinates being 1:

a=pi/2 , 5pi/2 , 9pi/2 , ....

List thinking of the x-coordinates being 1:

a=0, 2pi, 4pi,...

So let's come up with a pattern for these because there are infinite number of solutions that continue in this way.

If you notice in the first list the number next to pi is going up by 4 each time.

So for the first list we can say a=(4pi*n+pi)/2 where n is an integer.

The next list the number in front of pi is just even.

So for the second list we can say a=2*n*pi where n is an integer.

So the solutions is a=2*n*pi   ,    a=(4pi*n+pi)/2

We really should make sure if this is okay for our original equation.

We don't have to worry about the second fraction because sin(a)=cos(a) only when a is pi/4 or pi/4+2pi*n OR (pi+pi/4) or (pi+pi/4)+2pi*n.

Now the second fraction we have 1-tan(a) in the denominator, and it is 0 when:

tan(a)=1

sin(a)/cos(a)=1                       =>              sin(a)=cos(a)

So the only thing we have to worry about here since we said sin(a)=cos(a) doesn't hurt our solution is the division by the cos(a).

When is cos(a)=0?

cos(a)=0 when a=pi/2 or any rotations that stop there (+2npi thing) or at 3pi/2 (+2npi)

So the only solutions that work is the a=2*n*pi where n is an integer.

Answer:

[tex]\large\boxed{a=2k\pi\ for\ k\in\mathbb{Z}}[/tex]

Step-by-step explanation:

[tex]\bold{a=x}[/tex]

[tex]\text{The domain:}\\\\1-\tan x\neq0\ \wedge\ \sin x-\cos x\neq0\ \wedge\ x\neq\dfrac{\pi}{2}+k\pi\ (from\ \tan x)\\\\\tan x\neq1\ \wedge\ \sin x\neq\cos x\\\\x\neq\dfrac{\pi}{4}+k\pi\ \wedge\ x\neq\dfrac{\pi}{4}+k\pi\ for\ k\in\mathbb{Z}[/tex]

[tex]\dfrac{\cos^2x}{1-\tan x}+\dfrac{\sin^3x}{\sin x-\cos x}=\sin x+\cos x[/tex]

[tex]\text{Left side of the equation:}[/tex]

[tex]\text{use}\ \tan x=\dfrac{\sin x}{\cos x}\\\\\dfrac{\cos^2x}{1-\tan x}=\dfrac{\cos^2x}{1-\frac{\sin x}{\cos x}}=\dfrac{\cos^2x}{\frac{\cos x}{\cos x}-\frac{\sin x}{\cos x}}=\dfrac{\cos^2x}{\frac{\cos x-\sin x}{\cos x}}=\cos^2x\cdot\dfrac{\cos x}{\cos x-\sin x}\\\\=\dfrac{\cos^3x}{\cos x-\sin x}\\\\\dfrac{\cos^2x}{1-\tan x}+\dfrac{\sin^3x}{\sin x-\cos x}=\dfrac{\cos^3x}{\cos x-\sin x}+\dfrac{\sin^3x}{\sin x-\cos x}\\\\=\dfrac{\cos^3x}{\cos x-\sin x}+\dfrac{\sin^3x}{-(\cos x-\sin x)}[/tex]

[tex]=\dfrac{\cos^3x}{\cos x-\sin x}-\dfrac{\sin^3x}{\cos x-\sin x}\\\\=\dfrac{\cos^3x-\sin^3x}{\cos x-\sin x}\qquad\text{use}\ a^3-b^3=(a-b)(a^2+ab+b^2)\\\\=\dfrac{(\cos x-\sin x)(\cos^2x+\cos x\sin x+\sin^2x)}{\cos x-\sin x}\qquad\text{cancel}\ (\cos x-\sin x)\\\\=\cos^2x+\cos x\sin x+\sin^2x\qquad\text{use}\ \sin^2x+\cos^2x=1\\\\=\cos x\sin x+1[/tex]

[tex]\text{We're back to the equation}[/tex]

[tex]\cos x\sin x+1=\sin x+\cos x\qquad\text{subtract}\ \sin x\ \text{and}\ \cos x\ \text{from both sides}\\\\\cos x\sin x+1-\sin x-\cos x=0\\\\(\cos x\sin x-\sin x)+(1-\cos x)=0\\\\\sin x(\cos x-1)-1(\cos x-1)=0\\\\(\cos x-1)(\sin x-1)=0\iff \cos x-1=0\ or\ \sin x-1=0\\\\\cos x=1\ or\ \sin x=1\\\\x=2k\pi\in D\ or\ x=\dfrac{\pi}{2}+2k\pi\notin D\ for\ k\in\mathbb{Z}[/tex]

Answer:

Parallel

Step-by-step explanation:

Instead of putting this and slope intercept form.  I'm going determine the

x-intercept and the y-intercept of both.

The x-intercept can be found by setting y to 0 and solving for x.

The y-intercept can be found by setting x to 0 and solving for y.

So let's look at 3x-2y=5.

x-intercept?

Set y=0.

3x-2(0)=5

3x       =5

 x       =5/3

y=intercept?

Set x=0.

3(0)-2y=5

     -2y=5

        y=-5/2

So we are going to graph (5/3,0) and (0,-5/2) and connect it with a straightedge.

Now for 6y-9x=6.

x-intercept?

Set y=0.

6(0)-9x=6

     -9x=6

        x=-6/9

        x=-2/3

y-intercept?

Set x=0.

6y-9(0)=6

6y        =6

 y        =1

So we are going to graph (-2/3,0) and (0,1) and connect it what a straightedge.

After graphing the lines by hand you can actually do an algebraic check to see if they are parallel (same slopes), perpendicular (opposite reciprocal slopes), or neither.

Let's find the slope by lining up the points and subtracting then putting 2nd difference over 1st difference.

So the points on line 1 are: (5/3,0) and (0,-5/2)

 (5/3 ,  0  )

- (0   ,-5/2)

-----------------

5/3       5/2

The slope is (5/2)/(5/3)=(5/2)*(3/5)=3/2.

The points on line 2 are: (-2/3,0) and (0,1)

 (-2/3 , 0)

- (   0  ,  1)

-----------------

 -2/3      -1

The slope is -1/(-2/3)=-1*(-3/2)=3/2.

The slopes are the same so they are parallel. The line lines are definitely not the same; if you multiply the top equation by -3 you get -9x+6y=-15 which means the equations are not the same.  Also they had different x- and y-intercepts.  So these lines are parallel.

This is what we should see in our picture too.

The lines 3x - 2y = 5 and 6y - 9x = 6 are parallel lines

The lines 3x - 2y = 5 and 6y - 9x = 6 are parallel lines if their their slopes are equal

Lets rewrite the equation to be in slope intercept form: y = mx + b. where m is the slope

3x - 2y = 5

2y = 3x - 5

y = 3x/2 - 5/2. slope is 3/2

6y - 9x = 6

6y = 9x + 6

y = 3x/2 + 1. slope is 3/2

Comparing the equations with equal slope of 3/2, and from the graph shows that they are parallel lines

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Answer:

1/6

Step-by-step explanation:

There is x candy in total and y students in total.

Group A consists of 0.8y students, and group B consists of 0.2y students.

Group A gets 40% of x, or 0.4x. Each student in group A gets 0.4x/(0.8y) = 0.5x/y candy per person.

Group B gets 60% of x, or 0.6x. Each student in group B gets 0.6x/(0.2y) = 3x/y candy per person.

The ratio of candy each student in group A has to the amount of candy each student in group B has is:

(0.5x/y)/(3x/y) = 0.5/3 = 1/6

The ratio of the amount of candy a student in group A has to the amount of candy a student in group B has is equal to 1/6

percentage, a relative value indicating hundredth parts of any quantity.

Given that Eighty percent of the students in a class (group A) share 40% of the candy equally.

The remaining 20% of the students (group B) share the other 60% of the candy equally.

We need to find the common fraction between Group A and group B if ratio of the amount of candy a student in group A has to the amount of candy a student in group B has is equal

Let the total number of candy's are x

The total number of students are y.

The number of students in group A=0.8y

The number of students in group B=0.2y

Group A gets 40% of x, or 0.4x.

Each student in group A gets 0.4x/(0.8y) = 0.5x/y candy per person.

Group B gets 60% of x, or 0.6x. Each student in group B gets 0.6x/(0.2y) = 3x/y candy per person.

The ratio of candy each student in group A has to the amount of candy each student in group B has is:

(0.5x/y)/(3x/y) = 0.5/3 = 1/6

Hence, the ratio of the amount of candy a student in group A has to the amount of candy a student in group B has is equal to 1/6

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Answer:

230.

Step-by-step explanation:

[tex] \: \: \: \: 300 - 25 - \frac{15}{100} \times 300 = \\ = 275 - 45 = \\ = 230[/tex]

Hope this helps!

Answer:

230 blue marbles

Step-by-step explanation:

In order to answer this question, we need to find how many marbles are red.

We know that 25 marbles are green

15% of the marbles are red.

There are a total of 300 marbles.

We would multiply 300 by 0.15 to find how many marbles are red.

[tex]300*0.15=45[/tex]

45 marbles are red.

In the question, it says the rest of the marbles are blue.

This means that we would subtract 300 by 25 and 45 to find how many marbles are blue.

[tex]300-25-45=230[/tex]

This means that 230 marbles are blue.

The function that represents the given scenario is; f(x) = 6(4)ˣ

We know the general formula for this population function is;

f(x) = abˣ

where;

a is initial population

b is the common ratio

x is number of years

We are given;

a = 6

b = 4

Thus;

f(x) = 6(4)ˣ

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Answer:

5

Step-by-step explanation:

7 ducks=7/7 ducks

7/7-?=5/7

7/7-2/7=5/7

5/7=5 ducks

Answer:

y=5

x=7

Step-by-step explanation:

When given an expression in the form a⁻ᵇ the expression is the same as 1/aᵇ where  a and b are integers.

Solving the question by giving the equivalents of the expression:

Therefore (-j)⁻⁷ is the same as 1/j⁷

The value of x=7

k⁻⁵+m⁻¹⁰=1/k⁵+1/m¹⁰

Therefore the value of y is 5.

Question 1:

For this case we have that by definition of properties of powers it is fulfilled that:

[tex]a ^ {- 1} = \frac {1} {a ^ 1} = \frac {1} {a}[/tex]

So, if we have the following expression:

[tex](-j) ^ {- 7}[/tex]

We can rewrite it as:

[tex]\frac {1} {(- j) ^ 7}[/tex]

So we have to:

[tex]x = 7[/tex]

ANswer:

[tex]x = 7[/tex]

Question 2:

For this case we have that by definition of properties of powers it is fulfilled that:

[tex]a ^ {- 1} = \frac {1} {a ^ 1} = \frac {1} {a}[/tex]

If we have the following expression:

[tex]k^{-5}+m^{-10}[/tex]

We can rewrite it as:

[tex]\frac {1} {k ^ 5} + \frac {1} {m ^ {10}}[/tex]

So we have[tex]y = 5[/tex]

Answer:

[tex]y = 5[/tex]

Answer:

The value of y = 4.3

Step-by-step explanation:

It is given that,

5/7 = y/6

To find the value of y

We have, 5/7 = y/6

(5 * 6) = (y * 7)

y * 7 = 5 * 6

y = (5 * 6)/7

 = 30/7

 = 4.29

 ≈ 4.3

Therefore the correct answer, the value of y = 4.3

Option C ,

40 degrees

Because angle Z was 70 (isosceles triangle)

Then 180-140=40 degrees

Answer:

x = ±3

Step-by-step explanation:

x^2 +3 = 12

Subtract 3 from each side

x^2 +3-3 = 12-3

x^2 = 9

Take the square root of each side

sqrt(x^2) =±sqrt(9)

x = ±3

Answer:

B

Step-by-step explanation:

1/2x < -3

Multiply each side by 2

1/2x *2 < -3*2

x < -6

Since x is less than -6, there is an open circle at -6

Less than means the line goes to the left

Answer:

58%

Step-by-step explanation:

you write it as

308= ? (because you dont know what percent of 530 is yet) x 530

so it should look like 308=?x530

then divide on both sides and then turn your answer into a percentage.

Answer: Dividing 185 by 4 is the first step in finding out whether the owner will be able to use a whole number of panels. Hope this helps!

Step-by-step explanation:

The owner will not be able to use a whole number of panels btw

Answer:

D one diagonal bisects opposite angles and the other diagonal does not

Option D, 'one diagonal bisects opposite angles and the other diagonal does not,' does NOT guarantee that a quadrilateral is a kite.

To determine if a quadrilateral is a kite or not based on its properties.

A quadrilateral is a kite if and only if its diagonals are perpendicular and exactly one of them bisects the other diagonal. Let's analyze the given options:

Therefore, option D, 'one diagonal bisects opposite angles and the other diagonal does not,' does NOT guarantee that a quadrilateral is a kite.

Answer:

$3.49

12 quarters,4 dimes,1 nickle, 4 pennies

Step-by-step explanation:

First remember that;

Quarter=25 pennies

Dime=10 pennies

Nickel =4 pennies

I cent =0.01$

Given the amount as $3.50

Assume you use 4 quarters, this means a whole dollar, so it will be;

1 dollar = 4quarters

3 dollars=?

cross multiply

3×4=12 quarters

The remaining $0.49

Here identify 40 cents, which is 40 pennies

But 10 pennies=1 dime

so 40 pennies=?

cross-multiply

(40×1)÷10=4 dimes

and a nickel for 4 pennies.

Total area is 110 square meter

Answer:

140

Step-by-step explanation:

Answer:

27

Step-by-step explanation:

The range is the greatest value subtract the smallest value

greatest value = 50 and smallest value = 23, so

range = 50 - 23 = 27

Answer:

27

Step-by-step explanation:

23,32,37,40,41,41,41,42,43,48,49,49,50.

Range=50-23=27

Answer:

Option A There is not enough information to determine

Step-by-step explanation:

we know that

If two lines are perpendicular, then the product of their slopes is equal to -1

You must calculate the slope of the AB segment and the BC segment, however the coordinates of the vertices are not known.

therefore

There is not enough information to determine

Answer:

C 15, 205.

Step-by-step explanation:

That would be C because 15^2 = 225. The others are all correct.

The whole number not followed by its square is (c) 15, 205

The square of a number x can be represented as:

y = x^2

Using the above equation, we have:

121  = 11^2 --- true

169  = 13^2 --- true

205  = 15^2 --- false

64  = 8^2 --- true

4  = 2^2 --- true

Hence, the whole number not followed by its square is (c) 15, 205

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Answer:

The correct answer option is A. 1397.46 ft^2.

Step-by-step explanation:

We are to find the area of the shaded region. For that, we will divide the figure into smaller shapes, find their areas separately and then add them up.

From the given figure, we can see that there are two semi circles )or say one whole circle if we combine them) at the ends while 2 rectangles at the top and bottom.

Radius of circle = [tex]\frac{20+7+7}{2} [/tex] = 17

Area of circle = [tex]\pi r^2 = \pi \times 17^2[/tex] = 907.92 square ft

Area of rectangles = [tex]2(l \times w) = 2(35 \times 7)[/tex] = 490 square ft

Area of shaded region = 907.9 + 490 = 1397.46 ft^2

Answer:

A.) 1,397.46 ft²

Step-by-step explanation:

I got it correct on founders edtell

Answer:

B. to the right of 0

Step-by-step explanation:

the opposite of -10 would be 10 and out of all the answer choices the number 10 only fits into B.

Answer:

B to the right of zero why? ITS COMMON SENSE!

Step-by-step explanation:

Answer:

a

Step-by-step explanation: