What is 6532 rounded to the nearest thousand

Answer:

8.987 (Approximate)

Step-by-step explanation:

We have to find the sum of a G.P. series up to sixth terms.

The first term of the series is 6 and common ratio is [tex]\frac{1}{3}[/tex].

So, the sum is  

[tex]6 + 2 + \frac{2}{3} + \frac{2}{9} + \frac{2}{27} + \frac{2}{81}[/tex]

= [tex]6 \times \frac{1 - (\frac{1}{3})^{6}}{1 - \frac{1}{3} }[/tex]

= 8.987 (Approximate) (Answer)

We know the sum of a G.P.

a + ar + ar² + ar³ + ......... up to n terms = [tex]a\frac{1 - r^{n}}{1 - r}[/tex]  

where -1 < r < 1.

Answer:

[tex]\sqrt{85}[/tex]

Step-by-step explanation:

Given

x - [tex]\frac{1}{x}[/tex] = 9 ← square both sides

(x - [tex]\frac{1}{x}[/tex])² = 9²

x² - 2 + [tex]\frac{1}{x^2}[/tex] = 81 ( add 2 to both sides )

x² + [tex]\frac{1}{x^2}[/tex] = 83

Now

(x + [tex]\frac{1}{x}[/tex])² = x² + [tex]\frac{1}{x^2}[/tex] + 2, thus

x² + [tex]\frac{1}{x^2}[/tex] = 83 + 2 = 85

(x + [tex]\frac{1}{x}[/tex] )²= 85 ( take the square root of both sides)

x + [tex]\frac{1}{x}[/tex] = [tex]\sqrt{85}[/tex]

Answer:

Parallel lines

Step-by-step explanation:

Parallel lines have equal slopes

The product of the slopes of perpendicular lines equals - 1

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y-intercept )

Rearrange the 2 equations into this form and compare slopes

5y - x = 5 ( add x to both sides )

5y = x + 5 ( divide all terms by 5 )

y = [tex]\frac{1}{5}[/tex] x + 1 ← in slope- intercept form

with slope m = [tex]\frac{1}{5}[/tex]

5y - x = - 5 ( add x to both sides )

5y = x - 5 ( divide all terms by 5 )

y = [tex]\frac{1}{5}[/tex] x - 1 ← in slope- intercept form

with slope m = [tex]\frac{1}{5}[/tex]

Since slopes are equal then the lines are parallel

Answer:

Nine less than two-thirds of number is less than the number plus one.

Step-by-step explanation:

I jus got it right on edge.

Nine less than two-thirds of the number is less than the number plus one.

Algebra is the study of mathematical symbols, and the rule is the manipulation of those symbols.

The expression is given below.

2/3y − 9 < y + 1

Then complete the sentence.

Then we have

Nine less than two-thirds of the number is less than the number plus one.

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Answer:

Therefore the distance across the patch of swamp water is 50 ft

Step-by-step explanation:

Given:

VW = 100 ft

WX = 60 ft

XZ = 30 ft

To Find:

ZY = l = ?

Solution:

In  Δ VWX and Δ YZX

∠W ≅ ∠ Z    …………..{measure of each angle is 90° given}

∠VXW ≅ ∠YXZ      ..............{vertically opposite angles  are equal}

Δ ABC ~ Δ DEC ….{Angle-Angle Similarity test}

If two triangles are similar then their sides are in proportion.

[tex]\frac{VW}{YZ} =\frac{WX}{ZX} =\frac{VX}{YX}\ \textrm{corresponding sides of similar triangles are in proportion}\\[/tex]  

On substituting the given values we get

[tex]\frac{100}{l} =\frac{60}{30}\\\\l=\frac{3000}{60}=50\ ft[/tex]

Therefore the distance across the patch of swamp water is 50 ft

Answer:

11

Step-by-step explanation:

2x+3y=27

3x-2y=8

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

3(2x+3y)=3(27)

-2(3x-2y)=-2(8)

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

6x+9y=81

-6x+4y=-16

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

13y=65

y=65/13

y=5

2x+3(5)=27

2x+15=27

2x=27-15

2x=12

x=12/2

x=6

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

x+y=6+5=11

Answer:

a) [tex]8^8\sqrt{8}[/tex]

Step-by-step explanation:

Given,

[tex]\sqrt{8^1^7[/tex]

We have to simplify the expression by using "The Law of Indices".

[tex]x^m\times x^n=x^m^+^n[/tex]

So we can rewrite the expression as,

[tex]\sqrt{8^1^7[/tex]=[tex]\sqrt{8^1^+^1^6} =\sqrt{8}\times \sqrt{8^1^6}[/tex]

Now according to law of indices, which is;

[tex](x^m)^n=x^m^n[/tex]

So we can rewrite the expression as

[tex]\sqrt{8}\times \sqrt{8^1^6}=\sqrt{8}\times (8^1^6)^\frac{1}{2} \ \ \ \ Or\ \ \ \sqrt{8}\times 8^{16\times\frac{1}{2}} = 8^8\sqrt{8[/tex]

Hence the final Answer is [tex]8^8\sqrt{8[/tex].

Answer:

Step-by-step explanation:

8h – 3 = 11h + 12

Collecting like terms

8h - 11h = 12 + 3

- 3h = 15

h = 15/-3

h = -5

Answer:

$6770.65

Step-by-step explanation:

The purchase price is $20,300

The depreciation rate is 9.5% per year, in decimal, that is:

9.5/100 = 0.095

We want to find the value of car after 11 years. We will use the compound decay formula:

[tex]F=P(1-r)^t[/tex]

Where

F is the future value (what we want to find after 11 years)

P is the present value, purchase price (20,300 given)

r is the rate of depreciation (r = 0.095)

t is the time in years ( t = 11)

Substituting, we solve for F:

[tex]F=P(1-r)^t\\F = 20,300(1-0.095)^{11}\\F=20,300(0.905)^{11}\\F=6770.65[/tex]

Thus,

The value of the car would be around $6770.65, after 11 years

Answer:

Step-by-step explanation:

Use Newton's Law of Cooling for this one.  It involves natural logs and being able to solve equations that require natural logs.  The formula is as follows:

[tex]T(t)=T_{1}+(T_{0}-T_{1})e^{kt}[/tex] where

T(t) is the temp at time t

T₁ is the enviornmental temp

T₀ is the initial temp

k is the cooling constant which is different for everything, and

t is the time (here, it's in minutes)

If we are looking first for the temp after 20 minutes, we have to solve for the k value.  That's what we will do first, given the info that we have:

T(t) = 80

T₁ = 30

T₀ = 100

t = 5

k = ?

Filling in to solve for k:

[tex]80=30+(100-30)e^{5k}[/tex] which simplifies to

[tex]50=70e^{5k}[/tex] Divide both sides by 70 to get

[tex]\frac{50}{70}=e^{5k}[/tex] and take the natural log of both sides:

[tex]ln(\frac{5}{7})=ln(e^{5k})[/tex]

Since you're learning logs, I'm assuming that you know that a natural log and Euler's number, e, "undo" each other (just like taking the square root of something squared).  That gives us:

[tex]-.3364722366=5k[/tex]

Divide both sides by 5 to get that

k = -.0672944473

Now that we have a value for k, we can sub that in to solve for T(20):

[tex]T(20)=30+(100-30)e^{-.0672944473(20)}[/tex] which simplifies to

[tex]T(20)=30+70e^{-1.345888946}[/tex]

On your calculator, raise e to that power and multiply that number by 70:

T(20)= 30 + 70(.260308205) and

T(20) = 30 + 18.22157435 so

T(20) = 48.2°

Now we can use that k value to find out when (time) the temp of the object cools to 35°:

T(t) = 35

T₁ = 30

T₀ = 100

k = -.0672944473

t = ?

[tex]35=30+100-30)e^{-.0672944473t}[/tex] which simplifies to

[tex]5=70e^{-.0672944473t}[/tex]

Now divide both sides by 70 and take the natural log of both sides:

[tex]ln(\frac{5}{70})=ln(e^{-.0672944473t})[/tex] which simplifies to

-2.63905733 = -.0672944473t

Divide to get

t = 39.2 minutes

The temperature of the object after 20 minutes is 48.2° and temperature of body will be 35° after 39.2 minutes.

The formula can be expressed as:

[tex]\[ \frac{dT}{dt} = -k(T - T_a) \][/tex]

where:

[tex]\( T \)[/tex] is the temperature of the object at time [tex]\( t \)[/tex],

[tex]\( T_a \)[/tex] is the ambient temperature,

[tex]\( k \)[/tex] is a positive constant that depends on the characteristics of the object and the environment.

First, we need to find the constant [tex]\( k \)[/tex]. We have the following data:

Initial temperature of the object, [tex]\( T_0 = 100^\circ \)[/tex],

Temperature of the object after 5 minutes, [tex]\( T_1 = 80^\circ \)[/tex],

Ambient temperature, [tex]\( T_a = 30^\circ \)[/tex],

Time [tex]\( t_1 = 5 \)[/tex] minutes.

Using the integrated form of Newton's law of cooling, we have:

[tex]\[ T = T_a + (T_0 - T_a)e^{-kt} \][/tex]

Plugging in the values for [tex]\( T_1 \)[/tex] and [tex]\( t_1 \)[/tex], we get:

[tex]\[ 80 = 30 + (100 - 30)e^{-k \cdot 5} \][/tex]

Solving for [tex]\( k \)[/tex], we find:

[tex]\[ 50 = 70e^{-5k} \][/tex]

[tex]\[ e^{-5k} = \frac{50}{70} \][/tex]

[tex]\[ -5k = \ln\left(\frac{50}{70}\right) \][/tex]

[tex]\[ k = -\frac{1}{5}\ln\left(\frac{50}{70}\right) \][/tex]

Now that we have [tex]\( k \)[/tex], we can find the temperature after 20 minutes [tex]\( t_2 = 20 \)[/tex] minutes:

[tex]\[ T_2 = 30 + (100 - 30)e^{-k \times 20} \][/tex]

Substituting [tex]\( k \)[/tex] into the equation, we get:

[tex]\[ T_2 = 30 + (100 - 30)e^{\frac{1}{5}\ln\left(\frac{50}{70}\right) \times 20} \][/tex]

[tex]\[ T_2 = 30 + 70e^{\frac{20}{5}\ln\left(\frac{50}{70}\right)} \][/tex]

[tex]\[ T_2 = 30 + 70e^{4\ln\left(\frac{50}{70}\right)} \][/tex]

[tex]\[ T_2 = 30 + 70\left(\frac{50}{70}\right)^4 \][/tex]

[tex]\[ T_2 = 48.2^\circ[/tex]

Now, we need to solve for the time [tex]\( t_3 \)[/tex] when the temperature of the object is [tex]\( 35^\circ \)[/tex]:

[tex]\[ 35 = 30 + (100 - 30)e^{-kt_3} \][/tex]

[tex]\[ 5 = 70e^{-kt_3} \][/tex]

[tex]\[ e^{-kt_3} = \frac{5}{70} \][/tex]

[tex]\[ -kt_3 = \ln\left(\frac{5}{70}\right) \][/tex]

[tex]\[ t_3 = -\frac{1}{k}\ln\left(\frac{5}{70}\right) \][/tex]

Substituting [tex]\( k \)[/tex] into the equation, we get:

[tex]\[ t_3 = -\frac{5}{\ln\left(\frac{50}{70}\right)}\ln\left(\frac{5}{70}\right) \][/tex]

[tex]\ln\left(\frac{50}{70}\right)} = -0.336[/tex]

[tex]\ln\left(\frac{5}{70}\right) = -2.639[/tex]

[tex]|\[ t_3 = -\frac{5}{(-0.336)\right)} \times\ -2.639}|[/tex]

[tex]\ t_3 = 39.2 \text{minutes}[/tex]

Answer:

$97

Step-by-step explanation:

First, what do we know? Client A payed 488, Client B payed 294. They were supposed to pay equal amounts, but clearly that hasn't happened. If they had been fair, they would have divided the total of each bill equally between them. There is a way for us to do this, simply add the two amounts, and then divide by two.

[tex]488+294=782[/tex]

[tex]\frac{782}{2} =391[/tex]

So, both clients A and B were each supposed to pay 391. How much did client A overpay? We can find this number by looking at the difference between (or subtracting) the amount due (391) and the amount paid (488)

[tex]488-391=97[/tex]

We can verify this is correct by adding 97 to 294, to see if client B will now have paid as much as client A.

[tex]294+97=391[/tex],

which is what client B should have payed, and will have payed once he pays client A the 97 dollars owed.

Thus, client A is owed $97.

Answer:

Parenthesis

Step-by-step explanation:

The parenthesis are always the first step in the order of operations.

:)

Answer:

The correct answer is D. 114

Step-by-step explanation:

There are 252 students of twelfth grade at Vista View High School.

99 walked to school

11 went by bicycle

28 used the school bus

To find the amount of twelfth graders that rode in a car, we do this calculation:

Amount of twelfth graders that rode in a car = Total of twelfth graders - those who walked - those who went by bicycle - those who used the bus

Replacing with the real values, we have:

Amount of twelfth graders that rode in a car = 252 - 99 - 11 - 28 = 252 - 138 = 114

The correct answer is D. 114

Answer:

114

Step-by-step explanation:

Answer:

Karoun bought 12n + 15 numbers of pencils.

40 pencils

Step-by-step explanation:

If there are n numbers of pencils in each box of the pencil then there are 4n numbers of pencils in each package of pencils.

Now, Karoun bought 3 packages of pencils and get (5 × 3) = 15 pencils free of cost.

Therefore, Karoun bought (4n × 3) + 15 = 12n + 15 numbers of pencils. (Answer)

Now, if n = 10 pencils then in each package of pencils there will be 4n i.e. (4 × 10) = 40 pencils. (Answer)

Line XD equals line XE and is also equal to line XF that proved by using AAS postulate of congruence ⇒ B

Step-by-step explanation:

Let us revise the cases of congruence

In Δ ABC

∵ Ray AL bisects ∠A ⇒ (divides it into two equal angles)

∴ m∠DAX = m∠EAX

∵ Ray BM bisects ∠B ⇒ (divides it into two equal angles)

∴ m∠EBX = m∠FBX

∵ XD ⊥ AC

∴ m∠XDA = 90°

∵ XE ⊥ AB

∴ m∠XEA = 90°

∵ XE ⊥ BC

∴ m∠XFB = 90°

Now lets prove that Δ ADX and ΔAEX are congruent

In Δs ADX and AEX

∵ m∠ADX = m∠AEX ⇒ (their measures are 90°)

∵ m∠DAX = m∠EAX ⇒ proved

∵ AX is a common side in both triangles

- By using the AAS postulate of congruence

∴ Δ ADX ≅ Δ AEX

∴ XD = XE

Let us do the same with Δ BEX and Δ BFX

In Δs BEX and BFX

∵ m∠BEX = m∠BFX ⇒ (their measures are 90°)

∵ m∠EBX = m∠FBX ⇒ proved

∵ BX is a common side in both triangles

- By using the AAS postulate of congruence

∴ Δ BEX ≅ Δ BFX

∴ XE = XF

∵ XE = XD

∵ XE = XF

- If one side is equal two other sides then the two other sides are

  equal, that means the three sides are equal

∴ XD = XF

∴ XD = XE = XF

Line XD equals line XE and is also equal to line XF that proved by using AAS postulate of congruence

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s = 3x - 15 is the required expression for perimeter of table top

Perimeter of square tabletop is 240 cm

Solution:

A square tabletop has an area given as:

[tex](9x^2 - 90x+225) cm^2[/tex]

The dimensions  of the tabletop have the form cx - di  ,where cand d are whole numbers

To find perimeter of tabletop when x = 25 centimeters

Let us first find the length of each side of square

Given area is:

[tex]area = (9x^2 - 90x+225)[/tex]

We know that,

[tex]area = (side)^2 = s^2[/tex]

Therefore,

[tex]s^2 = (9x^2 - 90x+225)\\\\s^2 = (3x - 15)(3x - 15)\\\\s^2 = (3x - 15)^2[/tex]

Taking square root on both sides,

s = 3x - 15

The above expression is the required expression for perimeter of table top

To find perimeter when x = 25 centimeter

The perimeter of square is given as:

[tex]perimeter = 4s[/tex]

perimeter = 4(3x - 15)

Substitute x = 25

perimeter = 4(3(25) - 15)

perimeter = 4(60) = 240

Therefore perimeter of square tabletop is 240 cm

Answer:

-5

Step-by-step explanation:

-7u+9=2

-7u=2-9

-7u=-7

7u=7

u=7/7

u=1

3(1)-8=3-8=-5

Answer:

  y = -3x -16

Step-by-step explanation:

For problems like this, I like to start with a variation of the point-slope form of the equation of a line:

  y = m(x -h) +k . . . . . for a line with slope m through point (h, k)

For your given values, this is ...

  y = -3(x +3) -7

  y = -3x -9 -7 . . . . eliminate parentheses; next, combine terms

  y = -3x -16

Answer:

-3x-16

Step-by-step explanation:

The polynomial equation with zeroes 2i, -2i, 2 is [tex]x^3 -2x^2 + 4x - 8 = 0[/tex]

Given that zeros of polynomial are 2i, -2i, 2

To find: polynomial equation in standard form

zeros of polynomial are 2i, -2i, 2. So we can say,

x = 2i

x = -2i

x = 2

Or x - 2i = 0 and x + 2i = 0 and x - 2 = 0

Multiplying the above factors, we get the polynomial equation

[tex](x - 2i)(x + 2i)(x - 2) = 0\\[/tex] ------- eqn 1

Using a algebraic identity,

[tex](a - b)(a + b) = a^2 - b^2[/tex]

Thus [tex](x - 2i)(x + 2i) = x^2 - (2i)^2[/tex]

We know that [tex]i^2 = -1[/tex]

[tex]Thus (x - 2i)(x + 2i) = x^2 - (2i)^2 = x^2 -4(-1) = x^2 + 4[/tex]

Substitute the above value in eqn 1

[tex](x^2 + 4)(x - 2) = 0[/tex]

Multiply each term in first bracket with each term in second bracket

[tex]x^3 -2x^2 + 4x - 8 = 0[/tex]

Thus the required equation of polynomial is found

Answer:

f(x) = (x - 2)²

Step-by-step explanation:

f(x) = x² - 4x + 4 is a perfect trinomial.

You know a trinomial is perfect when double the square root of the first term multiplied by the square root of the last term equals the middle term.

As an equation, a trinomial in the form ax² + bx + c = 0:

2√a√c = b is a perfect trinomial.

Perfect trinomial are factored in this form:

(√a ± √c )(√a ± √c ) = (√a ± √c )²

Whether the sign is + or - depends on if the middle term is positive or negative.

In f(x) = x²-4x+4, the middle term is a negative.

The square root of the first term is 1.

The square root of the second term is 2.

The factored form is (x - 2)²

Answer:

The whole number temperatures, in kelvin, at which Andrew can conduct his experiment are 287, 288, 289,290,291.

Step-by-step explanation:

The temperature in degrees Celsius is 273.13 less than the temperature in kelvin; mathematically this means:

[tex]k=c+273.15[/tex]

Where [tex]k[/tex] is the temperature in kelvin and [tex]c[/tex] is the temperature in Celsius.

From this relationship we convert Andrew's temperature range—

13.5°c to 18.5°c—to kelvin:

[tex]k_1=13.5^oc+273.15=286.65[/tex]

[tex]k_2=18.5^oc+273.15=291.65[/tex]

Thus Andrew can conduct his experiment between 286.65 and 291.65 kelvin, and the whole number temperatures between these extremes are 287, 288, 289,290,291.

Thus the whole number temperatures, in kelvin, at which Andrew can conduct his experiment are 287, 288, 289,290,291.

Final answer:

To convert the temperature range from Celsius to Kelvin, add 273.15 to each end of the Celsius range given. The range for Andrew's experiment in Kelvin is from 287 K to 291 K, inclusive.

Explanation:

When converting temperatures from Celsius to Kelvin, one applies the simple formula: K = °C + 273.15. For Andrew's experiment, the temperature must be kept between 13.5°C and 18.5°C. Converting these to the Kelvin scale, we get:

Andrew can conduct his experiment at whole number Kelvin temperatures that fall within this range. Therefore, the range in Kelvin is from 287 K to 291 K, inclusive.

Answer:

17

Step-by-step explanation:

Given

[tex]b(1)=16\\ \\b(n)=b(n-1)+1[/tex]

Finding the second term of the sequence means to find [tex]b(2).[/tex] To find [tex]b(2)[/tex] substitute [tex]n=2[/tex] into the second expression:

[tex]b(2)=b(2-1)+1\\ \\b(2)=b(1)+1\\ \\b(2)=16+1\\ \\b(2)=17[/tex]

This is the new equation obtained after performing the subtraction.

[tex]\[ 9x - 2y = 0 \][/tex]

When subtracting one equation from another, we subtract the corresponding elements of the equations. Here's the step-by-step process:

Given the two equations:

1. [tex]\( x - 3y = 6 \)[/tex] (First equation)

2. [tex]\( -8x - y = 6 \)[/tex] (Second equation)

We want to subtract the second equation from the first. We do this by subtracting each term of the second equation from the corresponding term in the first equation:

Step 1: Subtract the x-terms:

[tex]\[ x - (-8x) = x + 8x = 9x \][/tex]

Step 2: Subtract the y-terms:

[tex]\[ -3y - (-y) = -3y + y = -2y \][/tex]

Step 3: Subtract the constants:

[tex]\[ 6 - 6 = 0 \][/tex]

So after subtracting the second equation from the first, the result is:

[tex]\[ 9x - 2y = 0 \][/tex]

This is the new equation obtained after performing the subtraction.

To find the area of this rectangle, we first solve for the width using the given perimeter, yielding 4 inches. The length, three times this, is 12 inches. Multiplying these values together gives an area of 48 square inches.

We are dealing with a rectangle whose length is three times its width. If we call the width of the rectangle 'w', then its length is '3w'. The perimeter of a rectangle is 2 times the sum of its length and width.

So 2*(w+3w) = 32. This simplifies to 8w = 32. Solving for 'w', we find that the width of the rectangle is 4 inches. Then, the length would be three times the width, which is 12 inches.

Having determined these dimensions, we can find the area of the rectangle. The area of a rectangle is its length multiplied by its width. So in this case, it would be 4 inches (width) times 12 inches (length) to give us an area of 48 square inches.

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Answer:

8.

Step-by-step explanation:

If you mean the GCF of 16 and 24 , from the prime factors, it is 2*2*2

= 8.

Answer:

8

Step-by-step explanation:

Final answer:

The length of wire required to fence the circular field is found by calculating the circumference of the field. The number of times the field can be fenced with a given length of wire is found by dividing the total wire length by the circumference.

Explanation:

To find the length of wire required to fence the circular field, we need to find the circumference of the field. The formula to calculate the circumference of a circle is C = πd, where C is the circumference and d is the diameter. So, in this case, the circumference is C = π * 56 cm. Next, to find how many times the field can be fenced with 704m of wire, we divide the total length of wire by the circumference of the field. Therefore, the number of times the field can be fenced is 704 m / (π * 56 cm).

Step-by-step explanation:

I hope you are able to understand the answer

A student asked to verify a Pythagorean identity, which typically involves applying the Pythagorean Theorem, a² + b² = c², to trigonometric functions. However, the specific identity to be verified was not provided in the question.

The question involves verifying a Pythagorean identity, which likely refers to an equation involving the use of the Pythagorean Theorem. The Pythagorean Theorem states that for a right-angled triangle, the square of the length of the hypotenuse (c) is equal to the sum of the squares of the other two sides (a and b). The theorem is expressed mathematically as a² + b² = c².

However, the student's question seems incomplete as it does not provide the specific identity they wish to verify. Common Pythagorean identities include trigonometric identities that relate sine, cosine, and tangent to one another using the Pythagorean Theorem, such as sin²(x) + cos²(x) = 1 for any angle x. To verify a Pythagorean identity, one typically substitutes the trigonometric functions with their corresponding right triangle side ratios and demonstrates that the equation holds true.

Answer:

y-y/x-x

1--1/3-4

1+1/3-4

2/-1

= -2

then you plug in with any one of the points

y=mx+b

1= -2(3) + b

1 = -6 + b

b = 1+6

b = 7

so the equation is: y = -2x+7

Answer:

[tex]y=(x-3)^2-2[/tex]

Step-by-step explanation:

Given vertex of parabola [tex](3,-2)[/tex]

Where [tex](h,k)[/tex] is the vertex.

[tex](h,k)=(3,-2)\\h=3\ and\ k=-2[/tex]

Also parabola opens up.

The equation of parabola with vertex [tex](h,k)[/tex]

[tex]y= a(x - h)^2 + k[/tex]

If [tex]a>0[/tex] parabola opens up.

[tex]a<0[/tex] parabola opens down.

As the parabola opens up the value of [tex]a[/tex] will greater than zero.

Plugging vertex of parabola in equation [tex]y= a(x - h)^2 + k[/tex]

[tex]y= a(x - 3)^2 -2[/tex]

Let us plug [tex]a=1[/tex]

The equation will be [tex]y= a(x - 3)^2 -2[/tex]

Answer:

Step-by-step explanation:

The word "sum" means adding. So, we have 11 + .....

The thing that 11 is being added to is a "product" a product means two things multiplied together. The things being multiplied are the number 2 and the number "r".