What is the average rate of change in the area as the radius changes from 2.5 to 5.5 feet?

Answer: $1.1x

Step-by-step explanation:

At the end of year 1, it will increase by 10%:

[tex]x+0.1x=x(1+0.1)=1.1x[/tex]

Answer:

[tex]A' = (-2,2)[/tex]

[tex]B' = (2,2)[/tex]

[tex]C' = (2,-2)[/tex]

Step-by-step explanation:

Given

[tex]A = (-6,6)[/tex]

[tex]B = (6,6)[/tex]

[tex]C = (6,-6)[/tex]

[tex]k = \frac{1}{3}[/tex]

Required

The new coordinates

To do this, we simply multiply the coordinates of [tex]\triangle ABC[/tex] by the factor of dilation.

i.e.:

[tex]A' = A * k[/tex]

[tex]B' = B * k[/tex]

[tex]C' = C * k[/tex]

So, we have:

[tex]A' = (-6,6) * \frac{1}{3}[/tex]

[tex]A' = (-2,2)[/tex]

[tex]B' = (6,6) * \frac{1}{3}[/tex]

[tex]B' = (2,2)[/tex]

[tex]C' = (6,-6) * \frac{1}{3}[/tex]

[tex]C' = (2,-2)[/tex]

Answer:

x = - 1, x = - 0.5

Step-by-step explanation:

Given

2x(x + 1.5) = - 1 ← distribute parenthesis on left side

2x² + 3x = - 1 ( add 1 to both sides )

2x² + 3x + 1 = 0 ← in standard form

(2x + 1)(x + 1) = 0 ← in factored form

Equate each factor to zero and solve for x

x + 1 = 0 ⇒ x = - 1

2x + 1 = 0 ⇒ 2x = - 1 ⇒ x = - 0.5

Answer:

1.9

Step-by-step explanation:

You round when the numbers behind the decimal are 5 or greater. This being said, you need to round from the number furthest away from the decimal. In this case it would be 9. 9 is greater than 5 so we round 8 up to 9 and get rid of the number behind it. We are left with 1.9

Answer:

the answer is -299.808 well that's what I think it's

Answer:

-299.808

Step-by-step explanation:

= 10700/100 + 5192/100 - 412

= 107 + 5.192 - 412

= -299.808

Answer:

c on edge

Step-by-step explanation:

Answer:

It is 3/2, I got it right on my exam.

Step-by-step explanation:

Dilation is a transformation that allows you to resize an object. The dilation factor of the given point is 3/2.

Dilation is a transformation that allows you to resize an object. Dilation is a technique for making items bigger or smaller. This transformation yields a picture that is identical to the original shape. However, there is a size discrepancy in the form.

Given the dilation of the k around the origin as K = (2,4) → (3,6), therefore, the point k is dilated 1 unit to the right and 2 units upwards.

Let the dilation factor be represented by a, then,

2 × a = 3

a = 3/2

Similarly for the y-axis,

4 × a = 6

a = 6/4 = 3/2

Hence, the dilation factor of the given point is 3/2.

Learn more about Dilation:

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Answer:

a.2×1= 2 + 3×1=3 +4×2=8 now we have to put this ans and the ans is 13.

b. 5×1×2=10 + 3×1×2=6+ 7×2×2=28 now also we have to put this ans and this ans is 13

Answer:

hope it helps

Step-by-step explanation:

Answer:

Step-by-step explanation:

S is the midpoint of QR.

⇒ RS = QS    

QR ⊥ PS

⇒ ∠RSP = ∠QSP = 90°

In ΔPSR & ΔPSQ,

Statement                     Reason

RS ≅ QS                       S is midpoint of QR

∠RSP ≅ ∠QSP              QR ⊥ PS

PS ≅ PS                         Reflexive property

ΔPSR ≅ ΔPSQ               Side Angle Side - S A S Congruent

PR≅  PQ                        CPCTC

Answer:

29/14

Step-by-step explanation:

Start by working out the LCM of these (or any common denominator). A common denominator for 4 and 7 is 28 cuz 4 x 7 = 28.

3/4 = 21/28

4/7 = 16/28

3/4 = 21/28

(21+16+21)/28 = 58/28 which can simplify to 29/14

hi

Golden rule : you can only add or substract fraction if and only if they have the same denominator.

so two solutions :

you see that one denominator is a multiple of the second. so you multiply all the fraction by this multiple.

ex : 1/3 + 5/6

here I can easely convert 1/3 into a fraction with 6 as 6 is 3x2.

so : 1x2 = 2 and 3x2 = 6

1/3 is 2/6

so. 1/3 +5/6 = 2/6 +5/6

fraction have same denominator, I add numerators : 2/6 +5/6 =7/6

if it not that evident, you multiply fraction A up and down by denominator of fraction B.

and you multiply up and down fraction B by denomonator of fraction A

let' s see with your exemle :

3/4 +4/7+3/4 = ?

first I add fraction with same denominator :

3/4+3/4+4/7 = 6/4 +4/7

fraction A is 6/4.

I multiply up and down by denominator fractiok B which is " 7"

so. 6/4 = 6x7/4×7 = 42/28

let's apply method to fraction B whixh is 4/7 . let' multiply up and down by denominator of fraction A which is " 4"

so : 4/7 = 4x4 /7x4 = 16/28

now we have :

6/4 +4/7 = 42/28 +16/28

both my fractions have same denominator "28" so I add numerator :

42/28 +16/28 = 58/28

Now must wonder if 58/28 can be simplify. Good trick is to decompose numbers, and in prime number is the best way to do

58 = 2x29

28 = 7x2x2

so 58/28= (2×29)/ (7x2x2)

when there is same number up and down I can cross it out .

so here one "2" up and down can be rule out

so 58/28 = 2x29 /7×2×2 = 29/7x2 =29/14

I can not symplify more, so calculus is over.

in short :

3/4+4/7+3/4 = 6/4+4/7

= 42/28 +16/28

=58/28

= 29/14

Answer:

Step-by-step explanation:

Let the amount of gas in the tank before filling was x.

We have:

let the amount of gas be x

[tex]\\ \sf\longmapsto x+2\dfrac{4}{7}=4[/tex]

[tex]\\ \sf\longmapsto x+\dfrac{18}{7}=4[/tex]

[tex]\\ \sf\longmapsto x=4-\dfrac{18}{7}[/tex]

[tex]\\ \sf\longmapsto x=\dfrac{28-18}{7}[/tex]

[tex]\\ \sf\longmapsto x=\dfrac{10}{7}[/tex]

[tex]\\ \sf\longmapsto x=1\dfrac{3}{7}[/tex]

Answer:

The values of [tex]a[/tex] and [tex]b[/tex] are [tex]5[/tex] and [tex]-\frac{15}{7}[/tex], respectively.

Step-by-step explanation:

There are mistakes in the statement, correct form is presented below:

[tex]5+\frac{\sqrt{3}}{7} + 2\sqrt{3} = a - b\cdot \sqrt{3}[/tex].

By direct comparison we have the following system of equations:

[tex]a = 5[/tex] (1)

[tex]\frac{\sqrt{3}}{7}+2\sqrt{3} = -b\cdot \sqrt{3}[/tex] (2)

In (2) we solve for [tex]b[/tex]:

[tex]\left(\frac{1}{7}+2 \right)\cdot \sqrt{3} = -b\cdot \sqrt{3}[/tex]

[tex]b = -\frac{15}{7}[/tex]

The values of [tex]a[/tex] and [tex]b[/tex] are [tex]5[/tex] and [tex]-\frac{15}{7}[/tex], respectively.

Answer:

A, it's the only one with a number greater then 1

Answer:

b) 6

Step-by-step explanation:

x - 4 = 2

x = 2 + 4

x = 6

Step-by-step explanation:

0 10

0-6

this is the answer. I am sure

The value of numbers are 6 and 10 which are a distance of 2 units from 8 on a number line.

Number line is a horizontal line where numbers are marked at equal intervals one after another form smaller to greater.

We have to given that;

Number line is shown in figure.

And, We have to find the value of numbers which are a distance of 2 units from 8 on a number line.

Now,

We can find the value of numbers as;

⇒ | 8 - 2 | = 6

⇒ | 8 + 2 | = 10

Therefore, The value of numbers are 6 and 10 which are a distance of 2 units from 8 on a number line.

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Answer:

it will take 5days to reach a total of 180meters

Step-by-step explanation:

80m+20+20+20+20+20

=180meters

=5days

Answer: the answer to your question is 7

Step-by-step explanation:

Answer:

2nd option

Step-by-step explanation:

Given f(x) then f(x + a) is a horizontal translation of f(x)

• If a > 0 then a shift left of a units

• If a < 0 then a shift right of a units

Here g(x) = (x - 6)²

The base graph has been shifted right 6 units

Answer:

[tex]we \: know \: that \: commutative \: prop \\ = x + y = y + x \\ then \\ 4xy + 0 = 0 + 4xy \\ thank \: you[/tex]

Answer:

-13 1/2 < -6/8

Step-by-step explanation:

*Any negative in either the numerator or denominator will make the entire fraction negative.

1/2  = .5

-13 1/2 = -13.5

-6/8 = -3/4

-3/4 = -0.75

-6/8 = -0.75

-13 1/2 = -13.5

Which means:

-13 1/2 < -6/8

Hope this helped.

Answer:

[tex] < [/tex]

Step-by-step explanation:

[tex] \frac{6}{8} [/tex]

Is close to 0 (to positive) you can use a time line to determine which is a correct answer

Answer:

8 hours 10 minutes

I hope this answers your question

Answer: paul has been away from home for 7 hours and 50 minutes.

Step-by-step explanation:

Answer:

(-4, -1)

Step-by-step explanation:

The solution to the system of equations on a graph is the point in which they intersect. Therefore, the solution is (-4, -1).

Answer:

y= - 1.4

x= 0.2

Step-by-step explanation:

2x-Y= -1

x-2y=4

The depth of the mirror of the cross-section of a searchlight would be 8.33 cm if the light bub has a vertex at 3 cm and the mirror is 20 centimeters across at the origin.

A cross-section is perpendicular to the axis of the symmetry goes through the vertex of the parabola. The cross-sectional shape of the mirrored section of most searchlights or spotlights is parabolic.

Given:

a = 3

y = [tex]\frac{20}{2}[/tex] cm = 10 cm

Solution:

from the equation [tex]y^{2} = 4ax[/tex]

[tex]y^{2} = 4*3*x\\ y^{2} = 12x[/tex]

putting x, 10 cm in the equation

[tex]x=\frac{10^{2} }{12} \\\\x= \frac{100}{12} \\\\x= 8.33 cm[/tex]

thus, the depth of the mirror would be - 8.33 cm

Learn more about other problems of the parabola:

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Hi.

Let's call X number of visits.

so members pay for a year : membership 24 and reduce entrance at 3 dollars.

so you member pay : 3X+24

a non member pay entrance fee of 6 for each visit.

so 6X

Question : for how many visit would a membership will be more intresting than standar entrance ?

3X+24 = 6X

3X-6X = -24

-3X=-24

X= -24/-3

X =24/3

X= 8

Under 8 visit a year, it's cheaper non being a member.

at 8 entrance cost is the same for members and non member. After it better to choose membership as

9 entrances for member is : 3x9 +24 = 27+24= 51

non member : 9x6 = 54

And so on for each supplementary visits.

Answer:

D

Step-by-step explanation:

Answer:

It's D on there

Step-by-step explanation:

Answer:

The answer is 2.)

Step-by-step explanation:

Given initial velocity=135 ft/s

& cliff=76 foot

Given quadratic equation

⇒                      (let h=0 it is given)

⇒  

⇒  

⇒    t=8.96≈9 s (the other root is negative)

Hence, rocket will take 9 s to hit the ground after launched.

Answer:  Choice D

0 = 16t^2 + 135t + 76;  9 s

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

Explanation:

The equation we start with is

[tex]h = -16t^2 + vt + c\\\\[/tex]

where v is the starting or initial velocity, and c is the starting height.

We're told that v = 135 and c = 76

We let h = 0 to indicate when the object hits the ground, aka the height is 0 ft.

That means the equation updates to [tex]0 = -16t^2 + 135t + 76\\\\[/tex]

Based on that alone, the answer is between A or D

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

We'll use the quadratic formula to solve for t

We have

So,

[tex]t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\\\t = \frac{-135 \pm \sqrt{135^2 - 4(-16)(76)}}{2(-16)}\\\\t = \frac{-135 \pm \sqrt{23,089}}{-32}\\\\t \approx \frac{-135 \pm 151.9506}{-32}\\\\t \approx \frac{-135 + 151.9506}{-32} \ \text{ or } \ t \approx \frac{-135 - 151.9506}{-32}\\\\t \approx \frac{16.9506}{-32} \ \text{ or } \ t \approx \frac{-286.9506}{-32}\\\\t \approx -0.52971 \ \text{ or } \ t \approx 8.96721\\\\[/tex]

We ignore the negative t value because a negative time duration makes no sense.

The only practical solution here is roughly 8.96721 which rounds to 9.0 or simply 9 when we round to the nearest tenth (one decimal place).

In short, the object will hit the ground at the 9 second mark roughly. Or put another way: the object is in the air for about 9 seconds.

From this, we can see that the final answer is choice D.

Keep in mind that we aren't accounting for any wind resistance. Considering this variable greatly complicates the problem and requires much higher level mathematics. So we just assume that there is no wind at this moment.

Answer:

[tex]area \: of \: a \: squre = {a}^{2} \\ {a}^{2} = 176 \\ a = 13.26m \\ perimeter = 4a = 4 \times 13.26 = 53.04m^2 \\ thank \: you[/tex]

Answer:

Concept:

Here, we need to know how to find the area and perimeter of the square.

Area (square) = s²

Perimeter (square) = 4s

Solve:

Given information

Area = 1764 m²

Given formula

A = s²

Substitute values into the formula

1764 = s²

Square root both sides

√s² = √1764

[tex]\boxed{s=42m}[/tex]

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

Given information

s = 42 m

Given formula

P = 4s

Substitute values into the formula

P = 4 (42)

Simplify by multiplication

[tex]\boxed{P=168m}[/tex]

Hope this helps!! :)

Please let me know if you have any questions