A landscape contractor wants to make an exposed gravel border of uniform width around a rectangular pool in a garden. The pool is 10 feet long and 6 feet wide. There is enough material to cover 36 square feet. How wide should the border be

Answer:

A' = (-2, -3)

B' = (0, -3)

C' = (-1, 1)

I did this on edg

Step-by-step explanation:

Since it's a reflection over the x axis, the x coordinates will stay the same while the y coordinates will be opposite of what they are.

After reflected: A'(-2,-3), B'(0,-3), C'(-1,1)

Answer: 20

Step-by-step explanation:

Answer:

y = (1/4)x + 1

Step-by-step explanation:

The points mentioned are (-4,0) and (4,2).

We can find the slope m:

(2-0)/(4-(-4)) = 2/8 = 1/4

We now have the equation y = (1/4)x + b

Now we must find the y-intercept.

Plugging in the first point:

0 = (1/4)*(-4) + b

0 = -1 + b

1 = b

y = (1/4)x + 1

Answer:

[tex]\boxed{x = 2+2\sqrt{6} ~~or~~x = 2 - 2 \sqrt{6} }[/tex]

.

Step-by-step explanation:

To find x, use pythagoras form

[tex]c^2 = a^2 + b^2[/tex]

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

[tex]x^2 + 8x + 16 = x^2 + x^2 + 2x + 1[/tex]

[tex]2x^2 + 2x + 1 - x^2 - 8x - 16 = 0[/tex]

[tex]x^2 - 6x - 15 = 0[/tex]

[tex]x^2 - 6x = 15[/tex]

[tex](x - 3)^2 - 9 = 15[/tex]

[tex](x - 3)^2 = 15 + 9[/tex]

[tex](x - 3)^2 = 24[/tex]

[tex]x - 2 = \pm \sqrt{24}[/tex]

[tex]x = 2 \pm 2\sqrt{6}[/tex]

.

[tex]x_1 = 2 + 2\sqrt{6}[/tex]

[tex]x_2 = 2 - 2\sqrt{6}[/tex]

.

Happy to help :)

We're given the ODE,

y'' + 4y' + 3y = 8t exp(-t ) + 6 exp(-t ) - (9t + 6)

(where I denote exp(x) = eˣ )

First determine the characteristic solution:

y'' + 4y' + 3y = 0

has characteristic equation

r ² + 4r + 3 = (r + 1) (r + 3) = 0

with roots at r = -1 and r = -3, so the characteristic solution is

y = C₁ exp(-t ) + C₂ exp(-3t )

For the non-homogeneous equation, assume two ansatz solutions

y₁ = (at ² + bt + c) exp(-t )

and

y₂ = at + b

• y'' + 4y' + 3y = 8t exp(-t ) + 6 exp(-t ) … … … [1]

Compute the derivatives of y₁ :

y₁ = (at ² + bt + c) exp(-t )

y₁' = (2at + b) exp(-t ) - (at ² + bt + c) exp(-t )

… = (-at ² + (2a - b) t + b - c) exp(-t )

y₁'' = (-2at + 2a - b) exp(-t ) - (-at ² + (2a - b) t + b - c) exp(-t )

… = (at ² + (b - 4a) t + 2a - 2b + c) exp(-t )

Substitute them into the ODE [1] to get

→   [(at ² + (b - 4a) t + 2a - 2b + c) + 4 (-at ² + (2a - b) t + b - c) + 3 (at ² + bt + c)] exp(-t ) = 8t exp(-t ) + 6 exp(-t )

(at ² + (b - 4a) t + 2a - 2b + c) + 4 (-at ² + (2a - b) t + b - c) + 3 (at ² + bt + c) = 8t + 6

4at + 2a + 2b = 8t + 6

→   4a = 8   and   2a + 2b = 6

→   a = 2   and   b = 1

→   y₁ = (2t ² + t ) exp(-t )

(Note that we don't find out anything about c, but that's okay since it would have gotten absorbed into the first characteristic solution exp(-t ) anyway.)

• y'' + 4y' + 3y = -(9t + 6) … … … [2]

Compute the derivatives of y₂ :

y₂ = at + b

y₂' = a

y₂'' = 0

Substitute these into [2] :

4a + 3 (at + b) = -9t - 6

3at + 4a + 3b = -9t - 6

→   3a = -9   and   4a + 3b = -6

→   a = -3   and   b = 2

→   y₂ = -3t + 2

Then the general solution to the original ODE is

y(t) = C₁ exp(-t ) + C₂ exp(-3t ) + (2t ² + t ) exp(-t ) - 3t + 2

Use the initial conditions y (0) = 2 and y' (0) = 2 to solve for C₁ and C₂ :

y (0) = C₁ + C₂ + 2 = 2

→   C₁ + C₂ = 0 … … … [3]

y'(t) = -C₁ exp(-t ) - 3C₂ exp(-3t ) + (-2t ² + 3t + 1) exp(-t ) - 3

y' (0) = -C₁ - 3C₂ + 1 - 3 = 2

→   C₁ + 3C₂ = -4 … … … [4]

Solve equations [3] and [4] to get C₁ = 2 and C₂ = -2. Then the particular solution to the initial value problem is

y(t) = -2 exp(-3t ) + (2t ² + t + 2) exp(-t ) - 3t + 2

Answer:C

Step-by-step explanation:

click on the picture attached to get the answer

hope it helps!

..........

Answer:

2*9=18 14*13=182 18+182=200

Step-by-step explanation:

well you multiply 2 times 9 and you get 18 and then you multiply 14 times 13 and you get 182 and then you 18+182 and that will give you 382.

Answer:

72 square units

Step-by-step explanation:

The area of the rectangle in the middle: 6x8 = 48 square units

The area of the triangle on the left: (5x6)/2 = 30/2 = 15 square units

The area of the triangle on the right: (3x6)/2 = 18/2 =9

(The 6 for the height of the triangle on the right is from the left side of the rectangle in the middle, they have to be the same length)

48+15+9 = 72 square units

Answer:

Its C

Step-by-step explanation:

Step-by-step explanation:

No. of banana    Cost      Unit price

4                            2            0.50

6                            3             0.50

7                           3.5           0.50

20                         10            0.50

20                          10             0.50

33                          16.50        0.50

Cost = No of banana x unit price

Cost = 4 x 0.50 = 2

Cost = 6 x 0.50 = 3

Cost = 7 x 0.50 = 3.50

Cost = 20 x 0.50 = 10

No of banana = Cost / Unit price

No of banana = 10 / 0.50 = 20

No of banana = 16.50 / 0.50 = 33

Answer:

4h

2d

Step-by-step explanation:

Answer:

2 m/s

Step-by-step explanation:

Answer:

2m/s²

Step-by-step explanation:

acceleration= v-u/t

=3-1/1

=2m/s²

Answer:

144°

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Geometry

Step-by-step explanation:

Step 1: Define

Number of sides n = 10

Step 2: Find

Step 3: Find 1 angle

We have 10 sides.

Answer:

10

Step-by-step explanation:

give me brainly

am i correct? let me know

Answer:

B

Step-by-step explanation:

4x+2y=6

6x-4y=-12

2y=6-4x

y=3-2x

6x-4(3-2x)=-12

6x-12+8x=-12

2x-12=-12

2x=0

x=0

4(0)+2y=6

2y=6

y=3

(0,3)

Answer:

Carle is cutting pieces of string that are exactly 24 3/8 inches long. How many pieces can she cut from a ball of string that has 100 feet? 1 foot= 12 inches. A) 49 pieces.Oct 14, 2016

Step-by-step explanation:

Answer:

a

Step-by-step explanation:

because I said it was jejwjwjeiejehxueh ed uejrhcu

Answer:

580

Step-by-step explanation:

If 100 centimeters are in a meter, then we have to multiply 5.8 by 100 to find out the length of the string in centimeters.

(hope this helped)

Answer:

Her rate of change is 75 ft/h

Step-by-step explanation:

Rate of Change

It measures how one quantity varies with respect to another. The second variable usually is the time.

The climber is hiking. We are given two points of her progress: At 2 hours, she is at 400 ft. This corresponds to the point (2,400).

At 6 hours, she is at 700 ft. This is the point (6,700)

The rate of change is the slope between these two points or the quotient of the variations of each variable:

[tex]\displaystyle m=\frac{700\ ft-400\ ft}{6\ h-2\ h}[/tex]

[tex]\displaystyle m=\frac{300\ ft}{4\ h}[/tex]

[tex]m=75\ ft/h[/tex]

Her rate of change is 75 ft/h

Answer:

-9

Step-by-step explanation:

Answer:

The equation is y=2/3x-7

Step-by-step explanation:

-9-(-7) /-3-0

-2/-3

that's your answer

Answer: no? because the comma goes after every three numbers.

Answer:

$2.35 per quart (rounded to nearest cent)

Step-by-step explanation:

6 quarts --- $14.12

1 quart --- [tex]\frac{1}{6}[/tex] × $14.12 = $2.35 (rounded to nearest cent)

Answer:

angle 1 = 70 degree and angle 2=95 degree

Step-by-step explanation:

for the first angle

180 -70 = 110

and 180-110 = 70 ( opposite angles )

for the second angle

180-85 =95

and again 180-95 =85(opposite angles)

180-85=95 (the straight line property)

Answer:

Please check the explanation.

Step-by-step explanation:

Given that

|v|=38

Ф = 120°

Finding the horizontal component

The horizontal component can be obtained using the formula

Vx = |v| cos Ф

     = 38 cos 120°

     = 38 (-0.5)

      = -19

Thus, the  horizontal component is:

Vx = -19

Finding the vertical component

The vertical component can be obtained using the formula

Vy = |v| sin Ф

    = 38 sin 120°

    = 38 (0.86)

    = 32.68

Thus, the vertical component is:

Vy = -19

v = |v| cos Ф i + |v| sin Ф j

As

|v|=38

Ф = 120°

Thus, the vector is

v = 38 cos 120° i + 38 sin 120° j

or

v = -19 i + 32.68 j

Answer:

Step-by-step explanation:

(a)

From the given information:

Let An denote the amount in the account at the end of n month

Thus,

[tex]A_o[/tex] =1000

After the first month ends, interest = [tex]=A_o \times \dfrac{3}{100} \times \dfrac{1}{12}[/tex]

= [tex]0.0025A_o[/tex]

Additional deposit = $200

Now,

[tex]A_1 = A_o + 0.0025A_o + 200[/tex]

[tex]=1.0025A_o + 200 --- eqn(1)[/tex]

Interest after 2nd month, = [tex]A_1 \times \dfrac{3}{100} \times \dfrac{1}{2}[/tex]

Additional deposit = $200

[tex]A_2 = A_1 + 0.0025A_1 + 200[/tex]

[tex]=1.0025A_1 + 200 --- eqn(2)[/tex]

Thus:

[tex]A_n = 1.0025A_{n-1} +200[/tex]

(b)

From above;

[tex]A_1 = 1.0025A_ 0+ 200[/tex]

Thus;

[tex]A _2 = 1.0025A_ 1+ 200[/tex]

[tex]A_2 = 1.0025( 1.0025A_ 0+ 200)+ 200[/tex]

[tex]A_2 = 1.0025^2A_o +200 ( 1+1.0025)[/tex]

[tex]A _3 = 1.0025A_ 2+ 200[/tex]

[tex]A _3 = 1.0025( 1.0025^2A_o +200 ( 1+1.0025))+ 200[/tex]

[tex]A _3 = 11.0025^3A_o +200 ( 1+1.0025 + 1.0025^2)[/tex]

To continuity...

[tex]A_n = 1.0025^n A_o + 200 (1+1.0025 +1.0025^2 +...+ 1.0025^{n-1} ) ---- (3)[/tex]

Thus, we have a geometric term [tex]1 + 1.0025 +1.0025^2 +...+ 1.0025^{n-1}[/tex]

where:

a = 1 (first term)

r = 1.0025>1 (common ratio)

Thus;

[tex]1 + 1.0025 +1.0025^2 +...+ 1.0025^{n-1} = \dfrac{a(r^n -1)}{r-1}[/tex]

[tex]= \dfrac{1(1.0025^n -1)}{1.0025-1}[/tex]

[tex]= 400(1.0025^n -1)[/tex]

From equation(3)

[tex]A_n = (1.0025^n \times 1000) + 80000(1.0025^n -1)[/tex]

[tex]A_n = 1.0025^n \times 81000-80000[/tex]

Answer:

48

Step-by-step explanation:

80% = .8

60 x .8 = 48

Answer: 48

Step-by-step explanation:

80% of 60 or 80/100 x 60 = 48