Simple interest.

2700=2200(1+r)^5

Answers

Answer 1

Answer: solve your equation step-by-step.

2700=2200(1+r)^5

Flip the equation.

2200(r+1)^5 = 2700

Divide both sides by 2200.

2200(r+1)^5 / 2200 = 2700 / 2200

(r+1)^5=27/22

Solve Exponent.

(r+1)^5=27/22

((r+1)5) ^1/5 = (27/22)^1/5 (Raise both sides to power)

r+1=1.041809

r+1−1=1.041809−1(Subtract 1 from both sides)

r = 0.041809

Check answers. (Plug them in to make sure they work.)

r = 0.041809 (Works in original equation)

Answer 2

Answer: $2700=2200(r+1)^5\n(27)/(22)=(r+1)^5\n\sqrt[5]{(27)/(22)}=\sqrt[5]{(r+1)^5}\n(27)/(22)^(1)/(5)=r+1\nr=(27)/(22)^(0.2)-1\nr=1.0418-1\nr=0.0418$

r = 0.0418

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1. Classify the figure in as many ways as possible.rectangle; square; quadrilateral;
parallelogram; rhombus
rectangle; square; parallelogram
rhombus; quadrilateral; square
square; rectangle; quadrilateral

Answers

Answer:

(A)

Step-by-step explanation:

From the figure, we can see that all the sides of the given figure are congruent, all the four angles are right angled.

Thus, following the properties of quadrilateral,it has four sides (edges) , four vertices (corners) and the interior angles that add to 360 degrees.

Thus, the given figure is a quadrilateral.

Following the properties of parallelogram, A parallelogram has opposite sides equal and also the opposite angles equal. In the given figure, opposite sides are equal and also the opposite angles are equal.

Thus, the given figure is parallelogram.

Following the properties of rectangle, A rectangle has four sides, all the angles are right angled and has opposite sides equal,the given figure satisfies these properties.

Thus, the given figure is rectangle.

Following the properties of square, A square has all the sides equal and all the angles are of 90°, the given figure satisfies these properties.

Thu, the given figure is square.

Following the properties of rhombus, A rhombus has four sides, all the angles are right angled and has opposite sides equal, the given figure satisfies these properties.

Thus, the given figure is rhombus.

Answer:

1. classify the figure in as many ways as possible

rectangle, square, quadrilateral, parallel, rhombus.

2. which statement is true

All rectangles are quadrilaterals.

3. Lucinda wants to build a square sandbox

arrange four equal-length sides so the diagonals are also equal lengths.

4. j and m are base angles of isosceles trapezoid jklm.

151 degrees

5. determine whether the statement below is always, sometimes, or never true.

Never true.

Find the slope of the line containing the points (-1,6) and (2,1).

Answers

$(x_1;\ y_1);\ (x_2;\ y_2)\n\nm=(y_2-y_1)/(x_2-x_2)\n==========================\n\n(-1;\ 6);\ (2;\ 1)\n\nm=(1-6)/(2-(-1))=(-5)/(2+1)=(-5)/(3)=-1(2)/(3)$

Write an equation for the line below. Show work

Answers

x=3 is your answer. This is because it only crosses the x-axis and this occurs at positive 3. y=3 on the other hand would be a horizontal line that only touches positive 3 on the y-axis.

The library is to be given 5 books as a gift. the books will be selected from a list of 21 titles. if each book selected must have a different title, how many possible selections are there?

Answers

This is a combination problem. Order does not matter and the books are all different. You would write this as 21C5 or $\tbinom{21}{5}$ . Combinations are not fractions.

The answer is $(21 * 20 * 19 * 18 * 17)/(1 * 2 * 3 * 4 * 5)$

which = 20349 different ways these books can be chosen.

Answer:

20,349.

Step-by-step explanation:

We have been that a library is to be given 5 books as a gift. the books will be selected from a list of 21 titles.

We will combinations to solve our given problem.

$_(r)^(n)\textrm{C}=(n!)/(r!(n-r)!)$

Different ways to choose 5 books from 21 tiles:

$_(5)^(21)\textrm{C}=(21!)/(5!(21-5)!)$

$_(5)^(21)\textrm{C}=(21!)/(5!(16)!)$

$_(5)^(21)\textrm{C}=(21*20*19*18*17*16!)/(5*4*3*2*1*(16)!)$

$_(5)^(21)\textrm{C}=(21*19*18*17)/(3*2)$

$_(5)^(21)\textrm{C}=(21*19*3*17)/(1)$

$_(5)^(21)\textrm{C}=20,349$

Therefore, there are 20349 possible selections.

Two pound bags of candy are being used to fill containers that hold 1 1/2 pounds each. How many containers can be filled by each two pound bag of candy?

Answers

Answer:

4 Containers Can Be Filled By Each Two Pound Bag Of Candy .

Step-by-step explanation:

GIven Two Pound Bag Of Candy Used To Fill Container That hold $1(1)/(2)$ Pound Each .

Now There are 4 Container of Capacity $1(1)/(2)$ Pound Needed To fill Exactly 3 two Pound Bags .

Thus 4 container have Total capacity Of 6 Pound . Which Is Easily Filled By The 3 two Pound Bags . ( Without Any Loss )

The right rectangular prism is made up of 100 cubes. Each cube has edge lengths of 2 inches. What is the volume of the right rectangular prism, in cubic inches

Answers

Answer:

800 inches³

Step-by-step explanation:

∵ The volume of each cube is s³

∵ The length of the side is 2 inches

∴ The volume of the cube = 2³ = 8 inches³

∵ The prism consists of 100 cubes

∴ The volume of the prism = 100 × 8 = 800 inches³

Another method:

We divide the number of cubes to form the rectangular prism

5 cubes as a length , 2 cubes as a width and 10 cubes as height

∵ The length of the side of each cube is 2 inches

∴ The length = 5 × 2 = 10 inches

∴ The width = 2 × 2 = 4 inches

∴ The height = 10 × 2 = 20 inches

∵ The volume of the prism = L × W × H

∴ V = 10 × 4 × 20 = 800 inches³

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Simple interest.2700=2200(1+r)^5

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Simple interest.

2700=2200(1+r)^5