In the figure, polygon ABCD is transformed to create polygon A’B’C’D’. This transformation is a

Solve the following addition and subtraction problems. a. 3 km 9 hm 9 dam 19 m + 7 km 7 dam b. 5 sq.km 95 ha 8,994 sq.m + 11 sq.

kipiarov [429]

As a general rule to solve the problem we are going to transform all values to the lower unit.
a. 3 km 9 hm 9 dam 19 m + 7 km 7 dam
3,000 m 900 m 90 m 19 m + 7,000 m 70 m = 4,009 + 7,070 = 11,079 m
b. 5 sq.km 95 ha 8,994 sq.m + 11 sq. km. 11 ha 9,010 sq. m.
5,000,000 sq m 95,0000 sq m 8,994 sq m + 11,000,000 sq m 110,000 sq
9,010 sq m
5,103,994 sq m + 11,119,010 sq m = 16,223,004 sq m
c. 44 m â€“ 5 dm
44 m - 0.5 m = 43.5 m
d. 73 km 47 hm 2 dam - 11 km 55 hm

73,000 m 4,700 m 20 m - 11,000 m 5,500 m
77,720 m - 16,500 m = 61,220 m

5 0

2 years ago

6 questions = 30 points someone please help me

eduard

Answer:

1)D

2)D

3)A

4)B

5)Cannot Answer - No data

6)C

7)C

6 Questions answered for 30 points.

Step-by-step explanation:

Answer for the 1st Question,

Area of a cylinder can be calculated by,

Area of the cylindrical part = $2\pi rL$

Area of the parts that cover the open parts of cylinder = $2\pi r^2$

Therefor Total are of the cylinder = $2\pi rL+2\pi r^2$

Area of the total cylinder = $2*3.14*6.8*14.2+2*3.14*6.8^2$

= $896.784$

Therefor best estimate will be D) 923 in^2

Answer for the 2nd Question,

ABC is a triangle.

By Pythagorean theorem it says,

$(AC)^2+(CB)^2=(AB)^2$

The distance from A to C = 6

The distance from C to B = 5

$(AC)^2=6^2=36$

$(CB)^2=5^2=25$

$(AC)^2+(CB)^2=(AB)^2=36+25=61$

Therefor $(AB)^2=61\\(AB)=\sqrt{61}$

Therefor Answer is D) Sqr 61

Answer to Question 3

To find the trend you can draw an imaginary line that fits the scatter plots.(I have attached a image drawing the imaginary lines)

Then the revenue can be calculated by calculating the area covered by the imaginary line drawn or area "under" the curve.

Area of the Store 1 = $\frac{1}{2} *(110+40)*110=8250$

Area of the Store 2= $\frac{1}{2} *110*80=4400$

Therefore Revenue from Store 1 = $8250

Revenue from Store 2 = $4400

Answer is A) Store 1 has more sales revenue

Answer for Question 4

$\frac{5}{12}=0.4167\\\frac{5}{11}=0.4545$

You can straight away discard answer D because it is less than both 5/12 and 5/11.

Also you can discard Answer C because 25/264 is very small (less than 1/10=0.1)

$\frac{11}{23} =0.4782$ , therefor this is not in between those two numbers.

Then it has to be $\frac{115}{264} =0.4356$

Answer is B)

Answer for question 6

The equation of a straight line is of the form $y=mx+c$

here m is the gradient and c is the intercept.

Find the intercept(c) by finding the y value when x value is 0, here it is +1.

Gradient (m) of the graph can be found by,

$\frac{y1-y2}{x1-x2} =\frac{7-3}{3-1} =2$

Therefor the equation of the graph is,

$y=2x+1$

Answer is C

Answer for question 7

$\frac{7}{8} =0.875$

$\frac{6}{9} =0.67$

We can discard answer A because the value is close to 0.5

We can discard answer B because 3/17 is a very small value (0.17)

$\frac{53}{72} =0.73$ , This falls between the above 2 numbers.

Therefor answer is C

5 0

2 years ago

Jose wants to start a restaurant and is writing a business proposal. His rent is $1,000 per month. The monthly operating costs a

Harman [31]

Answer:

Revenue = 5( 1000 + w+ k)

Step-by-step explanation:

Lets calculate expenses first

rent = $1000

wages = w

overhead cost = k

Total cost = rent + wage+ overhead

Total cost = 1000 + w+ k

Given Jose wants his monthly sales revenue, x, to be 4 times greater than the operating costs

Let revenue be R

Thus, putting the statement mathematically

revenue - total cost = 4 * operating cost

R - ( 1000 + w+ k) = 4( 1000 + w+ k)

R = 4( 1000 + w+ k) + ( 1000 + w+ k)

R = 5( 1000 + w+ k)

Thus, revenue should be 5( 1000 + w+ k).

4 0

2 years ago

Find the smallest number to be added to 11650 so that the sum will be a perfect square

sergejj [24]

Answer:

14

Step-by-step explanation:

Take the square root of 11650+14:

%:11650+14

108

So it’s a perfect square (108^2=11664).

3 0

2 years ago

A fruit basket contains oranges and grape fruits. One-third of the oranges and one-fourth of the grapefruits were spoiled. You t

den301095 [7]

So if 1/3 0f the oranges equals 4, than you times it by three to get 3/3, so there were 12 oranges and 28 grapefruits, so the answer is 40

5 0

2 years ago