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2 years ago

What is the area of the composite figure?

Mathematics

1 answer:

dangina [55]2 years ago

7 0

Answer:

i think its 96cm

Step-by-step explanation:

i hope lol

You might be interested in

Y varies directly as x and y =18 when x=1/2 write an equation that relates y and x

marusya05 [52]

Answer:

y = 36x

Step-by-step explanation:

Given that y varies directly as x, then the equation relating them is

y = kx ← k is the constant of variation

To find k use the condition y = 18 when x = $\frac{1}{2}$ , thus

k = $\frac{y}{x}$ = $\frac{18}{\frac{1}{2} }$ = 36

y = 36x ← equation of variation

6 0

2 years ago

The sum of two numbers is 19. The difference is 9. What are the two numbers?

Nostrana [21]

Answer:

the two numbers are 8 and 8

7 0

2 years ago

Read 2 more answers

I need help. Work has to be shown but i don’t know how to do them

mote1985 [20]

Step-by-step explanation:

1% of 500 is 5 so multiply 5 by 8 for one year. which is 40 so multiply that by 2.

1% of 500 = 5

5 × 8 = 40

40 × 2 = 80

1% of 1000 = 10

10 × 5 = 50

50 × 3 = 150

12 months = 6%

6/12 = 1 month

0.5 = 1 month

0.5 × 9 = 4.5 (the amount of interest for 9 months.)

800 × 4.5% = 36

12 months = 7%

7/12 = 1 month ( i'll leave this in fraction form because of the decimal points.)

7/12 x 8 = 4.666667 or 4.7 rounded.

1200 × 4.7% = 56.40

8 0

2 years ago

Pls show full working out

sweet-ann [11.9K]

Answer:

Question 1a. i) $x=1.8m$

Question 1a. ii) $Volume=27.6m^3$

Question 1b) $Volume=65.9m^3$

Explanation:

Question 1 a. i) Find the value of x.

$tan(\theta )=\dfrac{opposite\text{ }leg}{adjacent\text{ }leg}$

For the smalll triangle you can write:

$tan(\theta )=\dfrac{x}{1m}$

For tthe big triangle:

$tan(\theta )=\dfrac{x+2.7m}{2.5m}$

Substitute:

$\dfrac{x}{1m}=\dfrac{x+2.7m}{2.5m}$

Solve for x:

$2.5x=x+2.7m\\\\2.5x-x=2.7m\\\\1.5x=2.7m\\\\x=2.7m/1.5\\\\x=1.8m$

Question 1a ii) Find the volume of the frustrum

Find the volume of a cone with height = 2.7m + 1.8m = 4.5m, and radius = 2.5m

Formula:

$Volume=(1/3)\pi \times radius^2\times height$

Substitute:

$Volume=(1/3)\pi \times (2.5m)^2\times 4.5m=9.375\pi m^3$

Find the volume of a cone with heigth = 1.8m and radius = 1m

$Volume=(1/3)\pi \times (1m)^2\times 1.8m=0.6\pi m^3$

Subtract the volume of the small cone from the volume of the big cone

$Volume\text{ }of\text{ }frustrum=9.375\pi m^3-0.6\pi m^3=8.775\pi m^3\approx 27.6m^3$

Question 1b. Calculate the volume of the bin

i) Upper frustrum

This is the same frustrum from the equation of above, thus ist volume is 27.6m³.

ii) Lower frustrum

$\dfrac{x}{2.0m}=\dfrac{x+2.4m}{2.5m}$

$2.5x=2(x+2.4m)\\\\2.5x=2x+4.8m\\\\0.5x=4.8m\\\\x=9.6m$

$Volume=(1/3)\pi \times (2.5m)^2\times (9.6m+2.4m)-(2.0m)^2\times (9.6m)$

$Volume=38.3m^3$

iii) Add the volume of the two frustrums

$Volume=27.6m^3+38.3m^3=65.9m^3$

6 0

3 years ago

How many 1-centimeter cubes does it need to fill the rectangular prism below.

7nadin3 [17]

Answer:

12 cubes

Step-by-step explanation:

To find out how many cubes can fit into the rectangular prism, we need to find the volume.

Volume is Length × width × height

$6$ × $2$ × $1=12$

The volume is 12 units cubed.

We can fit 12 1-centimeter cubes in the rectangular prism.

Hope this helps :)

6 0

2 years ago