Need help asap ————

Solve the equation a + b - c = 180 for a. *

spin [16.1K]

Answer:

a = 180 - b + c

Step-by-step explanation:

a + b - c = 180

Move b by subtracting b from both sides.

a - c = 180 - b

Move c by adding c to both sides.

a = 180 - b + c

Therefore a equals 180 - b + c.

4 0

3 years ago

BRAINLIEST! HELP! Please help me! I NEED PART D AND E! I’m so lost :(

Zinaida [17]

Part A

Given:

Volume of square pyramid = 73.5 cubic inches

height of square pyramid = 4.5 inches

To find:

1)Side length of square pyramid.

2)if side length of square pyramid is rational or irrational

Steps:

1) Side length = $\sqrt{\frac{3v}{h} }$

Side length = $\sqrt{\frac{3*73.5}{4.5} }$

Side length = $\sqrt{\frac{3*73.5*10}{4.5*10} }$ (multiplying both sides to make it a whole number)

Side length = $\sqrt{\frac{3*735}{45} }$

Side length = $\sqrt{\frac{735}{15} }$ (3 gets canceled out)

Side length = $\sqrt{49}$

Side length = 7 inches

Therefore, the side length of the square pyramid is 7 inches long

2) 7 is a rational number, so the side length of the square pyramid is a rational number.

It is a rational number because,

it can be written in the form p/q, where

1) q $\neq$ 0

2) p and q are co-prime

3) p and q are integers

Part B

Given:

Volume of Khafre's pyramid = 9/10 of Khufu's pyramid's volume

Volume of Kristina's first pyramid = 73.5 cubic inches

To find:

Volume of Kristina's second pyramid

Steps:

Volume of Kristina's second pyramid = 9/10 the volume of Kristina first pyramid

Volume of second pyramid = $\frac{9}{10}*73.5$

Volume of second pyramid = 9 $*$ 7.35

Volume of second pyramid = 66.15 cubic inches

The volume of the second pyramid should be 66.15 cubic inches

Part C

Given:

Height of pyramid = 4.41 inches

Volume of pyramid = 66.15 cubic inches

To find:

1) Side length of pyramid

2) if the side length is rational or irrational

Steps:

1) Side length = $\sqrt{\frac{3v}{h} }$

Side length = $\sqrt{\frac{3*66.15}{4.41} }$

Side length = $\sqrt{\frac{3*66.15*100}{4.41*100} }$ (multiplying both sides to make it a whole number)

Side length = $\sqrt{\frac{3*6615}{441} }$

Side length = $\sqrt{\frac{6615}{147} }$ (3 gets canceled)

Side length = $\sqrt{45}$

Side length of pyramid is $\sqrt{45}$ inches

2) $\sqrt{45}$ cant be written in the form of a fraction, so it is an irrational number

(if u want i can explain this more)

Part D

Given

Height of model pyramid = 4.5 inches

Height of Khufu's pyramid = 755.75 feet (had to search the internet for the answer, so I am not sure if it is correct)

Side length of model pyramid = 7 inches

Side length of Khufu's pyramid = 481.4 feet (had to search the internet for the answer)

To find:

1)If the model is to scale with Khufu's pyramid

2) If the model is close to scale with Khufu's pyramid

Steps:

1) First lets find the ratio of the height to the side length of the model

Ratio = 4.5 in : 7 in

= 45 : 70

= 9 : 14

= 0.6428

Now lets find the ratio of the height to the side length of Khufu's pyramid

Ratio = 481.4 feet : 755.75 feet

= 5776.8 in : 9069 in

= 0.6369

Therefore the model is not to scale with Khufu's pyramid

2) Yes the scale is close

Part E

Given:

Height of model pyramid = 4.41 inches

Height of Khafre's pyramid = 448 feet

Side length of model pyramid = $\sqrt{45}$ inches

Side length of Khafre's pyramid = 706 feet

To find:

1) If the model is to scale with Khafre's pyramid

2) If the model is close to scale with Khafre's pyramid

Steps:

1) First lets find the ratio of the height to the side length of the model pyramid,

Ratio = 4.41 in : $\sqrt{45}$ in

= 4.41 : 6.7082

= 0.6574

Now lets find the ratio of the height to the side length of Khafre's pyramid

Ratio = 448 ft : 706 ft

= 5376 in : 8472 in

= 0.6345

Therefore the model is not to scale with Khafre's pyramid

2) No, the model is not close to scale, (it depends by what is meant by close, for me its 0.01)

Happy to help, and I hope you learnt something from this answer that will help you in the future

If you didn't understand any topic pls ask

3 0

3 years ago

the length of a soccer pitch is 20 m less than twice its width. the area of the pitch is 6000m^2. find its dimensions.

Setler79 [48]

X=width of a coccer pitch
2x-20=length of a soccer pitch
Area (rectangule)=length x width
We suggest this equation:
x(2x-20)=6000
2x²-20x=6000
2x²-20x-6000=0
x²-10x-3000=0
We solve this quadratic equation:

x=[10⁺₋√(100-4*1*-3000)]/2=[10⁺₋√(100+12000)]/2=
=(10⁺₋110)/2
we have two solutions:
x₁=(10-110)/2=-50, invalid solution.
x₂=(10+110)/2=60

x=60
2x-20=2(60)-20=120-20=100

Solution: the length is 100 m, and the width is 60 m.

To check:
Area=100 m*60 m=6000 m²
The twice of width is =2(60 m)=120 m,
20 m less than twice its width is: 120 m-20 m=100 m=the length.

8 0

3 years ago

Find the area of the figure below.

Fudgin [204]

Answer:

Area = 42 in^2

Step-by-step explanation:

here's your solution

=> divide above picture in two parts

=> rectangle and traingle

=> area of rectangle = length* width

=> area of rectangle = (5*6)in.

=> area of rectangle = 30 in^2

=> now, area of traingle = 1/2*base *height

=> area of traingle = 1/2*4*6

=> area of traingle = 12 in^2

=> area of figure = 30 in^2 + 12 in^2

=> Area = 42 in^2

hope it helps

3 0

3 years ago

Read 2 more answers

Select the equivalent expression

svetlana [45]

Answer:

x^-28 / 7^56

Step-by-step explanation:

6 0

3 years ago

Need help asap —————-