# A giraffe is 5 m 20cm tall. An Elephant is 1m 77cm shorter than the giraffe. A rhinoceros is 1m 58 cm shorter than the elephant. How tall is the rhinoceros

## Related Questions

Answer the questions in the picture please

Step-by-step explanation:

1) angle 2 and 4

2)angle 2 and 3

3)angle 1 and 4

Hope it helps

Write a decimal that is less than 3.45 but greater than 3.0. helppp!

any decimal between 3.0 and 3.45. You can,use 3.25 if u don't wanna think about it

Answer:There are mutipal answers but here’s one 3.1

Step-by-step explanation:

What is the solution to the equation One-fourth x minus one-eighth = Start Fraction 7 Over 8 End Fraction + one-half x? x = negative 5 x = negative 4 x = 4 x = 5.

Answer:x = negative 4

Step-by-step explanation:

The given equation is expressed as

1/4 × x - 1/8 = 7/8 + 1/2 × x

x/4 - 1/8 = 7/8 + x/2

First step is to find the lowest common multiple of the left hand side of the equation and the right hand side of the equation. Then, we would multiply both sides of the equation by the lowest common multiple. The lowest common multiple is 8. Therefore

x/4 × 8 - 1/8 × 8 = 7/8 × 8 + x/2 × 8

2x - 1 = 7 + 4x

7 + 4x = 2x - 1

Subtracting 2x and 7 from the left hand side of the equation and the right hand side of the equation, it becomes

7 - 7 + 4x - 2x = 2x - 2x - 1 - 7

2x = - 8

x = - 8/2 = - 4

Step-by-step explanation:

-4

4a+2(b+5a)+7 Answer with work I would really like it if the work was written if no that is ok ANSWER ASAP

Step-by-step explanation:

4a+2(b+5a)+7

4a+2b+10a+7

4a+10a+2b+7

14a+2b+7

The answer is 14a+2b+7

CIO and MAP are supplementary MAP measures five times the other angle. What is the complement of the smaller anglo?
Provide the equation to solve for the unknown angle
Measure of Angle CIO
MAP
Complement of smaller angle

Measure of CIO = 30°

Measure of MAP = 150°

Complement of smaller angle = 60°

Step-by-step explanation:

Given that <CIO and <MAP are supplementary, therefore:

m<CIO + m<MAP = 180°

Let x be m<CIO

m<MAP = 5x

Thus:

x + 5x = 180

Solve for x

6x = 180

Divide both sides by 6

x = 30°

The smaller angle = m<CIO = x = 30°

The bigger angle = m<MAP) = 5x = 5(30) = 150°

Complement of the smaller angle = 90 - 30 = 60°

Find the radius and height of a cylindrical soda can with a volume of 256cm^3 that minimize the surface area.B: Compare your answer in part A to a real soda can, which has a volume of 256cm^3, a radius of 2.8 cm, and a height of 10.7 cm, to conclude that real soda cans do not seem to have an optimal design. Then use the fact that real soda cans have a double thickness in their top and bottom surfaces to find the radius and height that minimizes the surface area of a real can (the surface area of the top and bottom are now twice their values in part A.

New height=?

A) Radius: 3.44 cm.

Height: 6.88 cm.

B) Radius: 2.73 cm.

Height: 10.92 cm.

Step-by-step explanation:

We have to solve a optimization problem with constraints. The surface area has to be minimized, restrained to a fixed volumen.

a) We can express the volume of the soda can as:

This is the constraint.

The function we want to minimize is the surface, and it can be expressed as:

To solve this, we can express h in function of r:

And replace it in the surface equation

To optimize the function, we derive and equal to zero

The radius that minimizes the surface is r=3.44 cm.

The height is then

The height that minimizes the surface is h=6.88 cm.

b) The new equation for the real surface is:

We derive and equal to zero

The radius that minimizes the real surface is r=2.73 cm.

The height is then

The height that minimizes the real surface is h=10.92 cm.

The minimal surface area for a cylindrical can of 256cm^3 is achieved with radius 3.03 cm and height 8.9 cm under uniform thickness, and radius 3.383 cm and height 7.14 cm with double thickness at top and bottom. Real cans deviate slightly from these dimensions possibly due to practicality.

### Explanation:

For a cylinder with given volume, the surface area A, radius r, and height h are related by the formula A = 2πrh + 2πr^2 (if the thickness is uniform) or A = 3πrh + 2πr^2 (if the top and bottom are double thickness). By taking the derivative of A w.r.t r and setting it to zero, we can find the optimal values that minimize A.

For a volume of 256 cm^3, this gives us r = 3.03 cm and h = 8.9 cm with uniform thickness, and r = 3.383 cm and h = 7.14 cm with double thickness at the top and bottom. Comparing these optimal dimensions to a real soda can (r = 2.8 cm, h = 10.7 cm), we see that the real can has similar but not exactly optimal dimensions. This may be due to practical considerations like stability and ease of holding the can.