All categories

2 years ago

A girls height is 3 1/3 feet .a giraffes heights is 3 times than the girl.how many inches than the girl ?

Mathematics

1 answer:

serious [3.7K]2 years ago

8 0

Answer:

Girl's height: 3 1/3 ft

Giraffes height: 10 ft

Difference between the heights: 6 2/3

Step-by-step explanation:

Ok, so the question is bit vague, so I'm not sure which of the following answers your are looking for, that's for you to put.

We have the girl's height.

We multiply the girl's height by 2 to get the giraffe's height.

We subtract the girls height from the giraffe's height to get the difference.

You might be interested in

Use absolute value to express the distance between −10 and 16 on the number line.

Ede4ka [16]

I think 12,13,14 that's what I think

4 0

2 years ago

Read 2 more answers

A cooling tower for a nuclear reactor is to be constructed in the shape of a hyperboloid of one sheet. The diameter at the base

timurjin [86]

Answer:

Step-by-step explanation:

Equation for a hyperboloid of one sheet, with center at the origen and axis along z-axis is:

(x/a)² + (y/b)² - (z/c)² = 1 (1)

We have to find a , b, and c

We can express equation (1)

(x/a)² + (y/b)² = (z/c)² + 1 (2)

Now if we cut the hyperboloid with planes parallel to xy plane we get for z = k ( K = 1 , 2 , 3 and so on ) circles of different radius

(x/a)² + (y/b)² = (k/c)² + 1

at z = k = 0 at the base of the hyperboloid d = 300 or r = 150 m

we have

(x/a)² + (y/b)² = 1

x² + y² = a² a² = (150)² a = b = 150

and x² + y² = (150)²

Now the other condition is at 200 m above the base d = 500 m r = 250 m minimum diameter then in equation (2) we have:

(x/a)² + (y/b)² = (z/c)² + 1

(1/a)² [ x² + y² ] = (z/c)² + 1

but x² + y² = r² and in this case r = 250 m then

(250)²/(150)² = (z/c)² + 1 ⇒ (62500/ 22500) = (200/c)² + 1

2,78 = 40000/c² + 1

2.78c² = 40000 + c²

1.78c² = 40000

c² = 40000/1.78

c² = 22471.91

c = 149,91

Then we finally have the equation:

x²/(150)² + y² /(150)² - z²/149,91 = 1

7 0

2 years ago

Baby Amelia's parents measure her height every month.

jonny [76]

The statement H(30) = H(25) + 5 means that, "when Amelia was 30 months old, she was 5 centimeters taller than when she was 25 months old"

<h3><u>Solution:</u></h3>

Given that,

Baby Amelia's parents measure her height every month

It is given that H(t) models Amelia's height (in centimeters) when she was "t" months old

Given statement is H(30) = H(25) + 5

So, we can say that H(30) = Amelia's height when she was 30 months old

H(25) = Amelia's height when she was 25 months old

So H(30) = H(25) + 5 means that,

Height of Amelia when she was 30 months old is 5 centimeter more than Amelia height when she was 25 years old

Or we can say that,

When Amelia was 30 months old, she was 5 centimeters taller than when she was 25 months old

7 0

2 years ago

f1(x) = ex, f2(x) = e−x, f3(x) = sinh(x) g(x) = c1f1(x) + c2f2(x) + c3f3(x) Solve for c1, c2, and c3 so that g(x) = 0 on the int

eimsori [14]

Answer:

(C1, C2, C3) = (K, K, -2K)

For K in the interval (-∞, ∞)

Step-by-step explanation:

Given

f1(x) = e^x

f2(x) = e^(-x)

f3(x) = sinh(x)

g(x) = 0

We want to solve for C1, C2 and C3, such that

C1f1(x) + C2f2(x) + C3f3(x) = g(x)

That is

C1e^x + C2e^(-x) + C3sinh(x) = 0

The hyperbolic sine of x, sinh(x), can be written in its exponential form as

sinh(x) = (1/2)(e^x + e^(-x))

So, we can rewrite

C1e^x + C2e^(-x) + C3sinh(x) = 0

C1e^x + C2e^(-x) + C3(1/2)(e^x + e^(-x)) = 0

So we have

(C1 + (1/2)C3)e^x + (C2 + (1/2)C3)e^(-x) = 0

We know that

e^x ≠ 0, and e^(-x) ≠ 0

So we must have

(C1 + (1/2)C3) = 0...........................(1)

and

(C2 + (1/2)C3) = 0..........................(2)

From (1)

2C1 + C3 = 0

=> C3 = -2C1.................................(3)

From (2)

2C2 + C3 = 0

=> C3 = -2C2................................(4)

Comparing (3) and (4)

2C1 = 2C2

=> C2 = C1

Let C1 = C2 = K

C3 = -2K

(C1, C2, C3) = (K, K, -2K)

For K in the interval (-∞, ∞)

3 0

2 years ago

The marching band walks 1/15 miles in 1.5 minutes. At this rate, how many minutes will it take to complete 1 mile? Record your a

ipn [44]

Answer:

It will take 22.5 minutes to complete 1 mile

Step-by-step explanation:

From the question,

The marching band walks 1/15 miles in 1.5 minutes

First we will determine the miles cover in 1 minute

If the marching band walks 1/15 miles in 1.5 minutes

then, they will cover $x$ miles in 1 minute

$x = \frac{1/15 * 1}{1.5}\\$

Then,

$x = \frac{1/15 }{1.5}$

$x = 2/45$ miles

∴ 2/45 miles are covered in 1 minute

Now,

If 2/45 miles are covered in 1 minute

Then, 1 mile will be covered in $y$ minute

$y = \frac{1 * 1}{2/45}$

$y =\frac{1}{2/45} \\y = \frac{45}{2}\\$

∴ $y = 22.5 minutes$

Hence, it will take 22.5 minutes to complete 1 mile

6 0

2 years ago