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2 years ago

Astatine call is the ratio it to men to women is 5 to 4 if there are 2800 men how many women are there

Mathematics

1 answer:

Mumz [18]2 years ago

4 0

Answer: 2240

Step-by-step explanation: Hope it helps

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Convert 1 Iinto an improper fraction.

FromTheMoon [43]

Answer:

only mixed number can be changed into improper fraction according to my khowlage of grade7

Step-by-step explanation:

thank you

3 0

3 years ago

Need help please thank you

umka21 [38]

2 cakes - Theresa's and Joe's
Theresa's cake had 6 pieces after she cut it. (2 times the size of Joe's pieces)
Joe's cake had 12 pieces after he cut it. (1/2 the size of Theresa's pieces)

We know that 8/12ths of ONE cake were eaten and that Joe ate 2 of his pieces.
We want to know how many pieces Theresa ate of her cake. Keeping in mind that her pieces are equal to 2 of Joe's pieces we can solve this question.

8/12 eaten total
if 2/12 by Joe
then 8-2 = 6, 6/12 by Theresa
(BUT: Theresa's pieces were twice the size of Joe's so we will divide by 2)
6/12 = 3/6

Answer: Theresa ate 3 pieces of her cake

7 0

3 years ago

G find the area of the surface over the given region. use a computer algebra system to verify your results. the sphere r(u,v) =

Svetach [21]

Presumably you should be doing this using calculus methods, namely computing the surface integral along $\mathbf r(u,v)$ .

But since $\mathbf r(u,v)$ describes a sphere, we can simply recall that the surface area of a sphere of radius $a$ is $4\pi a^2$ .

In calculus terms, we would first find an expression for the surface element, which is given by

$\displaystyle\iint_S\mathrm dS=\iint_S\left\|\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

$\dfrac{\partial\mathbf r}{\partial u}=a\cos u\cos v\,\mathbf i+a\cos u\sin v\,\mathbf j-a\sin u\,\mathbf k$
$\dfrac{\partial\mathbf r}{\partial v}=-a\sin u\sin v\,\mathbf i+a\sin u\cos v\,\mathbf j$
$\implies\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}=a^2\sin^2u\cos v\,\mathbf i+a^2\sin^2u\sin v\,\mathbf j+a^2\sin u\cos u\,\mathbf k$
$\implies\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|=a^2\sin u$

So the area of the surface is

$\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=\pi}\int_{v=0}^{v=2\pi}a^2\sin u\,\mathrm dv\,\mathrm du=2\pi a^2\int_{u=0}^{u=\pi}\sin u$
$=-2\pi a^2(\cos\pi-\cos 0)$
$=-2\pi a^2(-1-1)$
$=4\pi a^2$

as expected.

6 0

3 years ago

The sum of three prime numbers (other than two) is always odd.

grandymaker [24]

Step-by-step explanation:

The statement in the above question is True.

Sum of three prime numbers (other than two) is always odd.

Going by Christian Goldbach number theory ,

Goldbach stated that every odd whole number greater than 5 can be written as sum of three prime numbers .

Lets take an example,

3 + 3 + 5 = 11
3 + 5 + 5 = 13
5 + 5 + 7 = 17

Later on in 2013 the Mathematician <u>Harald Helfgott</u> proved this theory true for all odd numbers greater than five.

6 0

3 years ago

Does it matter if you add azero when you divide a dcimal.

Brilliant_brown [7]

No, it won't change the outcome.

3 0

3 years ago

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