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3 years ago

Which is a better deal: A can of 3 tennis balls for $3 or a box of 10 balls for $9?

Mathematics

1 answer:

Artemon [7]3 years ago

7 0

Answer:

10 for 9$

Step-by-step explanation:

because say the each ball cpsts a dollar, and the total for 3 would be 3$ for the cost of 10 would be 9 so its less but more mo ey spent

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Using n for the variable, write an algebraic expression that represents a number that is divisible by the given number. (Hint: A

iris [78.8K]

Answer:

'5n' is the correct answer.

Step-by-step explanation:

Let 'n' be any integer i.e. a number from the set {....., -3,-2,-1,0,1,2,3, ..... }

so 'n' can be termed as the variable here because its value can change and can be any value from the above set.

A number 'q' that can be divided by a a given number 'p' can be written as:

$n \times p$

When divided by 'p' :

$\dfrac{q}p = \dfrac{n \times p}{p}\\\dfrac{q}p = n$

So, The number 'q' is completely divisible by 'p' leaving 'n' as the quotient.

Using this concept, let us solve the questions:

a) Using 'n' as the variable, a number that is divisible by 5 can be written as:

$5n$

Read more on Brainly.com - brainly.com/question/16919676#readmore

7 0

3 years ago

Orignsl price of a comb : $0.99<br> Discount: 10%

denis23 [38]

Answer:

90 cents

Step-by-step explanation

divide .99 by 10 and that gets you 9 cents minus that off and its 90 Cents

7 0

3 years ago

Read 2 more answers

Suppose that surface σ is parameterized by r(u,v)=⟨ucos(3v),usin(3v),v⟩, 0≤u≤7 and 0≤v≤2π3 and f(x,y,z)=x2+y2+z2. Set up the sur

Bad White [126]

Looks like you have most of the details already, but you're missing one crucial piece.

$\sigma$ is parameterized by

$\vec r(u,v)=\langle u\cos3v,u\sin3v,v\rangle$

for $0\le u\le7$ and $0\le v\le\frac{2\pi}3$ , and a normal vector to this surface is

$\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\left\langle\sin3v,-\cos3v,3u\right\rangle$

with norm

$\left\|\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right\|=\sqrt{\sin^23v+(-\cos3v)^2+(3u)^2}=\sqrt{9u^2+1}$

So the integral of $f(x,y,z)=x^2+y^2+z^2$ is

$\displaystyle\iint_\sigma f(x,y,z)\,\mathrm dA=\boxed{\int_0^{2\pi/3}\int_0^7(u^2+v^2)\sqrt{9u^2+1}\,\mathrm du\,\mathrm dv}$

6 0

3 years ago

Someone please help me with this lol… have no idea what I’m doing

Sholpan [36]

Given:

$\cos \theta =\dfrac{3}{5}$

$\sin \theta$

To find:

The quadrant of the terminal side of $\theta$ and find the value of $\sin\theta$ .

Solution:

We know that,

In Quadrant I, all trigonometric ratios are positive.

In Quadrant II: Only sin and cosec are positive.

In Quadrant III: Only tan and cot are positive.

In Quadrant IV: Only cos and sec are positive.

It is given that,

$\cos \theta =\dfrac{3}{5}$

$\sin \theta$

Here cos is positive and sine is negative. So, $\theta$ must be lies in Quadrant IV.

We know that,

$\sin^2\theta +\cos^2\theta =1$

$\sin^2\theta=1-\cos^2\theta$

$\sin \theta=\pm \sqrt{1-\cos^2\theta}$

It is only negative because $\theta$ lies in Quadrant IV. So,

$\sin \theta=-\sqrt{1-\cos^2\theta}$

After substituting $\cos \theta =\dfrac{3}{5}$ , we get

$\sin \theta=-\sqrt{1-(\dfrac{3}{5})^2}$

$\sin \theta=-\sqrt{1-\dfrac{9}{25}}$

$\sin \theta=-\sqrt{\dfrac{25-9}{25}}$

$\sin \theta=-\sqrt{\dfrac{16}{25}}$

$\sin \theta=-\dfrac{4}{5}$

Therefore, the correct option is B.

6 0

3 years ago

Please help Really confusied dont do the last 2 at the bottom

Helen [10]

Number one or 2? so i can help you

6 0

3 years ago