What are the first four multiples of the number 6

What is 2.26 (6 repeating) as a fraction?

Gekata [30.6K]

Answer:

34/15

Step-by-step explanation:

I recognized the form of this decimal as a number divided by 3. However, when I multiplied it by 3, I got 6.8. This means that the fraction was not fully simplified. Multiplying by 5, I got 34. Since I had to multiply by 3 and then by 5 to get there, I multiplied by a total of 15. Hope this helps!

3 0

3 years ago

The perimeter of UVW is 22 2/1. And VZ = 2 1/2. Find UW and VW.

Cerrena [4.2K]

UW and VW are the same so 9
22.5-4.5= 18 then divide the 18 by 2 so 9

4 0

3 years ago

Read 2 more answers

Robert slices a large loaf of bread to make 12 sandwiches. He made 3 turkey sandwiches and 5 veggie sandwiches. The rest are ham

lions [1.4K]

We already know that there are 12 sandwiches.

If there are 3 turkey and 5 veggie then there's 8 out of 12 sandwiches that aren't ham.

12 sandwiches - 8 non-ham sandwiches = 4 ham sandwiches

Now we know there are 4/12 ham sandwiches

We need to simplify this now.
4/12 can be divided by 4

4 goes into 4 1 time and 4 goes into 12 3 times making the answer 1/3

1/3 of the sandwiches are ham.

3 0

3 years ago

Which expression is equivalent to n+n+n? A. 3^n B. 3n C. n^3 D. n+3

Irina18 [472]

Answer:

B. 3n

Step-by-step explanation:

n + n + n = 3n

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3n = n + n + n

#IhopeThisHelp

5 0

2 years ago

1.) Find the length of the arc of the graph x^4 = y^6 from x = 1 to x = 8.

xxTIMURxx [149]

First, rewrite the equation so that y is a function of x :

$x^4 = y^6 \implies \left(x^4\right)^{1/6} = \left(y^6\right)^{1/6} \implies x^{4/6} = y^{6/6} \implies y = x^{2/3}$

(If you were to plot the actual curve, you would have both $y=x^{2/3}$ and $y=-x^{2/3}$ , but one curve is a reflection of the other, so the arc length for 1 ≤ x ≤ 8 would be the same on both curves. It doesn't matter which "half-curve" you choose to work with.)

The arc length is then given by the definite integral,

$\displaystyle \int_1^8 \sqrt{1 + \left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx$

We have

$y = x^{2/3} \implies \dfrac{\mathrm dy}{\mathrm dx} = \dfrac23x^{-1/3} \implies \left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2 = \dfrac49x^{-2/3}$

Then in the integral,

$\displaystyle \int_1^8 \sqrt{1 + \frac49x^{-2/3}}\,\mathrm dx = \int_1^8 \sqrt{\frac49x^{-2/3}}\sqrt{\frac94x^{2/3}+1}\,\mathrm dx = \int_1^8 \frac23x^{-1/3} \sqrt{\frac94x^{2/3}+1}\,\mathrm dx$

Substitute

$u = \dfrac94x^{2/3}+1 \text{ and } \mathrm du = \dfrac{18}{12}x^{-1/3}\,\mathrm dx = \dfrac32x^{-1/3}\,\mathrm dx$

This transforms the integral to

$\displaystyle \frac49 \int_{13/4}^{10} \sqrt{u}\,\mathrm du$

and computing it is trivial:

$\displaystyle \frac49 \int_{13/4}^{10} u^{1/2} \,\mathrm du = \frac49\cdot\frac23 u^{3/2}\bigg|_{13/4}^{10} = \frac8{27} \left(10^{3/2} - \left(\frac{13}4\right)^{3/2}\right)$

We can simplify this further to

$\displaystyle \frac8{27} \left(10\sqrt{10} - \frac{13\sqrt{13}}8\right) = \boxed{\frac{80\sqrt{10}-13\sqrt{13}}{27}}$

7 0

3 years ago