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3 years ago

What is the LCM of 3.6 and 4

Mathematics

2 answers:

ikadub [295]3 years ago

8 0

Answer:

I think the answer is 36

Step-by-step explanation:

Ira Lisetskai [31]3 years ago

6 0

Answer:

Step-by-step explanation:

36 / 3.6 = 10

36 / 4 = 9

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How many solutions does the equation 6a - 3a - 6 = 2 + 3 have?

Softa [21]

Answer:

One, a=11/3.

Step-by-step explanation:

6a-3a-6=2+3

3a-6=5

3a=5+6

3a=11

a=11/3

5 0

3 years ago

Round the number 15.39624 to 2 decimal places

hodyreva [135]

15.40 becauwe 6 makes 9to 0 and it makes 3 to 4

6 0

3 years ago

Read 2 more answers

2) Tara is trying to find the nth term of the arithmetic sequence 5,8,11, and 14…. Which expression can Tara use to find the nth

juin [17]

Answer:

Below in bold.

Step-by-step explanation:

Nth term an = a1 + d(n - 1) where a1 = term 1 and d = common difference.

Here a1 = 5 and d = 8-5 = 3.

So an = 5 + 3(n - 1)

15th term a15

= 5 + 3(15 - 1)

= 5 + 42

= 47.

4 0

3 years ago

Find the area of the regular figure below:

Mice21 [21]

Answer:

1,496.49 cm squared

Step-by-step explanation:

The regular figure below is a hexagon.

The formula for the area of a regular hexagon is $A = \frac{3\sqrt{3} }{2} a^2$ . Here a = 24. Substitute and simplify for the area.

$A = \frac{3\sqrt{3} }{2} a^2\\A = \frac{3\sqrt{3} }{2} 24^2\\A = \frac{3\sqrt{3} }{2} * 576\\A = 1496.49$

6 0

3 years ago

Find the area of the surface. The part of the surface z = xy that lies within the cylinder x2 + y2 = 36.

rewona [7]

Answer:

Step-by-step explanation:

From the given information:

The domain D of integration in polar coordinates can be represented by:

D = {(r,θ)| 0 ≤ r ≤ 6, 0 ≤ θ ≤ 2π) &;

The partial derivates for z = xy can be expressed as:

$y =\dfrac{\partial z}{\partial x} , x = \dfrac{\partial z}{\partial y}$

Thus, the area of the surface is as follows:

$\iint_D \sqrt{(\dfrac{\partial z}{\partial x})^2+ (\dfrac{\partial z}{\partial y})^2 +1 }\ dA = \iint_D \sqrt{(y)^2+(x)^2+1 } \ dA$

$= \iint_D \sqrt{x^2 +y^2 +1 } \ dA$

$= \int^{2 \pi}_{0} \int^{6}_{0} \ r \sqrt{r^2 +1 } \ dr \ d \theta$

$=2 \pi \int^{6}_{0} \ r \sqrt{r^2 +1 } \ dr$

$= 2 \pi \begin {bmatrix} \dfrac{1}{3}(r^2 +1) ^{^\dfrac{3}{2}} \end {bmatrix}^6_0$

$= 2 \pi \times \dfrac{1}{3} \Bigg [ (37)^{3/2} - 1 \Bigg]$

$= \dfrac{2 \pi}{3} \Bigg [37 \sqrt{37} -1 \Bigg ]$

3 0

2 years ago