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4 years ago

If you know the LCM of 4 and 5, how could you find the LCM of 40 and 50?

Mathematics

1 answer:

Archy [21]4 years ago

5 0

Answer:

200

Step-by-step explanation:

The LCM of 4 and 5 is 20. Since you know this, you would multiply this LCM by how much you would multiply to get 40 from 4 or 50 from 5.

4 * 10 = 40

5 * 10 = 50

Since you multiply these factors by 10, you would also multiply the LCM by 10.

20 (LCM of 4 and 5) * 10 = 200

The LCM of 40 and 50 is 200.

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A map has a scale of 1 : 20,000. The real-life distance between two schools is 8 km. What is the distance between the two school

levacccp [35]

Answer:

40 cm

Step-by-step explanation:

1 km = 10^5 cm, so 8 km = 8×10^5 cm

The map distance is 1/20,000 = 5×10^-5 of the real distance, so the map distance is ...

(5×10^-5)(8×10^5 cm) = 40 cm

3 0

4 years ago

Use vertical multiplication to find the product of:

Rom4ik [11]

Answer:

$x^6+x^4+4x^3-2x^2-x+3$

Step-by-step explanation:

$x^3+2x+3$

$\times(x^3-x+1)$

---------------------------------

First step multiply your terms in your first expression just to the 1 in the second expression like so:

$x^3+2x+3$

$\times(x^3-x+1)$

---------------------------------

$x^3+2x+3$ Anything times 1 is that anything.

That is, $(x^3+2x+3) \cdot 1=x^3+2x+3$ .

Now we are going to take the top expression and multiply it to the -x in the second expression. $-x(x^3+2x+3)=-x^4-2x^2-3x$ . We are going to put this product right under our previous product.

$x^3+2x+3$

$\times(x^3-x+1)$

---------------------------------

$x^3+2x+3$

$-x^4-2x^2-3x$

We still have one more multiplication but before we do that I'm going to put some 0 place holders in and get my like terms lined up for the later addition:

$x^3+2x+3$

$\times(x^3-x+1)$

---------------------------------

$0x^4+x^3+0x^2+2x+3$

$-x^4+0x^3-2x^2-3x+0$

Now for the last multiplication, we are going to take the top expression and multiply it to x^3 giving us $x^3(x^3+2x+3)=x^6+2x^4+3x^3$ . (I'm going to put this product underneath our other 2 products):

$x^3+2x+3$

$\times(x^3-x+1)$

---------------------------------

$0x^4+x^3+0x^2+2x+3$

$-x^4+0x^3-2x^2-3x+0$

$x^6+2x^4+3x^3$

I'm going to again insert some zero placeholders to help me line up my like terms for the addition.

$x^3+2x+3$

$\times(x^3-x+1)$

---------------------------------

$0x^6+0x^4+x^3+0x^2+2x+3$

$0x^6-x^4+0x^3-2x^2-3x+0$

$x^6+2x^4+3x^3+0x^2+0x+0$

----------------------------------------------------Adding the three products!

$x^6+x^4+4x^3-2x^2-x+3$

8 0

4 years ago

Consider the series.

Serga [27]

Put 1

$\\ \rm\Rrightarrow \left(\dfrac{3^{1+1}}{3(1)+1}\right)$

$\\ \rm\Rrightarrow 3^2/4=9/4=2.2$

Put 2

$\\ \rm\Rrightarrow \left(\dfrac{3^{2+1}}{3(2)+1}\right)=3^3/7=27/7=3.8$

R is

3.8/2.2=19/11

Its greater than 1 .

So diverge series.

8 0

2 years ago

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Snowcat [4.5K]

Answer:

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This system of inequalities models the scenario:

2x + y = <8

x + y > 4

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Part

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3 years ago

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den301095 [7]

Answer:

Step-by-step explanation:

6 0

3 years ago

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