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4 years ago

Work out the lowest common multiple of 42 and 63

Mathematics

2 answers:

Tomtit [17]4 years ago

4 0

The lowest common multiple, I think is 7?

Andru [333]4 years ago

3 0

7 would be your answer if im not mistaking

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X increased by 6 algebraic expression

ddd [48]

Step-by-step explanation:

x increased by 6 is :

= x+6

3 0

3 years ago

Read 2 more answers

Pls help answer that shows work will get brainliest

hichkok12 [17]

The anwser to the problem is C.

7 0

3 years ago

How can you undo the addition of 7 on the left side of this equation? 6x + 7 = 19

BaLLatris [955]

6x - 7 = 19
6x = 19 + 7 
6x = 26 (divide both sides by 6 to get x) 
6x/6 = 26/6 
x = 4.33333333333

8 0

4 years ago

What is the solution for x in the equation 0.5x+4+0.9x=x+5

lorasvet [3.4K]

Answer: x=0.4

Step-by-step explanation:

0.5x + 4+0.9x = x+5

1.4x-x= +1

0.4x=1

X=0.4

Hope this helps

3 0

3 years ago

We take a milk carton from the refrigerator and put it on the table at room temperature. We assume that the temperature of the c

Kisachek [45]

We can solve this problem using separation of variables.

Then apply the initial conditions

EXPLANATION

We were given the first order differential equation

$\frac{dT}{dt}=k(T-a)$

We now separate the time and the temperature variables as follows,

$\frac{dT}{T-a}=kdt$

Integrating both sides of the differential equation, we obtain;

$ln(T-a)=kt +c$

This natural logarithmic equation can be rewritten as;

$T-a=e^{kt +c}$

Applying the laws of exponents, we obtain,

$T-a=e^{kt}\times e^{c}$

$T-a=e^{c}e^{kt}$

We were given the initial conditions,

$T(0)=4$

Let us apply this condition to obtain;

$4-20=e^{c}e^{k(0)}$

$-16=e^{c}$

Now our equation, becomes

$T-a=-16e^{kt}$

or

$T=a-16e^{kt}$

When we substitute a=20,
we obtain,

$T=20-16e^{kt}$

b) We were also given that,

$T(5)=8$

Let us apply this condition again to find k.

$8=20-16e^{5k}$

This implied

$-12=-16e^{5k}$

$\frac{-12}{-16}=e^{5k}$

$\frac{3}{4}=e^{5k}$

We take logarithm to base e of both sides,

$ln(\frac{3}{4})=5k$

This implies that,

$\frac{ln(\frac{3}{4})}{5}=k$

$k=-0.2877$

After 15 minutes, the temperature will be,

$T=20-16e^{-0.2877\times 15}$

$T=20-0.21376$

$T=19.786$

After 15 minutes, the temperature is approximately 20°C

6 0

3 years ago