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3 years ago

What are the first 6 nonzero multiples of 15?

Mathematics

2 answers:

Sphinxa [80]3 years ago

8 0

5,10,15,20,25,30,35,40,45,50,55,60 or 6,12,18,24,30,36,42,48,54,60,66,72

Naya [18.7K]3 years ago

6 0

1x15=15
2x15=30
3x15=45
4x15=60
5x15=75

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Robin is twice as old as Earl, and in 6 years he will be three times as old as Adam, who is 8 years younger than Earl. Find thei

valkas [14]

Answer:

Since, Robin is twice as old as Earl,

Thus, the ratio of the age of robin and earl = 2 : 3

Let the age of robin = 2x and the age of earl = 3x

Where x is any number,

According to the question,

3×Age of Adam = 2x + 6

Age of Adam = 1/3(2x+6)

Also, Age of Adam = Age of earl - 8

⇒ 1/3(2x+6) = 3x - 8

⇒ 2x + 6 = 9x - 24

⇒ -7x = - 30

⇒ x = 30/7

Hence, the age of Robin = 2x = 2×30/7 = 60/7 years

The age of Earl = 3x = 3×30/7=90/7 years

The age of Adam = 34/7 years

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3 years ago

Find the equation of the line that is perpendicular to y=-2/3 and passes through (4,8)

Scilla [17]

Answer:

y -8=-2

x -4 3

3y - 24=-2 +8

y= -2/3x +32/3

Step-by-step explanation:

6 0

3 years ago

Find the value of x

anastassius [24]

It would be 67 because its the same decrease as the other.

4 0

3 years ago

Which of these strategies would eliminate a variable in the system of equations? \begin{cases} 6x + 5y = 1 \\\\ 6x - 5y = 7 \end

frez [133]

Answer:

Option A and B are correct

Step-by-step explanation:

We have to add here

6x+5y=1.

6x-5y=7

sow e will get 12x=8 , x=8/12=2/3

Option A is correct.

If We substract bottom from furst then

6x+5y=1

-6x+5y=-7

10y=-6

y=-6/10=-3/5

we put y=-3/5 in any equation and find x

6x+5(-3/5)=1

6x=4

x=4/6=2/3

So Option B also correct.

5 0

3 years ago

Show that ( 2xy4 + 1/ (x + y2) ) dx + ( 4x2 y3 + 2y/ (x + y2) ) dy = 0 is exact, and find the solution. Find c if y(1) = 2.

fredd [130]

$\dfrac{\partial\left(2xy^4+\frac1{x+y^2}\right)}{\partial y}=8xy^3-\dfrac{2y}{(x+y^2)^2}$

$\dfrac{\partial\left(4x^2y^3+\frac{2y}{x+y^2}\right)}{\partial x}=8xy^3-\dfrac{2y}{(x+y^2)^2}$

so the ODE is indeed exact and there is a solution of the form $F(x,y)=C$ . We have

$\dfrac{\partial F}{\partial x}=2xy^4+\dfrac1{x+y^2}\implies F(x,y)=x^2y^4+\ln(x+y^2)+f(y)$

$\dfrac{\partial F}{\partial y}=4x^2y^3+\dfrac{2y}{x+y^2}=4x^2y^3+\dfrac{2y}{x+y^2}+f'(y)$

$f'(y)=0\implies f(y)=C$

$\implies F(x,y)=x^2y^3+\ln(x+y^2)=C$

With $y(1)=2$ , we have

$8+\ln9=C$

$\boxed{x^2y^3+\ln(x+y^2)=8+\ln9}$

8 0

3 years ago