What is the equation of the line that passes through the point (6,3) and has a slope of -1/3

Alina and Asim share Rs 210 in the ratio 2:5. How much money did each of them get?

slavikrds [6]

Answer:

Step-by-step explanation:

Alina=2x---eq(1)

Asim=5x----eq---(2)

So,we get

2x+5x=210

7x=210

x=30

Now,Putting '30' in eq 1 and eq 2

2*30=60

5*30=150

Therefore Alina got 60 Rs while Asim got 150 Rs

5 0

2 years ago

Estimate the answer for 1 2/3 x 5 1/10 (type a whole number

Westkost [7]

Answer:

Exact Form:

17/2

Decimal Form:

8.5

Mixed Number Form:

8 1/2

6 0

2 years ago

Adele is 5 years older than Timothy. In three years, Timothy will be of Adele’s age. What is Adele’s current age?

Ede4ka [16]

Adele is going to be 12 years old

5 0

2 years ago

(x +y)^5<br> Complete the polynomial operation

Vesna [10]

Answer:

Please check the explanation!

Step-by-step explanation:

Given the polynomial

$\left(x+y\right)^5$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=x,\:\:b=y$

$=\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i$

so expanding summation

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

solving

$\frac{5!}{0!\left(5-0\right)!}x^5y^0$

$=1\cdot \frac{5!}{0!\left(5-0\right)!}x^5$

$=1\cdot \:1\cdot \:x^5$

$=x^5$

also solving

$=\frac{5!}{1!\left(5-1\right)!}x^4y$

$=\frac{5}{1!}x^4y$

$=\frac{5x^4y}{1}$

$=5x^4y$

similarly, the result of the remaining terms can be solved such as

$\frac{5!}{2!\left(5-2\right)!}x^3y^2=10x^3y^2$

$\frac{5!}{3!\left(5-3\right)!}x^2y^3=10x^2y^3$

$\frac{5!}{4!\left(5-4\right)!}x^1y^4=5xy^4$

$\frac{5!}{5!\left(5-5\right)!}x^0y^5=y^5$

so substituting all the solved results in the expression

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

$=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

Therefore,

$\left(x\:+y\right)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

6 0

2 years ago

ABCD is a parallelogram. Segment AC = 4x + 10. Find the value of x and y.

valkas [14]

Answer:

The values of x and y in the diagonals of the parallelogram are x=0 and y=5

Step-by-step explanation:

Given that ABCD is a parallelogram

And segment AC=4x+10

From the figure we have the diagonals AC=3x+y and BD=2x+y

By the property of parallelogram the diagonals are congruent

∴ we can equate the diagonals AC=BD

That is 3x+y=2x+y

3x+y-(2x+y)=2x+y-(2x+y)

3x+y-2x-y=2x+y-2x-y

x+0=0 ( by adding the like terms )

∴ x=0

Given that segment AC=4x+10

Substitute x=0 we have AC=4(0)+10

=0+10

=10

∴ AC=10

Now (3x+y)+(2x+y)=10

5x+2y=10

Substitute x=0, 5(0)+2y=10

2y=10

$y=\frac{10}{2}$

∴ y=5

∴ the values of x and y are x=0 and y=5

5 0

3 years ago