Jan is 5 years older than Tom. In 3 years, Tom will be as old as Jan. What is Tom’s age now?

Mathematics

1 answer:

Lynna [10]2 years ago

5 0

Answer:

I think it's TOM 6 years old now and JAN is 8 years old now

You might be interested in

At the School Fair, Grade 3 raises $84 more than Grade 4. Grade 4 raises $112 more than Grade 2. Grade 3 raises $360. How much m

LekaFEV [45]

360-84= 276
276-112=164
grade 2 raised $164

7 0

3 years ago

Read 2 more answers

What is the solution of 3x+y=70 and x+y=44 using elimination

Mkey [24]

Answer: y = 31 and x = 13

Step-by-step explanation: Look at the image from the top left to top right, then to bottom right to bottom left.

4 0

2 years ago

Write the partial fraction decomposition fro the ration expression below

zlopas [31]

$\dfrac{5x-2}{(x-1)^2}=\dfrac a{x-1}+\dfrac b{(x-1)^2}$
$5x-2=a(x-1)+b$

Since $5x-2=5x-5+3=5(x-1)+3$ , it follows that $a=5$ and $b=3$ , so

$\dfrac{5x-2}{(x-1)^2}=\dfrac5{x-1}+\dfrac3{(x-1)^2}$

$\dfrac{x^3+x^2+x+2}{x^4+x^2}=\dfrac{x^3+x+x+2}{x^2(x^2+1)}=\dfrac ax+\dfrac b{x^2}+\dfrac{cx+d}{x^2+1}$
$x^3+x^2+x+2=ax(x^2+1)+b(x^2+1)+(cx+d)x^2$
$x^3+x^2+x+2=(a+c)x^3+(b+d)x^2+ax+b$

When $x=0$ , you find that $b=2$ . It's also clear that $a=1$ . So the remaining constants must be $1+c=1\implies c=0$ and $2+d=1\implies d=-1$ , and so you get

$\dfrac{x^3+x^2+x+2}{x^4+x^2}=\dfrac1x+\dfrac2{x^2}-\dfrac1{x^2+1}$

6 0

3 years ago

HURRY!!!!!! Which expression is equivalent to ____? ^^^

Pepsi [2]

$24^{\frac 1 3} = \sqrt[3]{8 \cdot 3} = 2 \sqrt[3]{3}$