Here is the correct and complete question: The units digit of a two-digit number is twice the tens digit. If the digits are reversed, the new number is 9 less than twice the original number. What is the original number?

Lets assume the original number be"10y+x". (x is unit digit and y is 10th digit)

∴ if number is reversed then resulting number be "10x+y".

As given: x= 2y

and $10x+y= 2(10y+x)-9$

Now, solving the equation to get original number.

$10x+y= 2(10y+x)-9$

Distributing 2 to 10y and x, then opening the parenthesis.

⇒ $10x+y= 20y+2x-9$

subtracting by (2x+y) on both side.

⇒ $8x= 19y-9$

subtituting the value of "x", which is equal to 2y.

∴ $8\times 2y= 19y-9$

⇒ $16y=19y-9$

subtracting both side by (16y-9)

⇒ $3y= 9$

cross multiplying

We get, $y= 3$

y=3

∵x= 2y

$x=2\times 3= 6$

∴ x= 6

Therefore, the original number will be 36 as x is the unit number and y as tenth number.

6 0

3 years ago

Add the two expressions. 32 – 4 and 2z + 5 Pls help quick!

alexgriva [62]

Answer:

2z + 33

Step-by-step explanation:

Add like terms like below:

(32 - 4) + (2z + 5) = 28 + 2z + 5

Combine like terms again:

(28 + 5) + 2z = 33 + 2z or 2z + 33

7 0

2 years ago

Cos4theta+cos2theta/ cos4theta-cos2theta= _____

vovangra [49]

$\bf \textit{Sum to Product Identities} \\\\ cos(\alpha)+cos(\beta)=2cos\left(\cfrac{\alpha+\beta}{2}\right)cos\left(\cfrac{\alpha-\beta}{2}\right) \\\\\\ cos(\alpha)-cos(\beta)=-2sin\left(\cfrac{\alpha+\beta}{2}\right)sin\left(\cfrac{\alpha-\beta}{2}\right) \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf \cfrac{cos(4\theta )+cos(2\theta )}{cos(4\theta )-cos(2\theta )}\implies \cfrac{2cos\left( \frac{4\theta +2\theta }{2} \right)cos\left( \frac{4\theta -2\theta }{2} \right)}{-2sin\left( \frac{4\theta +2\theta }{2} \right)sin\left( \frac{4\theta -2\theta }{2} \right)} \implies \cfrac{cos\left( \frac{6\theta }{2} \right)cos\left( \frac{2\theta }{2} \right)}{-sin\left( \frac{6\theta }{2} \right)sin\left( \frac{2\theta }{2} \right)}$

$\bf \cfrac{cos(3\theta )cos(\theta )}{-sin(3\theta )sin(\theta )}\implies -\cfrac{cos(3\theta )}{sin(3\theta )}\cdot \cfrac{cos(\theta )}{sin(\theta )}\implies -cot(3\theta )cot(\theta )$

8 0

3 years ago

hiii I don't need the answer but do you know what this is about and may you please explain? thank youuu ​

hiii I don't need the answer but do you know what this is about and may you please explain? thank youuu