All categories

2 years ago

Write the linear equation into Standard Form. 8+3x=-9y

Mathematics

1 answer:

Travka [436]2 years ago

6 0

Hello! :p

Original: 8+3x=-9y

standard form: 3x + 9y = -8

I hope this helps! Have a wonderful day! :)

You might be interested in

A man spends 8/5 of his salary<br> rent. If his left with & 2940. Find his<br> salary:

Ainat [17]

Answer:

A man spends 8/5 of his salary rent. If his left with & 2940. Find his salary:

Step-by-step explanation:

4 0

2 years ago

At the Tucson Zoo, Jason spends $6.00 to enter the zoo and $2.00 per animal to feed them. Meanwhile, at the Phoenix Zoo, Kim spe

shtirl [24]

Answer:

3 animals

Step-by-step explanation:

combine like terms

6 + 2x = 15 + x
-6 + x - 6 - x

--------------------------

3x = 9

2. divide both sides by 3

3x/ 3 = 9/3
x = 3

8 0

3 years ago

The probability of event A occurring is 3/5, while the probability of event B occurring is 2/7. If the probability of events A a

Harman [31]

Answer:

I think they are independent

Step-by-step explanation:

7 0

3 years ago

Ifx =y at two points what can you coclude about =y

AlekseyPX

Answer:

hello bro

Step-by-step explanation:

6 0

3 years ago

In quadratic drag problem, the deceleration is proportional to the square of velocity

Mars2501 [29]

Part A

Given that $a= \frac{dv}{dt} =-kv^2$

Then,

$\int dv= -kv^2\int dt \\ \\ \Rightarrow v(t)=-kv^2t+c$

For $v(0)=v_0$ , then

$v(0)=-kv^2(0)+c=v_0 \\ \\ \Rightarrow c=v_0$

Thus, $v(t)=-kv(t)^2t+v_0$

For $v(t)= \frac{1}{2} v_0$ , we have

$\frac{1}{2} v_0=-k\left( \frac{1}{2} v_0\right)^2t+v_0 \\ \\ \Rightarrow \frac{1}{4} kv_0^2t=v_0- \frac{1}{2} v_0= \frac{1}{2} v_0 \\ \\ \Rightarrow kv_0t=2 \\ \\ \Rightarrow t= \frac{2}{kv_0}$

Part B

Recall that from part A,

$v(t)= \frac{dx}{dt} =-kv^2t+v_0 \\ \\ \Rightarrow dx=-kv^2tdt+v_0dt \\ \\ \Rightarrow\int dx=-kv^2\int tdt+v_0\int dt+a \\ \\ \Rightarrow x=- \frac{1}{2} kv^2t^2+v_0t+a$

Now, at initial position, t = 0 and $v=v_0$ , thus we have

$x=a$

and when the velocity drops to half its value, $v= \frac{1}{2} v_0$ and $t= \frac{2}{kv_0}$

Thus,

$x=- \frac{1}{2} k\left( \frac{1}{2} v_0\right)^2\left( \frac{2}{kv_0} \right)^2+v_0\left( \frac{2}{kv_0} \right)+a \\ \\ =- \frac{1}{2k} + \frac{2}{k} +a$

Thus, the distance the particle moved from its initial position to when its velocity drops to half its initial value is given by

$- \frac{1}{2k} + \frac{2}{k} +a-a \\ \\ = \frac{2}{k} - \frac{1}{2k} = \frac{3}{2k}$

7 0

3 years ago