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3 years ago

Solve for y. 3X+2y=12

Mathematics

1 answer:

Montano1993 [528]3 years ago

3 0

3x + 2y = 12

2y = -3x + 12
y = -1.5x + 6

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Please help me

dusya [7]

3(4x - 2) = 2(6x - 3)

12x - 6 = 12x - 6

Add 6 on both sides

12x = 12x

Divide by 12 on both sides

x = x

That means this has infinite solutions. Your answer is B.

4 0

3 years ago

Read 2 more answers

How do you solve this can someone help me pls

horrorfan [7]

Answer:

x-8°+20°= 90°

=>x+12°=90°

=>x=90°-12°

=>x= 78°

5 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

What is the value of k? k = 28 k = 29 k = 31 k = 42

julia-pushkina [17]

Answer:

k=29

Step-by-step explanation:

we know that

-------> by supplementary angles

Solve for k

Combine like terms in the left side

Substract both sides

Divide by both sides

therefore

the answer is the option

6 0

2 years ago

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HELPP PLS ITS URGENT!! Write an equation for this graph pictured below!

Ganezh [65]

Answer:

y=9sin(theta)

Step-by-step explanation:

I'm assuming it's in degrees. The amplitude of the graph is 9, and so you multiply the base sine equation y=sin(theta) by nine. I don't see any shifts and the period remains at 2pi (360 degrees) so I don't think anything else is really needed.

6 0

2 years ago