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irinina [24]
3 years ago
11

It easy i just don't pay attention in class so I don't know the answer

Mathematics
1 answer:
Natasha_Volkova [10]3 years ago
5 0
I think the answer is J
Y = 7
You could also use
Geo calculator to make it easier.
You’ll just have to write the point and the choices.
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6 m<br> 2 m<br> 6 m<br> 2 m<br> 2 m<br> 6 m
Rina8888 [55]

Answer:

24 im pretty sure.

Step-by-step explanation:

7 0
3 years ago
write an equation in slope intercept form for the line described. perpendicular to y = -1/2x + 2/3, passes through (2,3)
Alex17521 [72]

Step-by-step explanation:

Hey there!

Follow the steps to get answer.

  • Use one point formula and find 1st equation.
  • After that you find the slope of second equation.
  • Use the condition of perpendicular lines and find the slope of first equation.
  • Put slope value of equation in equation (i) and simplify them to get equation.

The equation of a line passing through point (2,3) is;

(y-3)= m1(x-2).......(i).

Another equation is;

y =  \frac{ - 1}{2} x +  \frac{2}{3}

2nd equation..

Now, From equation (ii)

We have;

Comparing equation (ii) with y = mx+c.

We get;

Slope = -1/2.

For perpendicular lines,

m1 \times m2 =  - 1

m1 \times  \frac{ - 1}{2}  =  - 1

Therefore the slope is 2.

Put value of slope (m1) in equation (i). We get;

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

Simplify them to get equation.

(y - 3) = 2x - 4

y = 2x - 1

Therefore the required equation is y = 2x-1.

<em><u>Hope it helps</u></em><em><u>.</u></em><em><u>.</u></em>

4 0
3 years ago
A ski resort has 3 lifts, each with access to 6 ski trails. Explain how you can find the number of possible outcomes when choosi
svetlana [45]
You could multiply 6 times 3 to get your answer. The answer would be 18

4 0
3 years ago
The staff at the community center make a rectangular skating pit every year. They use 100 meters of boarding to enclose it. So t
Sliva [168]
For a width of 25, the length will also be 25, so the area is 25² = ...
  625.

_____
The rule is
  area = width×(50 -width)
3 0
3 years ago
"Suppose an object falling in the atmosphere has mass m=15kg and the drag coefficient is γ=9kg/s. Recall that the differential e
Art [367]

Answer:

a. v(t)= -6.78e^{-16.33t} + 16.33 b. 16.33 m/s

Step-by-step explanation:

The differential equation for the motion is given by mv' = mg - γv. We re-write as mv' + γv = mg ⇒ v' + γv/m = g. ⇒ v' + kv = g. where k = γ/m.Since this is a linear first order differential equation, We find the integrating factor μ(t)=e^{\int\limits^  {}k \, dt } =e^{kt}. We now multiply both sides of the equation by the integrating factor.

μv' + μkv = μg ⇒ e^{kt}v' + ke^{kt}v = ge^{kt} ⇒ [ve^{kt}]' = ge^{kt}. Integrating, we have

∫ [ve^{kt}]' = ∫ge^{kt}

    ve^{kt} = \frac{g}{k}e^{kt} + c

    v(t)=   \frac{g}{k} + ce^{-kt}.

From our initial conditions, v(0) = 9.55 m/s, t = 0 , g = 9.8 m/s², γ = 9 kg/s , m = 15 kg. k = y/m. Substituting these values, we have

9.55 = 9.8 × 15/9 + ce^{-16.33 * 0} = 16.33 + c

       c = 9.55 -16.33 = -6.78.

So, v(t)=   16.33 - 6.78e^{-16.33t}. m/s = - 6.78e^{-16.33t} + 16.33 m/s

b. Velocity of object at time t = 0.5

At t = 0.5, v = - 6.78e^{-16.33 x 0.5} + 16.33 m/s = 16.328 m/s ≅ 16.33 m/s

6 0
3 years ago
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