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mr_godi [17]
3 years ago
6

Plz look at the pic !!!!!!!

Mathematics
1 answer:
kicyunya [14]3 years ago
5 0

Answer:

The expression 5y * 3/5 is equivalent to: 3y

Step-by-step explanation:

Simplify 5 x 3y / 5 and you're left with 3y

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What is the function rule for the line?
11Alexandr11 [23.1K]
So with these equations we need to find the y-intercept and the slope because:

y = mx+b

y = f(×)

b = y-intercept

m = slope (slope = rise/run, rise over run)

rise is y-axis
run is x-axis

The y-intercept is (0,-2)
So now we know that one of the answers must have -2 in it

b = -2

m= rise/run = -2/3

The answer is:

f(x) = - \frac{2 }{3 } x - 2
5 0
3 years ago
Read 2 more answers
In the xy, y-plane, the parabola with equation y = ( x− 11)^2y=(x−11) 2 y, equals, (, x, −, 11, ), squaredintersects the line wi
IRINA_888 [86]

Answer:

10 units

Step-by-step explanation:

Allow me to revise your question for a better understanding.

<em>"In the xy-plane, the parabola with equation y = (x − 11) ² intersects the line with equation y = 25 at two points, A and B. What is the length of AB"  </em>

Here is my answer

Because the  parabola intersects the line with equation y = 25  Substituting y = 25 in the equation of the parabola y = (x - 11)², we get

25 = (x - 11)²

<=>x - 11 = ± 5  

<=> \left \{ {{x=6} \atop {x=16}} \right.

Thus A(16, 25) and B(6, 25) are the points of intersection of the given parabola and the given line.

So the length of AB = √[(16 - 6)² + (25 - 25)²]

= √100 = 10 units

5 0
3 years ago
Read 2 more answers
Find a formula for the inverse of the function f(x)=3x^3 + 2sinx + 4cosx
vitfil [10]
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5 0
3 years ago
Suppose an object is launched from ground level directly upward at 57.4 f/s Write a function to represent the object’s height ov
Semmy [17]

Answer: p(t) = (-16 ft/s^2)*t^2 + (57.4 ft/s)*t

Step-by-step explanation:

We can suppose that the only force acting on the object is the gravitational force, then the acceleration of the object will be equal to the gravitational acceleration.

Then we can write:

a(t) = -32 ft/s^2

Where the negative sign is because this acceleration is downwards.

Now, to get the vertical velocity of the object, we need to integrate over time to get:

v(t) = (-32 ft/s^2)*t + v0

where t represents time in seconds and v0 is the constant of integration, and in this case, is the initial vertical velocity.

In this case, the initial velocity is 57.4 ft/s upwards, then the velocity equation is:

v(t) = (-32 ft/s^2)*t + 57.4 ft/s

To get the position equation we need to integrate over time again, to get:

p(t) = (1/2)*(-32 ft/s^2)*t^2 + (57.4 ft/s)*t + p0

Where p0 is the initial height of the object, as it was launched from the ground, then the initial position is p0 = 0ft.

then the position equation (that is the function that represents the height of the object as a function over time) is:

p(t) = (1/2)*(-32 ft/s^2)*t^2 + (57.4 ft/s)*t

p(t) = (-16 ft/s^2)*t^2 + (57.4 ft/s)*t

3 0
3 years ago
Rewrite the following using the GCF and distributive properly : <br><br> 63+ 27
Yuki888 [10]
The answer is 90 i think
explain: none
3 0
3 years ago
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