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Ulleksa [173]
3 years ago
11

Find the length of the hypotenuse of a right triangle whose legs are 3 and square root of 2.

Mathematics
1 answer:
madreJ [45]3 years ago
6 0
Square root of 11 because:
sqrt2^2 = 2
3^2 = 9 
so C^2 or the hypotenuse would be the square root of 11 
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Point P(−3, −4) is rotated 90° counterclockwise about the origin. What are the coordinates of its image after this transformatio
Neporo4naja [7]
Counter clockwise means it goes backwards so -90°

After the 90° counter clockwise rotation the coordinates are (4,-3)

7 0
3 years ago
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What is the equation of the following graph in vertex form?
anygoal [31]

Answer:

I think the answer is 0,8

6 0
3 years ago
Please help!!!! <br><br> If f(1) = 2 and f(n+1) = f(n)^2 – 3 then find the value of f(3).
Black_prince [1.1K]

Answer:

The value of f(3) is -2.

Step-by-step explanation:

This is a recursive function. So

f(1) = 2

Now, we find f(2) in function of f(1). So

f(1+1) = f(1)^2 - 3

f(2) = f(1)^2 - 3 = 2^2 - 3 = 1

Now, with f(2), we can find the value of f(3).

f(2+1) = f(2)^2 - 3

f(3) = f(2)^2 - 3 = 1^2 - 3 = -2

The value of f(3) is -2.

4 0
3 years ago
What is the property of the equation 3x+5 - 5 = 20-5
Shalnov [3]

Answer:

x=5

Step-by-step explanation:

We simplify the equation to the form, which is simple to understand

3x+5-5=20-5

We move all terms containing x to the left and all other terms to the right.

+3x=+20-5-5+5

We simplify left and right side of the equation.

+3x=+15

We divide both sides of the equation by 3 to get x.

x=5

6 0
3 years ago
HI PLEASE HELP ME WITH MY CALCULUS 1 HW? I AM REALLY STUCK. I need help with parts d,e,g.
asambeis [7]

(d) The particle moves in the positive direction when its velocity has a positive sign. You know the particle is at rest when t=0 and t=3, and because the velocity function is continuous, you need only check the sign of v(t) for values on the intervals (0, 3) and (3, 6).

We have, for instance v(1)\approx-0.91 and v(4)\approx0.91>0, which means the particle is moving the positive direction for 3, or the interval (3, 6).

(e) The total distance traveled is obtained by integrating the absolute value of the velocity function over the given interval:

\displaystyle\int_0^6|v(t)|\,\mathrm dt=\int_0^3-v(t)\,\mathrm dt+\int_3^6v(t)\,\mathrm dt

which follows from the definition of absolute value. In particular, if x is negative, then |x|=-x.

The total distance traveled is then 4 ft.

(g) Acceleration is the rate of change of velocity, so a(t) is the derivative of v(t):

a(t)=v'(t)=-\dfrac{\pi^2}9\cos\left(\dfrac{\pi t}3\right)

Compute the acceleration at t=4 seconds:

a(t)=\dfrac{\pi^2}{18}\dfrac{\rm ft}{\mathrm s^2}

(In case you need to know, for part (i), the particle is speeding up when the acceleration is positive. So this is done the same way as part (d).)

6 0
3 years ago
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