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yaroslaw [1]
3 years ago
14

What simple machine is similar to a pulley?

Mathematics
1 answer:
jek_recluse [69]3 years ago
5 0
A lever in like a pully becasue it can pull just as so. Hope that helped. It is what my teacher taught me. 

You might be interested in
42. Use the Quadratic Equation to find the Roots of x2 + 3x - 3=0
almond37 [142]

Answer:

Step-by-step explanation:

The coefficients of the x terms are {1, 3, -3}, so the discriminant, b^2 - 4ac, is 3^2 - 4(1)(-3), or 9 + 12, or 21.  The positive nature of the discriminant tells us that there are two real, unequal roots.  Following the quadratic formula, we get:

       -3 ± √21

x = -----------------

              2

8 0
3 years ago
How to find x and also giving reason . Thanku
Colt1911 [192]
1a. x = 70 
     Alternate Interior Angles

1b. x = 120
     Alternate Interior Angles

1c. x = 110
     Corresponding Angles

2a. x = 100
      Alternate Interior Angles
      y = 80
      Supplementary Angles (with x)

2b. x = 75
      Corresponding Angles
      y = 105
      Supplementary Angles (with x)

2c. x = 70
      Same-Side Interior Angles 
      y = 110
      Supplementary Angles (with x)

Try the rest on your own! 


7 0
3 years ago
Please help me! I have to sumbmit in 15 minutes!
kompoz [17]

Answer:

The answer is 1030 ml

Hope this helps

5 0
3 years ago
Read 2 more answers
What is the solution to8(5-r)/5=-2r
Dmitry [639]

Expand 8(5-r)/5=-2r as indicated: 40 - 8r

----------- = - 2r

5

Multiply both sides by 5, obtaining 40 - 8r = - 10r

Combine like terms: 40 = - 2r

Divide both sides by -2: -20 = r

The solution to 8(5-r)/5=-2r is -20. Verify this by subbing -20 for r in the given equation.

4 0
3 years ago
Read 2 more answers
N=4; 2i and 3i are zeros <br> f(-1)=50
Nonamiya [84]

Solution:- \text{Let f(x) be any nth degree polynomial with n=4}\\

\text{Given that 2i and 3i are the zeroes of f(x)}

\text{so (x-2i) and (x-3i) are factors of f(x)}

\text{Since 2i is a zero of f(x) then its conjugate -2i is also a zero of f(x)}

\Rightarrow(x+2i) \text{is a factor}\\\text{Similarly, conjugate of 3i is -3i is also a zero of f(x) }\\\Rightarrow(x+3i)\text{is a factor}\\\text{So , }\\f(x)=k(x-2i)(x+2i)(x-3i)(x+3i)\\=k(x^2-(2i)^2)(x^2-(3i)^2)\\=k(x^2-4i^2)(x^2-9i^2)\\=k(x^2+4)(x^2+9)\\=k(x^4+13x^2+36)\\\text{As given}\\f(-1)=50\\\Rightarrow k((-1)^4+13(-1)^2+36)=50\\\Rightarrow k(1+13+36)=50\\\Rightarrow k(50)=50\\\Rightarrow k=1\\\text{So by substituting k=1 in f(x) we get ,}\\f(x)=(x^4+13x^2+36)

6 0
3 years ago
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