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Lady bird [3.3K]
3 years ago
6

Help Please(pre algebra)

Mathematics
1 answer:
miv72 [106K]3 years ago
7 0
Where’s the problem?
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I need help, It is a written problem
ratelena [41]
As long as the new rectangle is not moved, just dilated, line EF will be parallel.
3 0
3 years ago
If f(x)= |4x - 1|, find the value of x if(x) = 7
vaieri [72.5K]
F(x) = | 4x - 1 | .....when f(x) = 7
7 = | 4x - 1 |

4x - 1 = 7               -(4x - 1) = 7
4x = 7 + 1               -4x + 1 = 7
4x = 8                      -4x = 7 - 1
x = 8/4                     -4x = 6
x = 2                          x = -6/4
                                  x = - 3/2

solutions are x = -3/2, 2
6 0
3 years ago
Read 2 more answers
Find one positive and one negative coterminal angle to 2.7 Radians:
Pavlova-9 [17]

the two coterminal angles to the given one are:

8.98 rad and -3.58 rad

<h3>How to find coterminal angles?</h3>

For a given angle A in radians, the family of coterminal angles is defined as:

B = A + n*(2*pi)

Where pi = 3.14 rad

Where n can be any integer number.

In this case, we have an angle of 2.7 radians, then the coterminal angles are:

B = 2.7 rad + n*(6.28 rad)

One positive is what we get if we select n = 1, then:

B = 2.7 rad + 1*(6.28 rad) = 8.98 rad.

And if we select n = -1, we get the negative coterminal angle:

B' = 2.7 rad - 1*(6.28 rad) = -3.58 rad

Then the two coterminal angles to the given one are:

8.98 rad and -3.58 rad

If you want to learn more about coterminal angles:

brainly.com/question/19891743

#SPJ1

4 0
2 years ago
Esther drank 2/3 of water monday before going jogging . she drank 5/7 liter of water after her jog . how much water did esther d
Kitty [74]
2/3 + 5/7 = 14/21 + 15/21 = 29/21 = 1 8/21
4 0
3 years ago
Use the shell method to find the volume of the solid generated by revolving the regions bounded by the curves and lines about th
Nezavi [6.7K]

The two curves intersect when

-x^2=6-5x\implies x^2-5x+6=(x-3)(x-2)=0\implies x=2,x=3

The region bounded by x=0 and the two curves "ends" with a vertex where x=2. So over the interval [0, 2], we have 6-5x\ge-x^2, so that the volume is

\displaystyle2\pi\int_0^2x(6-5x+x^2)\,\mathrm dx=\boxed{\frac{16\pi}3}

That is, each cylindrical shell has a radius of x and height (6-5x)-(-x^2)=6-5x+x^2, and their contribution to the total volume is 2\pi times the radius and height.

5 0
3 years ago
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