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dybincka [34]
2 years ago
7

Solve the equation 23 = h - 52 for h.

Mathematics
2 answers:
UkoKoshka [18]2 years ago
4 0

1234567891011121314151617181920


Firdavs [7]2 years ago
3 0

are you looking for the h?

h=75

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What was edwin average rate of jogging in miles per hour​
Svetach [21]

Answer:Edwin average rate of jogging in miles per hour is

2 miles per hour

At the end of 30 min Edwin completed the track 3 times before Jose.

Step-by-step explanation:

Hope this helps

6 0
3 years ago
Two similar pyramids have lateral areas 8ft^2 and 18 ft^2. the volume of the large pyramid is 108 ft^2. find the volume of the s
ddd [48]
I assume the volume of the larger pyramid is 108 ft^3, not 108 ft^2.

The scale factor of edges of two solids is x.
The scale factor of their areas is x^2.
The scale factor of their volumes is x^3

The areas have a scale factor of 18/8 = 2.25

The length have a scale factor of sqrt(2.25) = 1.5

The volumes have a scale factor of 3.375

108/3.375 = 32

Answer: 32 ft^3
4 0
3 years ago
Can anyone help with this problem??
marishachu [46]

Answer:

blue line : y = 3

black horizontal line : y = 0

black vertical line : x = 0

Step-by-step explanation:

it is really simple.

blue line : no matter what value we pick for x, the functional value for y is always 3. and that defines the equation y = 3 + 0*x or simply y = 3.

the same with the black horizontal line. the difference is that y is now always 0. so, y = 0.

7 0
3 years ago
I will give brainliest<br> -50-c=5c-2<br> Show work pleasee
Lilit [14]

Answer:

- 8

Step-by-step explanation:

Step 1:

- 50 - c = 5c - 2     Equation

Step 2:

- 50 = 6c - 2    Add c on both sides

Step 3:

- 48 = 6c   Add 2 on both sides

Step 4:

- 48 ÷ 6     Divide

Answer:

c = - 8

Hope This Helps :)

6 0
3 years ago
Read 2 more answers
Find all the solutions of the equation in the interval [0,2pi). 4sin^(2)x=5-4cosx
gulaghasi [49]
                       4sin²(x) = 5 - 4cos(x)
        4{¹/₂[1 - cos(2x)]} = 5 - 4cos(x)
   4{¹/₂[1] - ¹/₂[cos(2x)]} = 5 - 4cos(x)
         4[¹/₂ - ¹/₂cos(2x)] = 5 - 4cos(x)
     4[¹/₂] - 4[¹/₂cos(2x)] = 5 - 4cos(x)
                2 - 2cos(2x) = 5 - 4cos(x)
              - 2                   - 2
                    -2cos(2x) = 3 - 4cos(x)
            -2[2cos²(x) - 1] = 3 - 4cos(x)
               -4cos²(x) + 2 = 3 - 4cos(x)
                               - 2  - 2
                     -4cos²(x) = 1 - 4cos(x)
-4cos²(x) + 4cos(x) - 1 = 0
 4cos²(x) - 4cos(x) + 1 = 0
               [2cos(x) - 1]² = 0
                  2cos(x) - 1 = 0
                              + 1 + 1
                       2cos(x) = 1
                            2         2
                         cos(x) = ¹/₂
                cos⁻¹[cos(x)] = cos⁻¹(¹/₂)
                                 x = 60, 300
                                 x = π/3, 5π/3

[0, 2π) = 0 ≤ x < 2π
[0, 2π) = 0 ≤ π/3 ≤ 2π or 0 ≤ 5pi/3 < 2π
5 0
2 years ago
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