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gogolik [260]
3 years ago
9

PLLLZZZZ HEEELLLPPPP!!!!!!!!!!! Questions are below! :D

Mathematics
1 answer:
V125BC [204]3 years ago
3 0

Answer:

12)  m∠BXD = 122°

13)  m∠GXE = 50°

Step-by-step explanation:

12)  m∠BXD = ∡BXC + ∡CXF + ∡FXD = 32° + 50° + 40° = 122°

13) m∠GXE = (90° - m∠AXG)  =  90° - 40° = 50°

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Solve 2 log(x) + log(5) = log (80) using the one-to-one property.
vodomira [7]

x=4

Step-by-step explanation:

2log(x)+log(5)=log(80)

log(x^2)+log(5)=log(80)

log(5x^2)=log(80)

5x^2 =80

x^2 =80/5

x^2=16 , taking square root both sides

x=√16

=4

5 0
2 years ago
A and B are sumplimenteey . m B =121 find m A
vichka [17]

Answer:

m∠A = 59°

Step-by-step explanation:

Supplementary angles mean that, when the two are combined, their total measurement will equal 180°.

It is given that one of the angles is 121°. Subtract 121 from 180:

180 - 121 = 59

m∠A = 59°

~

4 0
2 years ago
Please help me plsssssss! asappp. I'll mark brainlest and give points ​
Nookie1986 [14]

Answer:

Step-by-step explanation:

1)a) Abscissa of O   0

 Abscissa of A  4

Abscissa of B    -3     {opposite of 3}

Abscissa of C    3      {3 is the only positive integer like A}

Abscissa of D    -4.5

Abscissa of E   -6

Abscissa of F   -1           {Midpoint of AE = (6+4)/2 = 10/2 = 5th number from -6 or 4}

b) OB = 3 units

DA = 8.5    {4.5 +4 = 8.5}

7 0
3 years ago
Simplify the solution (2¼)½​
BlackZzzverrR [31]
I think the answer is 1 1/8
6 0
3 years ago
Read 2 more answers
Find the area of the shaded region. Round the nearest hundredth if necessary. YZ=14.2m
andreyandreev [35.5K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

1

   A_1  =  67.58 \ in^2

2

   A_2 =415.4 \ ft^2

3

   A_3  =  8.48 \ cm^2

4

  A_4 =  480.38 \ m^2

Step-by-step explanation:

Generally the area of a sector is mathematically represented as

         A =  \frac{\theta}{360} * \pi r^2

Now at r_1  = 11 in and  \theta_1 =  64^o

       A_1 =  \frac{64}{360} * 3.142  * 11^2

       A_1  =  67.58 \ in^2

Now at  r_2  = 20 ft in and  \theta_2  =  119 ^o

       A_2 =  \frac{119}{360} * 3.142 *  20^2

       A_2 =415.4 \ ft^2

Now at  r_3  = 6.5 cm   and  \theta_3 =  23 ^o

     A_3  =  \frac{23}{360} * 3.142 *  6.5 ^2

      A_3  =  8.48 \ cm^2

Now at  r_4  = 14.2 m   and  \theta_4 = 360 -87 =  273 ^o

         A_4 =  \frac{273}{360}  * 3.142 * 14.2^2

          A_4 =  480.38 \ m^2

6 0
3 years ago
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