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Gnoma [55]
3 years ago
14

Find the measure of /_ MFZ

Mathematics
1 answer:
marshall27 [118]3 years ago
7 0

Given:

MZ\cong PZ

m\angle MFZ=(x+9)^\circ

m\angle PFZ=(2x-1)^\circ

To find:

The measure of angle MFZ.

Solution:

In triangles MFZ and PFZ,

m\angle FMZ\cong \angle FPZ           (Right angle)

MZ\cong PZ           (Given)

FZ\cong FZ           (Common side)

\Delta MFZ\cong \Delta PFZ                  (HL congruent postulate)

\angle MFZ\cong \angle PFZ           (CPCTC)

m\angle MFZ=m\angle PFZ

x+9=2x-1

9+1=2x-x

10=x

The value of x is 10.

The measure of angle MFZ is:

m\angle MFZ=(x+9)^\circ

m\angle MFZ=(10+9)^\circ

m\angle MFZ=19^\circ

Therefore, the measure of angle MFZ is 19 degrees.

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Answer:

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Step-by-step explanation:

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Alex got 9 hits in his first 22 at bats. If his hitting rate remains constant, how many at bats would he have to have to reach 5
nikklg [1K]

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122.2222 repeating

Step-by-step explanation:

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3 years ago
How many different 5​-letter radio station call letters can be made a. if the first letter must be Upper C comma Upper X comma U
Lady_Fox [76]

Answer:

a) 1,518,000

b) 2,284,880

c) 60,720

Step-by-step explanation:

a) a. if the first letter must be Upper C comma Upper X comma Upper T comma or Upper M and no letter may be​ repeated?

We draw 5 boxes, and based on that we will see the total possible cases. There are 26 alphabets

The first box should have C or X or T or M .No letter may be repeated.

Any                    Any             Any                 Any                    Any

5 alphabets    of the            of the              of the                 of the

C,X , T , M      remaining     remaining       remaining          remaining

                   25 alphabets   24 alphabets   23 alphabets   22 alphabets

Therefore; total possible call letters = 5 × 25 × 24 × 23 × 22 = 1,518,000

b)

The first box should have C or X or T or M Repeats as allowed

Any                    Any             Any                 Any                    Any

5 alphabets    of the            of the              of the                 of the

C,X , T , M      remaining     remaining       remaining          remaining

                   26 alphabets   26 alphabets   26 alphabets   26 alphabets

Therefore Total possible call letters = 5 × 26 × 26 × 26 × 26 = 2,284,880

c)   The first box should have   C,X , T , M  and end with S

So the last place if fixed, and we now have 25 alphabets. The first box can go in 5 ways. The next box then will have only 24 letters to choose from, as the first box has taken a letter and the last box already has S in it. Repetition not allowed

Any                    Any             Any                 Any                       S

5 alphabets    of the            of the              of the                  is fixed

C,X , T , M      remaining     remaining       remaining           here

                   24 alphabets   23 alphabets   22 alphabets  

Therefore Total possible call letters = 5 × 24 × 23 × 22  × 1 =  60,720

3 0
3 years ago
I need help on this one
e-lub [12.9K]

The answer is B. :PPPPPPPPPP

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zhuklara [117]
145 blue paint are needed to make 28 pints of green paint
3 0
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