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Sphinxa [80]
2 years ago
7

Find the distance from A to C across the gorge illustrated in the figure

Mathematics
1 answer:
emmasim [6.3K]2 years ago
4 0

Applying the tangent ratio, the distance from A to C across the gorge, is:  60.6 ft.

<h3>What is Tangent Ratio?</h3>

The tangent ratio for solving a right triangle is given as, tan ∅ = opposite/adjacent.

Given the following:

  • ∅ = 25°
  • Adjacent length = 130 ft
  • Opposite length = AC

Apply the tangent ratio:

tan 25 = AC/130

AC = (130)(tan 25)

AC = 60.6 ft

Learn more about the tangent ratio on:

brainly.com/question/4326804

#SPJ1

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What is 64,005,874 in word and expanded form and 30,679,100
MakcuM [25]
64005874  
word form: sixty-four million, five thousand, eight hundred and seventy-four
expanded form: 60000000 + 4000000 + 5000 + 800 + 70 + 4
30679100
word: thirty million, six hundred seventy nine thousand, and one hundred
30000000 + 600000 + 70000 + 9000 + 100

hope this helps
8 0
3 years ago
What is <br> 5(2x+8)-5x=5(5x+6?
yanalaym [24]

5(2x+8)-5x=5(5x+6)

Distribute

10x+40-5x=25x+30

combine like terms

5x+40=25x+30

-5x       -5x

subtract 5x from both sides

40=20x+30

-30      -30

subtract 30 from both sides

10=20x

÷20   ÷20

divide both sides by 20

1/2=x is your final answer


4 0
3 years ago
Find the inverse of f(x)=2x-1
solong [7]

{ \qquad\qquad\huge\underline{{\sf Answer}}}

Let's find the inverse of given function ~

\qquad \sf  \dashrightarrow \: f(x) = 2x - 1

\qquad \sf  \dashrightarrow \: 2x = f(x) + 1

\qquad \sf  \dashrightarrow \: x =  \cfrac{f(x) + 1}{2}

[ now, replace f(x) and x with \sf{ {f}^{-1}(x)} ]

\qquad \sf  \dashrightarrow \: f {}^{ - 1} (x) =  \cfrac{x + 1}{2}

That's the required inverse function ~

8 0
1 year ago
Read 2 more answers
Help!! Screenshot below.
antiseptic1488 [7]
X=60 and if you plug in 60 to that equation, it equals 15.


1/4 x 60 =15

Hope this helps!
7 0
3 years ago
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If ( 3 y − 9 ) 2 + 7 = 43 , then what is(are) the values of y?
n200080 [17]

Answer:

y = 5 and y = 1

Step-by-step explanation:

\textsc {Subtract 7 from each side :}

⇒ (3y - 2)² + 7 - 7 = 43 - 7

⇒ (3y - 2)² = 36

\textsc {take the square root on each side :}

⇒ √(3y - 9)² = √36

⇒ 3y - 9 = ±6

\textsc {When equal to +6, add 9 on each side :}

⇒ 3y - 9 + 9 = 6 + 9

⇒ 3y = 15

\textsc {Divide by 3 on each side :}

⇒ 3y/3 = 15/3

⇒ y = 5

\textsc {When equal to -6, add 9 on each side :}

⇒ 3y - 9 + 9 = -6 + 9

⇒ 3y = 3

\textsc {Divide by 3 on each side :}

⇒ 3y/3 = 3/3

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4 0
2 years ago
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