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Aloiza [94]
3 years ago
12

I need help filling in the blank

Mathematics
1 answer:
Tcecarenko [31]3 years ago
6 0
Initial/Starting Point: 52.1 
Growth/Decay Constant: 1.023 

The graph is attached below. 

Possible word problems:
Bertha starts out with approximately 52.1 thousand mice. They constantly reproduce 1.023 mice every day. Model an equation

Hope I helped :) 

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Angle 3 and Angle 6 are an example of which type of angle pair?
Elena-2011 [213]

Answer:

Alternate Interior Angles

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3 years ago
If sin ∅ = a/b then prove that (sec ∅ + tan ∅) = √b+a/b-a​
nasty-shy [4]

Given that

Sin θ = a/b

LHS = Sec θ + Tan θ

⇛(1/Cos θ) + (Sin θ/ Cos θ)

⇛(1+Sin θ)/Cos θ

We know that

Sin² A + Cos² A = 1

⇛Cos² A = 1-Sin² A

⇛Cos A =√(1-Sin² A)

LHS = (1+Sin θ)/√(1- Sin² θ)

⇛ LHS = {1+(a/b)}/√{1-(a/b)²}

= {(b+a)/b}/√(1-(a²/b²))

= {(b+a)/b}/√{(b²-a²)/b²}

= {(b+a)/b}/√{(b²-a²)/b}

= (b+a)/√(b²-a²)

= √{(b+a)(b+a)/(b²-a²)}

⇛ LHS = √{(b+a)(b+a)/(b+a)(b-a)}

Now, (x+y)(x-y) = x²-y²

Where ,

  • x = b and
  • y = a

On cancelling (b+a) then

⇛LHS = √{(b+a)/(b-a)}

⇛RHS

⇛ LHS = RHS

Sec θ + Tan θ = √{(b+a)/(b-a)}

Hence, Proved.

<u>Answer</u><u>:</u> If Sinθ=a/b then Secθ+Tanθ=√{(b+a)/(b-a)}.

<u>also</u><u> read</u><u> similar</u><u> questions</u><u>:</u><u>-</u> i) sin^2 A sec^2 B + tan^2 B cos^2 A = sin^2A + tan²B..

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7 0
3 years ago
5 - 7(x + 1) + 10x + 3<br> Show step by step on how to solve for x
Svet_ta [14]
1+3x

Distribute -7 through the parenthesis
6 0
3 years ago
2+4447=8ii what is ii​
Anvisha [2.4K]

Answer:

ii = 556.125

Step-by-step explanation:

4449 = 8ii

divide both side by 8

4449 \div 8 = 8ii \div 8

ii = 556.125

8 0
3 years ago
I need to use Multi-Step function inverse to find the following:
dexar [7]

<u>Answer</u>:

f^{-1}(x) = \sqrt[5]{\frac{x+7}{9} -10}

<u>Explanation</u>:

given : f(x) = 9( x^5 + 10 ) - 7

replace x and y with each other.

  • x = 9( y^5 + 10 ) - 7

solve for y

  • x + 7 = 9( y^5 + 10 )
  • ( y^5 + 10 ) = \frac{x + 7}{9}
  • y^5 + 10 = \frac{x + 7}{9}
  • y^5 = \frac{x + 7}{9} - 10
  • y = \sqrt[5]{\frac{x+7}{9} -10}

Therefore,

  • f^{-1}(x) = \sqrt[5]{\frac{x+7}{9} -10}
5 0
2 years ago
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