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storchak [24]
3 years ago
14

What is the value of h? h = 20 h=35 h = 70 (2h) (2h)

Mathematics
1 answer:
postnew [5]3 years ago
7 0

Answer:

-7/2 or -3.5

Step-by-step explanation:

h = 20 = h = 35  = h = 70  (2h)  (2h)

19h = 35h = 70(2h)(2h)

16h = 70(2h)(2h)

-70 = 20h

-7/2 = h

-3.5 = h

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Need help with this one fast! Thanks.
OlgaM077 [116]
F(g(2) means to evaluate the g function for 2 then the result becomes the input for the f function.
g(2)= 5(2) - 6 = 10 - 6 = 4
f(4) = 4^(3/2) = (sqrt(4))^2 = 8
The answer is 8
3 0
3 years ago
Intersection point of Y=logx and y=1/2log(x+1)
GalinKa [24]

Answer:

The intersection is (\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2}).

The Problem:

What is the intersection point of y=\log(x) and y=\frac{1}{2}\log(x+1)?

Step-by-step explanation:

To find the intersection of y=\log(x) and y=\frac{1}{2}\log(x+1), we will need to find when they have a common point; when their x and y are the same.

Let's start with setting the y's equal to find those x's for which the y's are the same.

\log(x)=\frac{1}{2}\log(x+1)

By power rule:

\log(x)=\log((x+1)^\frac{1}{2})

Since \log(u)=\log(v) implies u=v:

x=(x+1)^\frac{1}{2}

Squaring both sides to get rid of the fraction exponent:

x^2=x+1

This is a quadratic equation.

Subtract (x+1) on both sides:

x^2-(x+1)=0

x^2-x-1=0

Comparing this to ax^2+bx+c=0 we see the following:

a=1

b=-1

c=-1

Let's plug them into the quadratic formula:

x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

x=\frac{1 \pm \sqrt{(-1)^2-4(1)(-1)}}{2(1)}

x=\frac{1 \pm \sqrt{1+4}}{2}

x=\frac{1 \pm \sqrt{5}}{2}

So we have the solutions to the quadratic equation are:

x=\frac{1+\sqrt{5}}{2} or x=\frac{1-\sqrt{5}}{2}.

The second solution definitely gives at least one of the logarithm equation problems.

Example: \log(x) has problems when x \le 0 and so the second solution is a problem.

So the x where the equations intersect is at x=\frac{1+\sqrt{5}}{2}.

Let's find the y-coordinate.

You may use either equation.

I choose y=\log(x).

y=\log(\frac{1+\sqrt{5}}{2})

The intersection is (\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2}).

6 0
3 years ago
Please please help <br><br> Given the function f(x)=1/3x*2-3x+5 determine the inverse relation
yuradex [85]

The inverse relation of the function f(x)=1/3x*2-3x+5 is f-1(x) = 9/2 + √(3x + 21/4)

<h3>How to determine the inverse relation?</h3>

The function is given as

f(x)=1/3x^2-3x+5

Start by rewriting the function in vertex form

f(x) = 1/3(x - 9/2)^2 -7/4

Rewrite the function as

y = 1/3(x - 9/2)^2 -7/4

Swap x and y

x = 1/3(y - 9/2)^2 -7/4

Add 7/4 to both sides

x + 7/4= 1/3(y - 9/2)^2

Multiply by 3

3x + 21/4= (y - 9/2)^2

Take the square roots

y - 9/2 = √(3x + 21/4)

This gives

y = 9/2 + √(3x + 21/4)

Hence, the inverse relation of the function f(x)=1/3x*2-3x+5 is f-1(x) = 9/2 + √(3x + 21/4)

Read more about inverse functions at:

brainly.com/question/14391067

#SPJ1

5 0
2 years ago
When six times a number is decreased by five, the result is 49. what is the number?
solong [7]
Put this into an equation. Make the missing value "x".
6x - 5 = 49
49 + 5 = 54
54 / 6 =  9 
x = 9.
The missing number is 9.

Hope This Helps You!
Good Luck Studying :)




3 0
3 years ago
Hey uh guys i need help pls thanks
vivado [14]

i don’t know for sure but i just figured it out....

7 0
3 years ago
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