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3 years ago

8

The blue are the two boards that you need to use.

1 answer:

ad-work [718]3 years ago

7 0

Answer:

for what?

Step-by-step explanation:

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Please help me with this .

Flauer [41]

Answer:

with what

Step-by-step explanation:

can you please post the picture

look at the picture to see how to add a pic to a question

7 0

2 years ago

Help. What’s the correct answer?

vesna_86 [32]

Answer:

f(g(x)) = -9(sqrt(x + 1) ) + 9

-36

Step-by-step explanation:

This is easier to see if you use the expanded notation.

f(g(x)) = ?

But what that means is that the left most function is f(x) and where every you see an x, you put in a g(x) like so

f(g(x)) = -9(g(x)) + 9 Now substitute the right side of g(x) into the equation for f(x)

f(g(x)) = -9(sqrt(x + 1) ) + 9

Edit to get the numerical answer.

This should be your answer or the expression that will give you the answer.

f(g(24)) = -9(sqrt(24+1) ) + 9

f(g(24)) = -9sqrt(25) +9

f(g(24)) = - 9*5 + 9

f(g(24)) = - 45 + 9

f(g(24)) = - 36

7 0

3 years ago

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you build four scale models of the Empire State Building. The smallest model is 3 centimeters tall. The height of each model is

WARRIOR [948]

The answer would be $3^4$ = 81

4 0

3 years ago

Read 2 more answers

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

Points H and F lie on circle c What is the length of line segment GH?

Nina [5.8K]

Answer:

6 units

Step-by-step explanation:

Given: Points H and F lie on circle with center C. EG = 12, EC = 9 and ∠GEC = 90°.

To find: Length of GH.

Sol: EC = CH = 9 (Radius of the same circle are equal)

Now, GC = GH + CH

GC = GH + 9

Now In ΔEGC, using pythagoras theorem,

$(Hypotenuse)^{2} = (Base)^{2} +(Altitude)^{2}$ ......(ΔEGC is a right triangle)

$(GC)^{2} = (GE)^{2} +(EC)^{2}$

$(GH + 9)^{2} = (9)^{2} +(12)^{2}$

$(GH )^{2} + (9)^{2} + 18GH = 81 + 144$

$(GH )^{2} + 81 + 18GH = 81 + 144$

$(GH )^{2} + 18GH = 144$

Now, Let GH = <em>x</em>

$x^{2} +18x = 144$

On rearranging,

$x^{2} +18 x - 144 = 0$

$x^{2} - 6x +24x + 144 = 0$

$x (x-6) + 24 (x - 6) =0$

$(x - 6) (x + 24) = 0$

So x = 6 and x = - 24

∵ x cannot be - 24 as it will not satisfy the property of right triangle.

Therefore, the length of line segment GH = 6 units. so, Option (D) is the correct answer.

3 0

3 years ago

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