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3 years ago

I need help with geometry one my question is construct AD congruent to JK

Mathematics

1 answer:

VARVARA [1.3K]3 years ago

7 0

Answer:

use your compass to copy the length to a new line

Step-by-step explanation:

To make a line segment that has the same length as another, set your compass width to the length of the one you want to copy, and use that to mark the length on the line where you want your copy.

Use a straightedge to draw a line longer than JK.

Label a point on that line A.
Set your compass to the length JK.
Mark that length from point A on the line you drew. Label the mark D.

AD is congruent to JK.

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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 answer correctly, if you don't know it - please just pass by.

Kazeer [188]

Answer: 7.50x500...try that and that’s your answer

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Your math teacher tells you that the next test is worth 100 points and contains 38 problems. Multiple-choice questions are worth

rusak2 [61]

If we assume that
x isthe number of multiple choice questions, and y is the number of word problems, we can write a system of equations like:
{ x + y = 38
{ 2x + 5y = 100

If we multiply the first equation by -2 we get:
{ -2x-2y = -76
{ 2x + 5y = 100

Now if we add both equations we get only equation with 1 unknown (y) :

3y = 24 ⇒ y = 8

Substituting the calculated value to the first equation we get:

x + 8 = 38 ⇒ x = 30

The solution:
{ x = 30
{ y = 8

means that:
The test had 30 multiple choice questions, and 8 word problems.

4 0

3 years ago

Find a report in a newspaper or magazine or in the internet that uses results from a survey

Radda [10]

I think you might need to be more specific..

8 0

3 years ago

Pls solve this for me :(

Rus_ich [418]

Answer:

33.83

Step-by-step explanation:

This is because if you trying to find h and the opposite side, you will have to use cos to get that answer. My using this equation

$\cos(31) = \frac{29}{x}$

You will find that x will equal 33.83

Hope this Helped

6 0

3 years ago

I need help with geometry one my question is construct AD congruent to JK​

I need help with geometry one my question is construct AD congruent to JK