Ask question

Ask question

All categories

3 years ago

14

Hi someone please help me I'm not sure how to do thissss

1 answer:

Otrada [13]3 years ago

6 0

Answer: (√10)*a

Step-by-step explanation:

In a triangle rectangle, we have 3 sides, in this case, the shorter base, a, the longer base, b and the hypotenuse h.

And the hypotenuse can be found with the Pythagorean's theorem.

a^2 + b^2 = h^2

now, we know that b = 3a, then we can replace it in the above equation:

a^2 + (3*a)^2 = h^2

a^2 + 9*a^2 = h^2

10*a^2 = h^2

now we can apply the square root to both sides and get:

√(10*a^2) = √h^2

(√10)*a = h

Then we have found the length of the hypotenuse in terms of the shorter side a.

You might be interested in

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

The graph shows the amount of money earned by two different workers.

Masteriza [31]

Answer:

The asnwer is B

Step-by-step explanation:

HOPE YOU DO WELL

3 0

3 years ago

Read 2 more answers

HURRY PLEASE!!!!!

tatuchka [14]

Answer:

1.32t

Step-by-step explanation:

Raise 1.08 to the 4th power. (1.08 ^1/4)^4t=1.36t

6 0

3 years ago

Question 6 of 15<br> Which of the following is a polynomial?

disa [49]

Answer:

A

Step-by-step explanation:

5 0

3 years ago

Can someone please help me solve and simplify this??

Viefleur [7K]

Answer:

The Answer is -x + 3.

Step-by-step explanation:

It's honestly one of the longest process'. Above is the right answer though. Hope it helped.

7 0

3 years ago