It’s due today and I have no idea what to do!! someone please help me ://

Mathematics

1 answer:

vfiekz [6]3 years ago

4 0

Answer:

$3 \sqrt{5}$

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The sides of a triangle are x,x+1,2x-1 and its area is x root 10. Find the value of x

Marianna [84]

You can use Heron's formula to solve this problem.

 $A= \sqrt{s(s-a)(s-b)(s-c)}$ , where A is the area; a,b,с are the sides; s is the semiperimeter of the triangle
 $s= \frac{a+b+c}{2}$

According to conditions:
a = x
b = x+1
с = 2x-1

Calculate the semiperimeter:
 $s= \frac{x+x+1+2x-1}{2} = \frac{4x}{2}=2x$

Now we can compose and solve the equation according to Heron's formula:
 $\sqrt{2x(2x-x)(2x-(x+1))(2x-(2x-1))}=x \sqrt{10} \\ \sqrt{2x(x)(2x-x-1)(2x-2x+1)}=x \sqrt{10} \\ \sqrt{2x^2(x-1)(1)}=x \sqrt{10} \\ \sqrt{2x^3-2x^2}=x \sqrt{10} \ \ \ |()^2 \\ 2x^3-2x^2=10x^2 \\ 2x^3-2x^2-10x^2=0 \\ 2x^3-12x^2=0 \ \ |:2 \\ x^3-6x^2=0 \\ x^2(x-6)=0 \\ \\ x^2=0 \\ x=0 \ \ \ \O \\ or \\ x-6=0 \\ x=6$

As a result, x = 6 is your answer.

I hope this helped.

7 0

3 years ago

PLEASE HELP ASAP!!!!!! Quadrilateral PQRS, with vertex P(-5, -3), undergoes a transformation to form quadrilateral P′Q′R′S′, wit

Nana76 [90]

Answer:

The vertex Q' is at (4,5)

Step-by-step explanation:

Given:

Quadrilateral PQRS undergoes a transformation to form a quadrilateral P'Q'R'S' such that the vertex point P(-5,-3) is transformed to P'(5,3).

Vertex point Q(-4,-5)

To find vertex Q'.

Solution: