Ivan used coordinate geometry to prove that quadrilateral EFGH is a square.

This is not nearly as threatening and scary as I first thought it was. You must be in the section in Geometry where you are taught that perimeter of similar figures exist in a one-to-one relationship while areas of similar figures exist in a squared-to-squared relationship. We will use that here.

The area formula for a regular polygon is

$A=\frac{1}{2}ap$ where a is the apothem and p is the perimeter. We are first asked for the area of the polygon, but it would make more sense to find the perimeter first, since we need it to find the area.

P = 5(8) so

P = 40

We are given that the area of the triangle inside that polygon is 22.022 units squared. Knowing that the area formula for a triangle is

$A=\frac{1}{2}bh$ we can sub in what we know and solve to find the height:

$22.022=\frac{1}{2}(8)h$ and

22.022 = 4h so

h = 5.5055 units

It just so happens that the height of that triangle is also the apothem of the polygon, so now we have what we need to find the area of the polygon:

$A=\frac{1}{2}(5.5055)(40)$

which gives us an area of

A = 110.11 units squared.

Here is where we can use what we know about similar figures and the relationships between perimeters and areas. We will set up a proportion with the smaller polygon info on top and the larger info on bottom. We know that the larger is 3 times the smaller, so the ratio of smaller to larger is

$\frac{s}{l}:\frac{1}{3}$

Since perimeter is one-to-one and we know the perimeter of the smaller, we can create a proportion to solve for the perimeter of the larger:

$\frac{s}{l}:\frac{1}{3}=\frac{40}{x}$

Cross multiply to get that the perimeter is 120 units. You could also have done this by knowing that if the larger is 3 times the size of the smaller, then the side measure of the larger is 24, and 24 * 5 = 120. But we used the way we used because now we have a means to find the area of the larger since we know the area of the smaller.

Area exists in a squared-to-squared relationship of the perimeter which is one-to-one. If the perimeter ratio is 1:3, then the area relationship is

$\frac{s}{l}:\frac{1^2}{3^2}$ which is, simplified:

$\frac{s}{l}:\frac{1}{9}$

Since we know the area for the smaller, we can sub it into a proportion and cross multiply to solve for the area of the larger.

$\frac{s}{l} :\frac{1}{9} =\frac{110.11}{x}$

A of the larger is 990.99 units squared

5 0

3 years ago

Find the zeros of the function f(x) = x2 + 5x + 6. A) y = 6 because the graph crosses the y-axis at 6. B) y = -0.25 because that

Fofino [41]

Answer:

D.)

Step-by-step explanation:

The zero's are referencing when y=0, note that when y=0 they are talking about the x-intercepts. You can graph the function and see when the graph crosses the x-axis or solve for the x-values. I will solve it via factoring and so:

$f(x)=x^2+5x+6$

Multiply the outer coefficients, in this case 1 and 6, and 1×6=6. Now let's think about all the factors of 6 we have: 6×1 and 2×3. Now is there a way that if we use any of these factors and add/subtract them they will return the middle term 5? Actually we can say 6-1=5 and 2+3=5. Let's try both.

First let's use 6 and -1 and so:

$x^2+5x+6\\\\x^2+6x-x+6\\\\x(x+6)-1(x-6)$

Notice how we have (x+6) and (x-6), these factors do not match so this is incorrect.

Now let's try 2 and 3 and so:

$x^2+5x+6\\\\x^2+3x+2x+6\\\\x(x+3)+2(x+3)\\\\(x+2)(x+3)$

Notice how the factors (x+3) matched up so this is a factor and so is (x+2), now to solve for the zero's let's make f(x)=0 and solve each factor separately:

Case 1:

$f(x)=x+2\\\\0=x+2\\\\x=-2$

Case 2:

$f(x)=x+3\\\\0=x+3\\\\x=-3$

So your zero's are when x=-2 and x=-3.

D.) x=-3 and x=-2 because the graph crosses the x-axis at -3 and -2.

~~~Brainliest Appreciated~~~

8 0

3 years ago