Benjamin wants to purchase a new bike, but he does not have enough money in his bank account to pay for one.

Use the dchnition of similarity to solve the following problem.

sveta [45]

Answer:

1063 ft

Step-by-step explanation:

For shadow problems, we assume the sun's rays are parallel, so the triangles formed are similar. That means the ratio of height to shadow length is the same for each object.

h/(580 ft) = (5.5 ft)/(3 ft)

h = (5.5 ft)(580/3) ≈ 1063 ft

The height of the monument is about 1063 feet.

8 0

3 years ago

Simplify.<br> Remove all perfect squares from inside the square root.<br> √180

miskamm [114]

Answer:

13.42

Step-by-step explanation:

$\sqrt{180} =6\sqrt{5} \\\\6\sqrt{5} =13.42$

3 0

3 years ago

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given the functions f(x)=4x^2-1, g(x)=x^2-8x+5, and h(x)=-3x^2-12x +1, rank them from least to greatest based on their axis of s

Lesechka [4]

The formula for axis of symmetry is -b/2a.

For f(x):
0/2(4)
0/8
Axis of symmetry= 0

For g(x)
-(-8)/2(1)
8/2
Axis of symmetry= 4

For h(x)
-(-12)/2(-3)
12/-6
Axis of symmetry= -2

Order from least to greatest: -2, 0, 4

Final answer: h(x), f(x), g(x)

7 0

3 years ago

The coordinates of the vertices of ΔPQR are (–2, –2), (–6, –2), and (–6, –5). The coordinates of the vertices of ΔPʹQʹRʹ are (−2

Minchanka [31]

No, ΔPʹQʹRʹ is a reflection of ΔPQR over the x-axis

5 0

3 years ago

Read 2 more answers

Consider the following function.

Kryger [21]

Answer:

See below

Step-by-step explanation:

I assume the function is $f(x)=1+\frac{5}{x}-\frac{4}{x^2}$

A) The vertical asymptotes are located where the denominator is equal to 0. Therefore, $x=0$ is the only vertical asymptote.

B) Set the first derivative equal to 0 and solve:

$f(x)=1+\frac{5}{x}-\frac{4}{x^2}$

$f'(x)=-\frac{5}{x^2}+\frac{8}{x^3}$

$0=-\frac{5}{x^2}+\frac{8}{x^3}$

$0=-5x+8$

$5x=8$

$x=\frac{8}{5}$

Now we test where the function is increasing and decreasing on each side. I will use 2 and 1 to test this:

$f'(2)=-\frac{5}{2^2}+\frac{8}{2^3}=-\frac{5}{4}+\frac{8}{8}=-\frac{5}{4}+1=-\frac{1}{4}$

$f'(1)=-\frac{5}{1^2}+\frac{8}{1^3}=-\frac{5}{1}+\frac{8}{1}=-5+8=3$

Therefore, the function increases on the interval $(0,\frac{8}{5})$ and decreases on the interval $(-\infty,0),(\frac{8}{5},\infty)$ .

C) Since we determined that the slope is 0 when $x=\frac{8}{5}$ from the first derivative, plugging it into the original function tells us where the extrema are. Therefore, $f(\frac{8}{5})=1+\frac{5}{\frac{8}{5}}-\frac{4}{\frac{8}{5}^2 }=\frac{41}{16}$ , meaning there's an extreme at the point $(\frac{8}{5},\frac{41}{16})$ , but is it a maximum or minimum? To answer that, we will plug in $x=\frac{8}{5}$ into the second derivative which is $f''(x)=\frac{10}{x^3}-\frac{24}{x^4}$ . If $f''(x)>0$ , then it's a minimum. If $f''(x)$ , then it's a maximum. If $f''(x)=0$ , the test fails. So, $f''(\frac{8}{5})=\frac{10}{\frac{8}{5}^3}-\frac{24}{\frac{8}{5}^4}=-\frac{625}{512}$ , which means $(\frac{8}{5},\frac{41}{16})$ is a local maximum.

D) Now set the second derivative equal to 0 and solve:

$f''(x)=\frac{10}{x^3}-\frac{24}{x^4}$

$0=\frac{10}{x^3}-\frac{24}{x^4}$

$0=10x-24$

$-10x=-24$

$x=\frac{24}{10}$

$x=\frac{12}{5}$

We then test where $f''(x)$ is negative or positive by plugging in test values. I will use -1 and 3 to test this:

$f''(-1)=\frac{10}{(-1)^3}-\frac{24}{(-1)^4}=-34$ , so the function is concave down on the interval $(-\infty,0)\cup(0,\frac{12}{5})$

$f''(3)=\frac{10}{3^3}-\frac{24}{3^4}=\frac{2}{27}>0$ , so the function is concave up on the interval $(\frac{12}{5},\infty)$

The inflection point is where concavity changes, which can be determined by plugging in $x=\frac{12}{5}$ into the original function, which would be $f(\frac{12}{5})=1+\frac{5}{\frac{12}{5}}+\frac{4}{\frac{12}{5}^2 }=\frac{43}{18}$ , or $(\frac{12}{5},\frac{43}{18})$ .

E) See attached graph

5 0

3 years ago