1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
yawa3891 [41]
3 years ago
14

A flock of geese is flying north for the summer while a bird watcher is observing. He notices that they are flying in the shape

of an absolute value graph. The lead goose is 5 miles east and 3 miles north of the bird watcher. A second goose is flying 1 mile east and 2 miles north of the bird watcher. If the bird watcher is standing at the origin write an equation for the flight of the geese in the form of y=a|(x-h)|+k
Mathematics
1 answer:
nata0808 [166]3 years ago
4 0

Answer:

y = -¼│x − 5│+ 3

Step-by-step explanation:

y = a│x − h│+ k

(h, k) is the vertex of the absolute value graph.  In this case, it's (5, 3).

y = a│x − 5│+ 3

One point on the graph is (1, 2).  Plug in to find the value of a.

2 = a│1 − 5│+ 3

2 = 4a + 3

a = -¼

Therefore, the graph is:

y = -¼│x − 5│+ 3

You might be interested in
Ples help me find slant assemtotes
FrozenT [24]
A polynomial asymptote is a function p(x) such that

\displaystyle\lim_{x\to\pm\infty}(f(x)-p(x))=0

(y+1)^2=4xy\implies y(x)=2x-1\pm2\sqrt{x^2-x}

Since this equation defines a hyperbola, we expect the asymptotes to be lines of the form p(x)=ax+b.

Ignore the negative root (we don't need it). If y=2x-1+2\sqrt{x^2-x}, then we want to find constants a,b such that

\displaystyle\lim_{x\to\infty}(2x-1+2\sqrt{x^2-x}-ax-b)=0

We have

\sqrt{x^2-x}=\sqrt{x^2}\sqrt{1-\dfrac1x}
\sqrt{x^2-x}=|x|\sqrt{1-\dfrac1x}
\sqrt{x^2-x}=x\sqrt{1-\dfrac1x}

since x\to\infty forces us to have x>0. And as x\to\infty, the \dfrac1x term is "negligible", so really \sqrt{x^2-x}\approx x. We can then treat the limand like

2x-1+2x-ax-b=(4-a)x-(b+1)

which tells us that we would choose a=4. You might be tempted to think b=-1, but that won't be right, and that has to do with how we wrote off the "negligible" term. To find the actual value of b, we have to solve for it in the following limit.

\displaystyle\lim_{x\to\infty}(2x-1+2\sqrt{x^2-x}-4x-b)=0

\displaystyle\lim_{x\to\infty}(\sqrt{x^2-x}-x)=\frac{b+1}2

We write

(\sqrt{x^2-x}-x)\cdot\dfrac{\sqrt{x^2-x}+x}{\sqrt{x^2-x}+x}=\dfrac{(x^2-x)-x^2}{\sqrt{x^2-x}+x}=-\dfrac x{x\sqrt{1-\frac1x}+x}=-\dfrac1{\sqrt{1-\frac1x}+1}

Now as x\to\infty, we see this expression approaching -\dfrac12, so that

-\dfrac12=\dfrac{b+1}2\implies b=-2

So one asymptote of the hyperbola is the line y=4x-2.

The other asymptote is obtained similarly by examining the limit as x\to-\infty.

\displaystyle\lim_{x\to-\infty}(2x-1+2\sqrt{x^2-x}-ax-b)=0

\displaystyle\lim_{x\to-\infty}(2x-2x\sqrt{1-\frac1x}-ax-(b+1))=0

Reduce the "negligible" term to get

\displaystyle\lim_{x\to-\infty}(-ax-(b+1))=0

Now we take a=0, and again we're careful to not pick b=-1.

\displaystyle\lim_{x\to-\infty}(2x-1+2\sqrt{x^2-x}-b)=0

\displaystyle\lim_{x\to-\infty}(x+\sqrt{x^2-x})=\frac{b+1}2

(x+\sqrt{x^2-x})\cdot\dfrac{x-\sqrt{x^2-x}}{x-\sqrt{x^2-x}}=\dfrac{x^2-(x^2-x)}{x-\sqrt{x^2-x}}=\dfrac
 x{x-(-x)\sqrt{1-\frac1x}}=\dfrac1{1+\sqrt{1-\frac1x}}

This time the limit is \dfrac12, so

\dfrac12=\dfrac{b+1}2\implies b=0

which means the other asymptote is the line y=0.
4 0
3 years ago
Jack leaves school to go home, He walks 6 blocks North and then 8
inysia [295]
Jack is 10 blocks away from the school
3 0
3 years ago
What is the range and domain of (-3,-22) (-2,-8) (-1,2) (0,8)) (1,10) (2,8) (3,2)​
BartSMP [9]

Answer:

Step-by-step explanation:

domain is x component so

domain = {-3 , -2 , -1 , 0 , 1 , 2 , 3 }

range is y component so

range = { -22 , -8 , 2 , 8 , 10 , 2 }

note : repeated element can be written only one time. For eg here in range 8 is 2 times but we can write only one time because it is the rule for lisying range and domain.

3 0
3 years ago
Brainliest, what is the total measure of angles 8 and 5 of angle 7 equals 61
Nata [24]
Angle 6 = angle 7 = 61⁰, as vertical angles
Angle 5 = angle 8,   as vertical angles

angle 6+angle 7 + angle 5 + angle 8 =360
61+61 + angle 5 + angle 8 =360
angle 5 + angle 8 = 360 - 61*2 = 238⁰

Answer D.  238⁰
8 0
3 years ago
Can someone plz help me... it would be great if u did all three questions
Artyom0805 [142]
1: 48
2: 18
3: 73
Hope this helps you!
4 0
3 years ago
Read 2 more answers
Other questions:
  • Anyone know the answer to this?
    12·1 answer
  • Log2x16=log39<br>need help to figure out
    12·1 answer
  • Solve this equation please −5×(−2)=?
    13·2 answers
  • Find the area of the triangle below.
    10·1 answer
  • A line passes through the point (6,1)and has a slope of 3/2. write an equation in slope intercept form
    12·2 answers
  • Midpoint of (-1,-4)(2,-1)
    15·1 answer
  • I need help asap!! I’m lost so bad
    5·1 answer
  • What is the measure of &lt;2?
    15·1 answer
  • Peter increased the amount of crunches he did from 32 to 36. By what percentage did Peter increase the amount of crunches he com
    13·1 answer
  • The graph of the function f(x) = (x-3)(x + 1) is shown.
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!