Ask question

Ask question

All categories

3 years ago

10

Jarret walks his puppy outside every day for 30 minutes. Angela walked her puppy every day for 90 minutes. Find the number of mi

nutes that each puppy was walked for 1, 2, 3, and 4 days. Then graph the results as ordered pairs. How many more minutes does Angela spend walking her puppy over 2 days compared to Jarret walking his puppy over 2 days?

1 answer:

IceJOKER [234]3 years ago

8 0

Answer:

Read the answers below, including the graph.

Step-by-step explanation:

1. Let's find the number of minutes that each puppy was walked for 1, 2, 3, and 4 days.

Day 1 Day 2 Day 3 Day 4

Angela 90 180 270 360

Jarret 30 60 90 120

Ordered pairs for Angela's puppy: (1,90); (2,180); (3,270); (4,360)

Ordered pairs for Jarret's puppy: (1,30); (2,60); (3,90); (4,120)

2. How many more minutes does Angela spend walking her puppy over 2 days compared to Jarret walking his puppy over 2 days?

180 - 60 = 120

<u>Angela spend walking her puppy 120 more minutes over 2 days compared to Jarret </u>

You might be interested in

1. (1 pt) Which is a solution to the equation y=3x+1 ?A. (3, 10) B. (2, 6) C. (1, 5) D. (0, 3)

frozen [14]

Answer:

the easy way to figure this out is to just plug in answers.

start with a, (10)=3(3)+1

10=9+1

10=10

this makes this a true statement

feel free to try the other options but this is your answer

A. (3,10)

7 0

3 years ago

The sum of two numbers is 196. One number is 64 more than the other. Find the numbers.

ValentinkaMS [17]

Answer:

<h2><u><em>66,130</em></u></h2>

I just guessed random numbers until I got 66 and 130.

4 0

3 years ago

Read 2 more answers

Find the maximum and minimum values of the function f(x,y)=2x2+3y2−4x−5 on the domain x2+y2≤100. The maximum value of f(x,y) is:

ryzh [129]

First find the critical points of <em>f</em> :

$f(x,y)=2x^2+3y^2-4x-5=2(x-1)^2+3y^2-7$

$\dfrac{\partial f}{\partial x}=2(x-1)=0\implies x=1$

$\dfrac{\partial f}{\partial y}=6y=0\implies y=0$

so the point (1, 0) is the only critical point, at which we have

$f(1,0)=-7$

Next check for critical points along the boundary, which can be found by converting to polar coordinates:

$f(x,y)=f(10\cos t,10\sin t)=g(t)=295-40\cos t-100\cos^2t$

Find the critical points of <em>g</em> :

$\dfrac{\mathrm dg}{\mathrm dt}=40\sin t+200\sin t\cos t=40\sin t(1+5\cos t)=0$

$\implies\sin t=0\text{ OR }1+5\cos t=0$

$\implies t=n\pi\text{ OR }t=\cos^{-1}\left(-\dfrac15\right)+2n\pi\text{ OR }t=-\cos^{-1}\left(-\dfrac15\right)+2n\pi$

where <em>n</em> is any integer. We get 4 critical points in the interval [0, 2π) at

$t=0\implies f(10,0)=155$

$t=\cos^{-1}\left(-\dfrac15\right)\implies f(-2,4\sqrt6)=299$

$t=\pi\implies f(-10,0)=235$

$t=2\pi-\cos^{-1}\left(-\dfrac15\right)\implies f(-2,-4\sqrt6)=299$

So <em>f</em> has a minimum of -7 and a maximum of 299.

4 0

3 years ago

EB bisects ∠RED. Find the measure of ∠BED if ∠REB = 5x – 3 and ∠RED = 8x + 16<br><br>plzzzzzzzzzzz

Allushta [10]

Answer:

this is 52 degrees

Step-by-step explanation:

7 0

3 years ago

Security A rounds every 2 minutes and 20 seconds in the gate 1. Security B rounds every 1 minute and 40 seconds in the gate 2. I

Elden [556K]

Answer:

$Time = 11\frac{2}{3}\ min$

Step-by-step explanation:

Given

$Security\ A = 2\ mins, 20\ secs$

$Security\ B = 1\ mins, 40\ secs$

Required

After how many minutes, will they round together

First, convert the given time to minutes

$Security\ A = 2\ mins, 20\ secs$

$Security\ A = 2\ mins+ 20\ secs$

$Security\ A = 2\ mins+ \frac{20}{60}\ min$

$Security\ A = 2\ mins+ \frac{1}{3}\ min$

$Security\ A = 2\frac{1}{3}\ min$

$Security\ B = 1\ mins, 40\ secs$

$Security\ B = 1\ mins+ 40\ secs$

$Security\ B = 1\ mins+ \frac{40}{60}\ min$

$Security\ B = 1\frac{2}{3}\ min$

So, we have:

$Security\ A = 2\frac{1}{3}\ min$

$Security\ B = 1\frac{2}{3}\ min$

List out the multiples of the time of both security personnel take round.

$Security\ A = 2\frac{1}{3}min,\ 4\frac{2}{3}min,\ 7min,\ 9\frac{1}{3}min,\ 11\frac{2}{3}min...$

$Security\ B = 1\frac{2}{3}min,\ 3\frac{1}{3}min,\ 5min,\ 6\frac{2}{3}min,\ 8\frac{1}{3}min,\ 10min\ ,11\frac{2}{3}min,...$

In the above lists, the common time is:

$Time = 11\frac{2}{3}\ min$

<em>This implies that they go on round after </em> $11\frac{2}{3}\ min$ <em></em>

4 0

3 years ago