a swimmer swam 48 kilometers in d days. What is the value of the d if the swimmer swam an average of 3.2 kilometers daily?

A trapezoid and a rectangle are both quadrilaterals. This means they both have_____ sides and_______ angles. The angles in a rec

Yanka [14]

Answer:

both have equal sides and equal Angel's

the angels In a rectangle 4 sides

the angles in a trapezoid is 4 sides

4 0

2 years ago

50 POINTS SOMEONE PLEASE HELP WHATS 2+2 <br> A.) 4<br><br> B.) 2<br><br> C.)5<br><br> D.) 1230

Semenov [28]

Answer:

2+2=4

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Identify a number that results in a 4-digit-number when 156,350 is subtracted from it

jekas [21]

160,000 - 156,350 =3,650

7 0

2 years ago

Help. Lol 6th grade math.

balu736 [363]

Answer:

The first one is howard and the second one is peter

Step-by-step explanation:

Howard is at a higher elevation because he is above the sea level.

Peter is further from sea level because he is 1.1 meters away, whereas Howard is only 0.8 meters away.

5 0

2 years ago

Read 2 more answers

21. Who is closer to Cameron? Explain.

pickupchik [31]

Problem 21

<h3>Answer: Jamie is closer</h3>

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Explanation:

A = Arthur's location = (20,35)
J = Jamie's location = (45,20)
C = Cameron's location = (65,40)

To find out who's closer to Cameron, we need to compute the segment lengths AC and JC. Then we pick the smaller of the two lengths.

We use the distance formula to find each length

Let's find the length of AC.

$A = (x_1,y_1) = (20,35)\\\\C = (x_2,y_2) = (65,40)\\\\d = \text{Distance from A to C} = \text{length of segment AC}\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(20-65)^2 + (35-40)^2}\\\\d = \sqrt{(-45)^2 + (-5)^2}\\\\d = \sqrt{2025 + 25}\\\\d = \sqrt{2050}\\\\d \approx 45.2769257\\\\$

The distance from Arthur to Cameron is roughly 45.2769257 units.

Let's repeat this process to find the length of segment JC

$J = (x_1,y_1) = (45,20)\\\\C = (x_2,y_2) = (65,40)\\\\d = \text{Distance from J to C} = \text{length of segment JC}\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(45-65)^2 + (20-40)^2}\\\\d = \sqrt{(-20)^2 + (-20)^2}\\\\d = \sqrt{400 + 400}\\\\d = \sqrt{800}\\\\d \approx 28.2842712\\\\$

Going from Jamie to Cameron is roughly 28.2842712 units

We see that segment JC is shorter than AC. Therefore, Jamie is closer to Cameron.

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Problem 22

<h3>Answer: Arthur is closest to the ball</h3>

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Explanation:

We have these key locations:

A = Arthur's location = (20,35)
J = Jamie's location = (45,20)
C = Cameron's location = (65,40)
B = location of the ball = (35,60)

We'll do the same thing as we did in the previous problem. This time we need to compute the following lengths:

AB
JB
CB

These segments represent the distances from a given player to the ball. Like before, the goal is to pick the smallest of these segments to find out who is the closest to the ball.

The steps are lengthy and more or less the same compared to the previous problem (just with different numbers of course). I'll show the steps on how to get the length of segment AB. I'll skip the other set of steps because there's only so much room allowed.

$A = (x_1,y_1) = (20,35)\\\\B = (x_2,y_2) = (35,60)\\\\d = \text{Distance from A to B} = \text{length of segment AB}\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(20-35)^2 + (35-60)^2}\\\\d = \sqrt{(-15)^2 + (-25)^2}\\\\d = \sqrt{225 + 625}\\\\d = \sqrt{850}\\\\d \approx 29.1547595\\\\$

Segment AB is roughly 29.1547595 units.

If you repeated these steps, then you should get these other two approximate segment lengths:

JB = 41.2310563

CB = 36.0555128

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So in summary, we have these approximate segment lengths

AB = 29.1547595
JB = 41.2310563
CB = 36.0555128

Segment AB is the smallest of the trio, which therefore means Arthur is closest to the ball.

3 0

2 years ago