Write 2 different equations with a sum of 9/5.

Perimeter is the distance around a figure or region.<br> a. true<br> b. false

makkiz [27]

A. True

The sum of the distances round the shape is the Perimeter.

3 0

3 years ago

Read 2 more answers

Can someone plz help me with this and i need the drawing to show work

Burka [1]

Answer:

20.833 feet

Step-by-step explanation:

The light reflects off the mirror at the same angle that it hits it at. So the triangle formed by the beam and mirror is similar to the triangle formed by Kevin and the mirror.

Therefore, we can write and solve a proportion.

h / 10 = 6.25 / 3

h = 20.833

The beam is 20.833 feet tall.

4 0

3 years ago

11b-12<-100

ruslelena [56]

Answer:

ok wait which symbol? this (<)?

well it should stay the same since -100 is greater than the 11b-12

LHS=RHS

11b÷11b b must cancell b also do the same on -12. -12÷11 will give u a lesser value compared to -100

prove me wrong

4 0

3 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.

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