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3 years ago

Someone please help me answer this!!

Mathematics

1 answer:

steposvetlana [31]3 years ago

4 0

Answer:

10 units is the answer.

Step-by-step explanation:

points are (-3,2) and (-9,-6)

Pythagorean theorem can be written as,

$d=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \\\\d=\sqrt{(-9-(-3))^2 + (-6-2)^2}\\\\d=\sqrt{(-6)^2+(-8)^2} \\\\d=\sqrt{36+64} \\\\d=\sqrt{100} \\\\d=10$

∴ distance between the points = 10 units

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Jaylen and ken started jogging from the same place in the opposite directions along a straight path. jaylen's speed was 30 m/min

erica [24]

The average speed of Ken is 70 m/min. Using the relation between the speed and distance, the required speed for Ken is calculated.

<h3>What is the relation between speed and distance?</h3>

The relationship between speed S and distance D with a period T is formed as

S = D/T

Units: m/sec (basic unit)

<h3>Calculation:</h3>

Given that,

Jaylen and Ken started jogging from the same place in opposite directions along a straight path.

This means, that the sum of distances traveled by both of them in opposite directions = 13.6 km (since it is given that at the end of their jogging, they are 13.6 km apart)

Consider,

The distance traveled by Jaylen = x km

Then the distance traveled by Ken = x - 2.4 km (since Jaylen jogged 2.4 km more than Ken)

Step 1: Finding the distance traveled by them:

x + x - 2.4 = 13.6

⇒ 2x = 13.6 + 2.4

⇒ 2x = 16

⇒ x = 8 km

So, the distance travelled by Jaylen = 8 km and by Ken = 8 - 2.4 = 5.6 km

Step 2: Finding the average speed traveled by Ken:

Since it is given that,

Jaylen's speed was 30 m/min faster than Ken's speed.

So,

For 30 m, the time = 1 min

then for 2.4 km i.e., 2400 m, the time = 80 min.

Therefore,

the average speed of Ken S = D/t

⇒ S = 5600 m / 80 min

∴ S = 70 m/min.

So, the average speed of Jaylen will be 70 m/min + 30 m/min = 100 m/min.

Thus, the average speed of Ken is 70 m/min.

Learn more about the relation between speed, distance, and time here:

brainly.com/question/3004254

#SPJ2

5 0

2 years ago

you invest $1,000 in each of two accounts. account a earns simple interest at a rate of2.42% over 4 years. account b earns sim

Setler79 [48]

Answer:

Account A = $ 96.80

Account B = $ 48.40

Step-by-step explanation:

Account A

Principal = $1,000

rate = 2.42% = 0.0242

time = 4 years

To find the interest we will use the formula :

I = PTR

I = 1000 x 4 x 0.0242

I = $96.80

Account B

P = 1,000

t = 24 months = 2 years

r = 2.42% = 0.0242

I = PTR

I = 1000 x 2 x 0.0242

I = $ 48. 40

difference in interest = Interest a - Interest b

difference = 96.80 - 48.40

difference = $ 48 . 40

The interest on Account A doubles the interest on Account B

6 0

3 years ago

Read 2 more answers

the specification for a plastic handle calls for a length of 6.0 inches +- .2 inches. The standard deviation of the process is e

erastova [34]

Answer:

USL = 6.2 inches

LSL = 5.8 inches

b) Cp = 1.33

Cpk = 0.67

Yes it meets all specifications

Step-by-step explanation:

The specification for a plastic handle calls for a length of 6.0 inches ± .2 inches. The standard deviation of the process is estimated to be 0.05 inches. What are the upper and lower specification limits for this product? The process is known to operate at a mean thickness of 6.1 inches. What is the Cp and Cpk for this process? Is this process capable of producing the desired part?

Given that:

Mean (μ) = 6.1 inches, Standard deviation (σ) = 0.05 inches and the length of the plastic handle is 6.0 inches ± .2

a) Since the length of the plastic handle is 6.0 inches ± .2 = (6 - 0.2, 6 + 0.2)

The Upper specification limits (USL) = 6 inches + 0.2 inches = 6.2 inches

The lower specification limits (LSL) = 6 inches - 0.2 inches = 5.8 inches

b) The Cp is given by the formula:

$Cp=\frac{(USL-LSL)}{6\sigma} =\frac{(6.2-5.8)}{6*0.05} =1.33$

The Cpk is given by the formula:

The upper specification limit lies about 3 standard deviations from the centerline, and the lower specification limit is further away, so practically all units will meet specifications

$Cpk=min(\frac{USL-\mu}{3\sigma},\frac{\mu -LSL}{3\sigma})=min(\frac{6.2-6.1}{3*0.05},\frac{6.1-5.8}{3*0.05})=min(0.67,2)=0.67$