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3 years ago

Complete the table of values for y = 2x + 3

Mathematics

2 answers:

JulsSmile [24]3 years ago

6 0

To complete the table, plug in the values into the equation (y=2x+3) if you already have the y then solve for x using inverse operations, and if you have the x then just use order of operations like normal

babymother [125]3 years ago

5 0

Answer:

you submit the numbers where the is x the numbers on top

check the solutions

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Starting at home, Luis traveled uphill to the gift store for 50 minutes at just 6 mph. He then traveled back home along the same

Orlov [11]

Let S = his average speed.

Let D = distance traveled one way

Time = distance / speed

Set up an equation:

d/6 + d/12 = 2d/S

Multiply the equation by 12s:

12s(d/6) + 12s(d/12) = 12s(2d/s)

Simplify to get:

2sd + sd = 24d

Combine like terms:

3sd = 24d

Divide both sides by d:

3s = 24

Divide both sides by 3:

s = 24/3

s = 8

His average speed was 8 mph.

8 0

3 years ago

Bo Karah can run a race in 4 hours 33 minutes if he runs at 7 m/s. How long will it take in hours and minutes if

lidiya [134]

Answer:

7 hours and 9 minutes

Step-by-step explanation:

4 hours and 33 minutes is 273 minutes

7 m/s —> 273 minutes

11 m/s —> M

Criss cross to get M

11x273= 7M

3003=7M

M=429 minutes

==> 429 minutes is 7 hours and 9 minutes

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2 years ago

Two cargo ships are sailing a direct course into harbor. Ship A is 22 miles from harbor. Ship Bis sailing into harbor on a cours

s344n2d4d5 [400]

Answer:

The distance of ship B from the Harbor is 32.26 miles

Step-by-step explanation:

Hello, great question. These types are questions are the beginning steps for learning more advanced Equations.

I have added a visual aid of the problem. As we can see the courses of both ship to the harbor forms a triangle with ship B being the hypotenuse. Since we are given the distance of the course for ship A and the angle between both ships, we can use this information with the <u>COSINE operator</u> to solve for the length of the course of ship B.

$cos(x) = \frac{adjacent}{hypotenuse}$