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wel
2 years ago
12

PLEASE HELP: On the assembly line, Max can process 15 engines in 40 mins. How many engines can he process in 2 hours?

Mathematics
2 answers:
Olegator [25]2 years ago
8 0

Answer:

He can process 45 engines in 2 hours.

Step-by-step explanation:

2 hours = 120 minutes

You can multiply 3 to 40 because 40x 3 = 120, but you have to do the same to 15, so 15 x 3 = 45.

Or you can divide the engines (15) by minutes (40) to find out how many engines can be processed per minute.

15/40 = 0.375 an engine per minute.

Then multiply that by 120, which is two hours in minutes.

0.375 x 120 = 45

Mama L [17]2 years ago
6 0
45 engines in 2 hours
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just olya [345]

ANSWER

f(0) = g(0)

EXPLANATION

From the graph we

g(0) =  - 2

because this is where the line x= 0 meets the graph of g(x).

Also

f(0) =  - 2

because this is where the line x= 0 meets the graph of f(x).

This implies that,

f(0) = g(0)

The correct choice is A.

6 0
3 years ago
Please solve this in 2 parts as stated... thank you!
fredd [130]

Solve the equation we have that:

a. S = 3V²/ 4

b. S = 108 feet

<h3>How to solve the parameters?</h3>

We have the equation to be;

V = 2√3S

Where;

  • V is the speed
  • S is the length

Let's make S the subject from the formula, divide both sides by 2√2S = v/ 2. Take square of both sides:

(\sqrt{3S} )^2 = \frac{V^2}{4}\\3S = \frac{V^2}{4}

Now we need to divide both sides by three:

S = 3V²/ 4

b. If v = 12miles per hour, substituting in the equation found earlier:

S = 3 (12²)/ 4

S= 432/ 4

S = 108 feet

Thus, the formula for the length is S = 3V²/ 4

Learn more about speed here: brainly.com/question/7359669

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4 0
1 year ago
Define table represents a group frequency discretion of the number of hours spent on the computer per week for 49 student. What
Scilla [17]

Observe the given data distribution table carefully.

The 5th class interval is given as,

14.0-17.4

The upper limit (UL) and lower limit (LL) of this interval are,

\begin{gathered} UL=17.4 \\ LL=14.0 \end{gathered}

Thus, the upper-class limit of this 5th class is 17.4.

6 0
1 year ago
What is the longest line segment that can be drawn in a right rectangular prism that is 15 cm​ long, 11 cm​ wide, and 10 cm​ tal
sammy [17]

Answer:

  √446 ≈ 21.12 cm

Step-by-step explanation:

The longest dimension of a rectangular prism is the length of the space diagonal from one corner to the opposite corner through the center of the prism. The Pythagorean theorm tells you the square of its length is the sum of the squares of the dimensions of the prism:

  d² = (15 cm)² +(11 cm)² +(10 cm)² = (225 +121 +100) cm² = 446 cm²

  d = √446 cm ≈ 21.12 cm

The longest line segment that can be drawn in a right rectangular prism is about 21.12 cm.

_____

<em>Additional comment</em>

The square of the face diagonal is the sum of the squares of the dimensions of that face. The square of the space diagonal will be the sum of that square and the square of the remaining prism dimenaion, hence the sum of squares of all three prism dimensions.

8 0
3 years ago
Turn 5/7 into its smallest form
DiKsa [7]

Answer:

5/7

Step-by-step explanation:

5 0
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