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1 year ago

The sum of two numbers is 79. Three

Mathematics

2 answers:

Oksana_A [137]1 year ago

8 0

The 2 numbers are 56 and 23
Explanation:
56+23=79
56X3=168
23X5=115
168+115=283

Semmy [17]1 year ago

7 0

Answer:

The two numbers are 23 and 56

Step-by-step explanation:

Let's say the two numbers are A and B.

We are told:

1) A+B=79

2) 3A+5B=283

Let's take the first expression and solve for A:

A+B=79

A=79-B

Now use this value of A in the second expression:

3A+5B=283

3(79-B)+5B=283

237-3B+5B=283

2B = 46

B = 23

Since B=23, we know from 1) that

A+B=79

A+23=79

A = 56

CHECK:

Does A+B=79?

56+23 = 79? YES

Does 3A+5B=283?

3(56)+5(23)=283

168 + 115 = 283? YES

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3 years ago

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2 years ago

A student walk 60m on a bearing

Katena32 [7]

Answer:

d = 234.6 m

Step-by-step explanation:

You can consider a system of coordinates with its origin at the beginning of the walk of the student.

When she start to walk, she is at (0,0)m. After her first walk, her coordinates are calculated by using the information about the incline and the distance that she traveled:

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she is at the coordinates (52.97 , 28.16)m.

Next, when she walks 180m to the east, her coordinates are:

(52.97+180 , 28.16)m = (232.97 , 28.16)m

To calculate the distance from the final point of the student to the starting point you use the Pythagoras generalization for the distance between two points:

$d=\sqrt{(x-x_o)^2+(y-y_o)^2}\\\\x=232.97\\\\x_o=0\\\\y=28.16\\\\y_o=0\\\\d=\sqrt{(232.97-0)^2+(28.16-0)^2}m=234.6m$

The displacement of the student on her complete trajectory was of 234.6m

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2 years ago