find two numbers such that the sum of twice the first and thrice the second is 92 ,and four times the first exceeds seven times the second by 2

First number = 25

Second number  = 14

Step-by-step explanation:

Let the numbers be x & y

The sum of twice the first and thrice the second is 92

2x + 3y = 92   (I)

Four times the first exceeds seven times the second by 2

4x - 7y = 2  (II)

Multiply  equation (I) by -2 and then add the equations

(I)*-2          -4x - 6y = -184

(II)               4x - 7y = 2      { Now add, thus x will be eliminated}

                      - 13y = -182 {Divide both sides by -13}

y = -182/-13

y = 14

Plugin the value of y in equation (I)

2x + 3*14 = 92

2x + 42 = 92

       2x = 92 - 42

       2x = 50

        x  = 50/2

x = 25

First number = 25

Second number  = 14

the answer is b: 18

step-by-step explanation:

use the pythagorean theorem to .

a^2+b^2=c^2

plug 18 for "r"

the 2 legs are 18 and 24

18^2+24^2=c^2

324+576=c^2

900=c^2

c=30

30-12=18

18 is the radius

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44

step-by-step explanation:

the