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11Alexandr11 [23.1K]
3 years ago
11

When solving negative one over three (x − 15) = −4, what is the correct sequence of operations?

Mathematics
1 answer:
iogann1982 [59]3 years ago
7 0
-1/3(x - 15) = -4

step 1 : multiply both sides by -3, step 2 : add 15 to both sides

because when u multiply both sides by -3, this cancels out the -1/3 on the left side leaving u with : x - 15 = -4 * -3.....x - 15 = 12.....and then u would add 15 to both sides....giving u : x = 12 + 15.....x = 27
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Step-by-step explanation:

arc BE is the same as the interior angle, or 76°

BE = 76 °

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Suppose the lengths of two sides of a right triangle are represented by 2x and 3 (x + 1), and the longest side is 17 units. Find
densk [106]

Answer:

x=4

Step-by-step explanation:

<u>Step 1</u>:-

given the lengths of two sides of a right angle are represented by 2x and 3(x+1) and longest side is 17 units.

AB = 2x and BC = 3(x+1) and longest side AC= 17

by using Pythagoras theorem

AC^2 = AB^2 + BC^2

<u>step 2:-</u>

The hypotenuse is longest side is AC = 17 units

(17)^2 = 4x^2 +9(X+1)^2

on simplification, we will use formula

(a + b)^2 = a^2 +2ab+b^2

289 = 4x^2 +9(x^2+2x+1)

13x^2 +18x-280 = 0

finding factors  70 X 52 = 3640

13x^2 +70x-52x-280 = 0

13x^2 -52x+ 70x-280 = 0

Taking common , we get

13x(x-4)+70(x-4)=0

x-4=0 and 13x+70=0

x=4 and 13x =-70

x=4 and x=\frac{-70}{13}

we can not choose negative value so x value is 4

Final answer:- x = 4

<u>verification:-</u>

<u></u>AC^2 = AB^2 + BC^2<u></u>

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289 = 64 +9(25)

289=289

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3 years ago
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I cant help you on this if there is no coefficients.

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