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AysviL [449]
2 years ago
12

PLEASE I need help on these 4 problems you don’t have to do them all but if you can at least do one of them TYSM

Mathematics
1 answer:
Brrunno [24]2 years ago
4 0

Answer:

For right angle triangle,

we use Pythagoras theorem that is:

c^{2} =a^{2} +b^{2}

c = \sqrt{a^{2} +b^{2} }

For question 1:

c = ?

a = 40

b = 9

putting them in formula,

c = \sqrt{40^{2} + 9^{2} }

c = 41

For question 2:

c = ?

a = 12

b = 13

putting them in formula,

c = \sqrt{12^{2} + 13^{2} }

c = approximately 17.69181

For question 3:

c = 35

a = 20

b = ?

putting them in formula,

c^{2} =a^{2} +b^{2}

35^{2} = 20^{2} + b^{2}

1225 = 400 + b^{2}

b^{2} = 1225 - 400

b^{2} = 825

\sqrt{b^{2} } = \sqrt{825}

b = 5 \sqrt{33}

For question 4:

c = 37

a = 20

b = ?

putting them in formula,

c^{2} =a^{2} +b^{2}

37^{2} = 20^{2} + b^{2}

1369 = 400 + b^{2}

b^{2} = 1369 - 400

b^{2} = 969

Taking square root on both sides

b = 31.12

Hope it helps.

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Step-by-step explanation:

Example 1

Solve the equation x3 − 3x2 – 2x + 4 = 0

We put the numbers that are factors of 4 into the equation to see if any of them are correct.

f(1) = 13 − 3×12 – 2×1 + 4 = 0 1 is a solution

f(−1) = (−1)3 − 3×(−1)2 – 2×(−1) + 4 = 2

f(2) = 23 − 3×22 – 2×2 + 4 = −4

f(−2) = (−2)3 − 3×(−2)2 – 2×(−2) + 4 = −12

f(4) = 43 − 3×42 – 2×4 + 4 = 12

f(−4) = (−4)3 − 3×(−4)2 – 2×(−4) + 4 = −100

The only integer solution is x = 1. When we have found one solution we don’t really need to test any other numbers because we can now solve the equation by dividing by (x − 1) and trying to solve the quadratic we get from the division.

Now we can factorise our expression as follows:

x3 − 3x2 – 2x + 4 = (x − 1)(x2 − 2x − 4) = 0

It now remains for us to solve the quadratic equation.

x2 − 2x − 4 = 0

We use the formula for quadratics with a = 1, b = −2 and c = −4.

We have now found all three solutions of the equation x3 − 3x2 – 2x + 4 = 0. They are: eftirfarandi:

x = 1

x = 1 + Ö5

x = 1 − Ö5

Example 2

We can easily use the same method to solve a fourth degree equation or equations of a still higher degree. Solve the equation f(x) = x4 − x3 − 5x2 + 3x + 2 = 0.

First we find the integer factors of the constant term, 2. The integer factors of 2 are ±1 and ±2.

f(1) = 14 − 13 − 5×12 + 3×1 + 2 = 0 1 is a solution

f(−1) = (−1)4 − (−1)3 − 5×(−1)2 + 3×(−1) + 2 = −4

f(2) = 24 − 23 − 5×22 + 3×2 + 2 = −4

f(−2) = (−2)4 − (−2)3 − 5×(−2)2 + 3×(−2) + 2 = 0 we have found a second solution.

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First divide by x + 2

Now divide the resulting cubic factor by x − 1.

We have now factorised

f(x) = x4 − x3 − 5x2 + 3x + 2 into

f(x) = (x + 2)(x − 1)(x2 − 2x − 1) and it only remains to solve the quadratic equation

x2 − 2x − 1 = 0. We use the formula with a = 1, b = −2 and c = −1.

Now we have found a total of four solutions. They are:

x = 1

x = −2

x = 1 +

x = 1 −

Sometimes we can solve a third degree equation by bracketing the terms two by two and finding a factor that they have in common.

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