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3 years ago

If function g has the factors (x - 7) and (x + 6), what are the zeros of function g?

Mathematics

1 answer:

SVEN [57.7K]3 years ago

5 0

Answer:

7 and -6

Step-by-step explanation:

Given

Factors: (x - 7) and (x + 6)

Required

Determine the zeros

To do this, we simply equate the factors to 0

$x - 7 = 0$

$x = 7$

$x + 6 = 0$

$x = -6$

So, the zeros are -7 and 6

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If x = (√2 + 1)^-1/3 then the value of x^3 + 1/x^3 is

Shtirlitz [24]

Step-by-step explanation:

Given: x = {√(2) + 1}^(-1/3)

Asked: x³+(1/x³) = ?

Solution:

We have, x = {√(2) + 1}^(-1/3)

⇛x = [1/{√(2) + 1}^(1/3)]

[since, (a⁻ⁿ = 1/aⁿ)]

Cubing on both sides, then

⇛(x)³ = [1{/√(2) + 1}^(1/3)]³

⇛(x)³ = [(1)³/{√(2) + 1}^(1/3 *3)]

⇛(x)³ = [(1)³/{√(2) + 1}^(1*3/3)]

⇛(x)³ = [(1)³/{√(2) + 1}^(3/3)]

⇛(x * x * x) = [(1*1*1)/{√(2) + 1)^1]

⇛x³ = [1/{√(2) + 1}]

Here, we see that on RHS, the denominator is √(2)+1. We know that the rationalising factor of √(a)+b = √(a)-b. Therefore, the rationalising factor of √(2)+1 = √(2) - 1. On rationalising the denominator them

⇛x³ = [1/{√(2) + 1}] * [{√(2) - 1}/{√(2) - 1}]

⇛x³ = [1{√(2) + 1}/{√(2) + 1}{√(2) - 1}]

Multiply the numerator with number outside of the bracket with numbers on the bracket.

⇛x³ = [{√(2) + 1}/{√(2) + 1}{√(2) - 1}]

Now, Comparing the denominator on RHS with (a+b)(a-b), we get

a = √2
b = 1

Using identity (a+b)(a-b) = a² - b², we get

⇛x³ = [{√(2) - 1}/{√(2)² - (1)²}]

⇛x³ = [{√(2) - 1}/{√(2*2) - (1*1)}]

⇛x³ = [{√(2) - 1}/(2-1)]

⇛x³ = [{√(2) - 1}/1]

Therefore, x³ = √(2) - 1 → → →Eqn(1)

Now, 1/x³ = [1/{√(2) - 1]

⇛1/x³ = [1/{√(2) - 1] * [{√(2) + 1}/{√(2) + 1}]

⇛1/x³ = [1{√(2) + 1}/{√(2) - 1}{√(2) + 1}]

⇛1/x³ = {√(2) + 1}/[{√(2) - 1}{√(2) + 1}]

⇛1/x³ = [{√(2) + 1}/{√(2)² - (1)²}]

⇛1/x³ = [{√(2) + 1}/{√(2*2) - (1*1)}]

⇛1/x³ = [{√(2) + 1}/(2-1)]

⇛1/x³ = [{√(2) + 1}/1]

Therefore, 1/x³ = √(2) + 1 → → →Eqn(2)

On adding equation (1) and equation (2), we get

x³ + (1/x³) = √(2) -1 + √(2) + 1

Cancel out -1 and 1 on RHS.

⇛x³ + (1/x³) = √(2) + √(2)

⇛x³ + (1/x³) = 2

Therefore, x³ + (1/x³) = 2

Answer: Hence, the required value of x³ + (1/x³) is 2.

Please let me know if you have any other questions.

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In the isosceles △ABC m∠ACB=120° and AD is an altitude to leg BC . What is the distance from D to base AB , if CD=4cm?

snow_lady [41]

If ΔACB is an isosceles triangle, then ∠A ≅ ∠B and AC ≅ CB

Since ∠C = 120° and ∠A + ∠B + ∠C = 180°, then ∠A = 30° and ∠B = 30°

Next, look at ΔADB. ∠A + ∠D + ∠B = 180°, so ∠A + 90° + 30° = 180° ⇒ ∠A = 30°

Now look at ΔADC. Since ∠A = 30° in ΔACB, and ∠A = 60° in ΔADB, then ∠A = 30° in ΔADC per angle addition postulate.

Now that we have shown that ΔADB and ΔADC are 30-60-90 triangles, we can use that formula to calculate the side lengths.

CD = 4 cm (given) so AC = 2(4 cm) = 8 cm

Since AC ≅ BC, then BC = 8 cm. Therefore, BD = 4 + 8 = 12 by segment addition postulate.

Lastly, look at ΔBHD. Since ∠B = 30° and ∠H = 90°, then ∠D = 60°. So, ΔBHD is also a 30-60-90 triangle.

BD = 12 cm, so HD = $\frac{12}{2}cm$ = 6 cm

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