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3 years ago

Factorise the following quadratic equations

Mathematics

2 answers:

vesna_86 [32]3 years ago

5 0

Answer:

see explanation

Step-by-step explanation:

In 13 - 17

Consider the factors of the constant term which sum to give the coefficient of the x- term

x² - x - 42 = (x - 7)(x + 6)

x² + x - 6 = (x + 3)(x - 2)

x² - 27x + 50 = (x - 25)(x - 2)

r² - 25 ← is a difference of squares and factors in general as

a² - b² = (a - b)(a + b) , thus

r² - 25

= r² - 5² = (r - 5)(r + 5)

stich3 [128]3 years ago

4 0

1. (x+6)(x-7)

2. (x+3)(x-2)

3. (x-2)(x-25)

4. (x-5)(x+5)

I hope it helps

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Use vectors to find the interior angles of the triangle with the given vertices. (Enter your answers as a comma-separated list.

SOVA2 [1]

Answer:

23.92°, 78.503°, and 77.577°

Step-by-step explanation:

The coordinates of the vertices of the triangle are;

X(-3, -5), Y(2, 6), Z(6, 3)

The vectors are;

z = $\left \langle 2 - (-3), 6 - (-5) \right \rangle$ = $\left \langle 5, 11 \right \rangle$

y = $\left \langle 6 - (-3), 3 - (-5) \right \rangle$ = $\left \langle 9, 8 \right \rangle$

x = $\left \langle 2 - 6, 6 - 3 \right \rangle$ = $\left \langle -4, 3 \right \rangle$

$cos (\alpha ) = \dfrac{z \cdot y}{\left | z \right | \times \left | y \right |}$

Therefore, we get;

$cos (\alpha ) = \dfrac{5 \times 9 + 11 \times 8 }{\left | \sqrt{5^2 + 11^2} \right | \times \left | \sqrt{9^2 + 8^2} \right |} = \dfrac{133}{\sqrt{146} \times \sqrt{145} } \approx 0.91409$

α = arccos(0.91490) ≈ 23.92°

γ = The angle between -y and x

-y = $\left \langle -9, -8 \right \rangle$

We get;

$cos (\gamma) = \dfrac{-y \cdot x}{\left | -y \right | \times \left | x \right |}$

Therefore;

$cos (\gamma) = \dfrac{-9 \times -4 + -8 \times 3 }{\left | \sqrt{(-9)^2 + (-8)^2} \right | \times \left | \sqrt{(-4)^2 + 3^2} \right |} = \dfrac{12}{\sqrt{145} \times \sqrt{25} } \approx 0.199309$