1.(8x+3)(4x-5) 2.(10x-2)(3x+6) please show your work will mark brainliest

Mathematics

1 answer:

ANEK [815]3 years ago

8 0

Answer:

32x^2 - 28x - 15

30x^2 + 54x - 12

Step-by-step explanation:

(8x+3)(4x-5)

Using FOIL

= (8x)(4x) + (8x)(-5) + (3)(4x) + (3)(-5)

= 32x^2 - 40x + 12x - 15

= 32x^2 - 28x - 15

(10x-2)(3x+6)

Using FOIL

= (10x)(3x) + (10x)(6) + (-2)(3x) + (-2)(6)

= 30x^2 + 60x - 6x - 12

= 30x^2 + 54x - 12

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The line with equation −a+3b=0 coincides with the terminal side of an angle θ in standard position and sinθ<0 .

Anna11 [10]

If $a$ is the variable of the horizontal axis, then you can solve for $b$ to get the equation of the line in slope-intercept form in the $a,b$ plane:

$-a+3b=0\implies b=\dfrac a3$

i.e. a line with slope $\dfrac13$ through the origin, which means it is contained in the first and third quadrants. Since the terminal side of $\theta$ has a negative sine, the angle must lie in the third quadrant.

Because the slope of the line is $\dfrac13$ , you can choose any length along the line to make up the hypotenuse of a right triangle with reference angle $\theta$ . Any such right triangle will have $\tan\theta=\dfrac13$ , regardless of whether the angle is the first or third quadrant. But since $\theta$ is known to lie in the third quadrant, and so $\sin\theta$ and $\cos\theta$ are both negative, you have

$\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac13\implies \dfrac{\sin\theta}{\cos\theta}=\dfrac{-1}{-3}\implies \cos\theta=-3$