The linear model of this case takes the form:

y = a(x-b) + k

The cost of having a package delivered has a base fee of $9.70
this is "k" >>>>> k=9.7 (fixed amount of fee)

THEN


Every pound over 5 lbs cost an additional $0.46 per pound

that means: 0.46(x-5)
in other words, if the package weighs foe example 9 pounds, then 9-5=4, it will cost 0.46*4 for these 4 extra pounds

Finally we have the linear form of this: C = 0.46 (W - 5) + 9.7

3 0

2 years ago

The slope of the line containing the points (-5,3) and (-2,1) is A.-3/2 B.-2/3 C.2/3

ad-work [718]

The correct answer is B. -2/3 that is the slope

5 0

2 years ago

16) Please help with question. WILL MARK BRAINLIEST + 10 POINTS.

Katyanochek1 [597]

We will use the sine and cosine of the sum of two angles, the sine and consine of $\frac{\pi}{2}$ , and the relation of the tangent with the sine and cosine:

$\sin (\alpha+\beta)=\sin \alpha\cdot\cos\beta + \cos\alpha\cdot\sin\beta

\cos(\alpha+\beta)=\cos\alpha\cdot\cos\beta-\sin\alpha\cdot\sin\beta$

$\sin\dfrac{\pi}{2}=1,\ \cos\dfrac{\pi}{2}=0$

$\tan\alpha = \dfrac{\sin\alpha}{\cos\alpha}$

If you use those identities, for $\alpha=x,\ \beta=\dfrac{\pi}{2}$ , you get:

$\sin\left(x+\dfrac{\pi}{2}\right) = \sin x\cdot\cos\dfrac{\pi}{2} + \cos x\cdot\sin\dfrac{\pi}{2} = \sin x\cdot0 + \cos x \cdot 1 = \cos x$

$\cos\left(x+\dfrac{\pi}{2}\right) = \cos x \cdot \cos\dfrac{\pi}{2} - \sin x\cdot\sin\dfrac{\pi}{2} = \cos x \cdot 0 - \sin x \cdot 1 = -\sin x$

Hence:

$\tan \left(x+\dfrac{\pi}{2}\right) = \dfrac{\sin\left(x+\dfrac{\pi}{2}\right)}{\cos\left(x+\dfrac{\pi}{2}\right)} = \dfrac{\cos x}{-\sin x} = -\cot x$

3 0

2 years ago