Please answer this correctly

Avery is currently making $14 per hour at her job. Her boss gives her a 15% raise for showing good work ethic.

sattari [20]

Answer:

C) $84

Step-by-step explanation:

40 x 14 = 560

she made $560 prior to raise

14 x 1.15 = 16.10

her new hourly rate is $16.10

40 x 16.1 = 644

she now earns $644 per week

644 - 560 = 84

4 0

3 years ago

Help I’m so confused

velikii [3]

Answer:

5x: linear; monomial

-7: constant; monomial

x^2 + 4: quadratic, binomial

x^2 - 3x + 11: quadratic, trinomial

2x - 9: linear, binomial

Step-by-step explanation:

Name using degree: just count the highest number of variables in a single term. if it's 0, its a constant polynomial, if it's 1 it's linear, 2 quadratic, 3 cubic

Name using number of terms: just count the number of "things" added or subtracted. If there's one, it's a monomial, two: binomial, three: trinomial.

6 0

2 years ago

Derivative of tan(2x+3) using first principle

kodGreya [7K]

$f(x)=\tan(2x+3)$

The derivative is given by the limit

$f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h$

You have

$\displaystyle\lim_{h\to0}\frac{\tan(2(x+h)+3)-\tan(2x+3)}h$
$\displaystyle\lim_{h\to0}\frac{\tan((2x+3)+2h)-\tan(2x+3)}h$

Use the angle sum identity for tangent. I don't remember it off the top of my head, but I do remember the ones for (co)sine.

$\tan(a+b)=\dfrac{\sin(a+b)}{\cos(a+b)}=\dfrac{\sin a\cos b+\cos a\sin b}{\cos a\cos b-\sin a\sin b}=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$

By this identity, you have

$\tan((2x+3)+2h)=\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}$

So in the limit you get

$\displaystyle\lim_{h\to0}\frac{\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}-\tan(2x+3)}h$
$\displaystyle\lim_{h\to0}\frac{\tan(2x+3)+\tan2h-\tan(2x+3)(1-\tan(2x+3)\tan2h)}{h(1-\tan(2x+3)\tan2h)}$
$\displaystyle\lim_{h\to0}\frac{\tan2h+\tan^2(2x+3)\tan2h}{h(1-\tan(2x+3)\tan2h)}$
$\displaystyle\lim_{h\to0}\frac{\tan2h}h\times\lim_{h\to0}\frac{1+\tan^2(2x+3)}{1-\tan(2x+3)\tan2h}$
$\displaystyle\frac12\lim_{h\to0}\frac1{\cos2h}\times\lim_{h\to0}\frac{\sin2h}{2h}\times\lim_{h\to0}\frac{\sec^2(2x+3)}{1-\tan(2x+3)\tan2h}$

The first two limits are both 1, and the single term in the last limit approaches 0 as $h\to0$ , so you're left with

$f'(x)=\dfrac12\sec^2(2x+3)$

which agrees with the result you get from applying the chain rule.

7 0

3 years ago

14. The sun casts a shadow from a flag pole. The height of the flag pole is three times the length of its shadow. The distance b

m_a_m_a [10]

7
It would be about 6.7 ft

8 0

3 years ago

You want to rent a television. Company A charges $25 per week plus a one-time fee of $52. Company B charges $37 per week plus a

RUDIKE [14]

52+25=77+25=102+25=127
16+36=53+37=90+37=127
Answer 3 weeks

8 0

3 years ago