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Vedmedyk [2.9K]

2 years ago

9

You and three friends go out to eat and decide to split the bill evenly. The food you order comes to a total of $48. If you leav

e a 15% tip and pay 6.25% sales tax, how much should you each pay?

1 answer:

zloy xaker [14]2 years ago

4 0

Answer:

11.64

Step-by-step explanation:

100%-> 48

15%-> 7.20

6.25%-> 3

48+7.20+3=58.20

58.20÷5=11.64

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Given: 3x + y = 1<br> Solve for y.<br> A. y=-3x + 1<br> B. y=-3x - 1<br> C. y= 3x - 1

g100num [7]

The answer is c. y=3x -1

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2 years ago

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erastovalidia [21]

Answer:

1. KLP + PLM = 180 degrees (straight line)

2. 3x + angle PLM = 180 degrees

3. angle PLM = 180 - 3x

4. PMN = P + PLM (Exterior angle)

5. 2x + 72 = x + 180 - 3x

6. x = 27

Step-by-step explanation:

1. Notice that angle KLP + angle PLM is a straight line, so KLP + PLM = 180 degrees (straight line)

2. angle KLP = 3x, so

3x + angle PLM = 180 degrees

3. angle PLM = 180 - 3x

4. PMN = P + PLM (Exterior angle)

5. 2x + 72 = x + 180 - 3x

6. 5 gives 4x = 108, so x = 27

6 0

2 years ago

Read 2 more answers

1. (5pts) Find the derivatives of the function using the definition of derivative.

andreyandreev [35.5K]

2.8.1

$f(x) = \dfrac4{\sqrt{3-x}}$

By definition of the derivative,

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

We have

$f(x+h) = \dfrac4{\sqrt{3-(x+h)}}$

and

$f(x+h)-f(x) = \dfrac4{\sqrt{3-(x+h)}} - \dfrac4{\sqrt{3-x}}$

Combine these fractions into one with a common denominator:

$f(x+h)-f(x) = \dfrac{4\sqrt{3-x} - 4\sqrt{3-(x+h)}}{\sqrt{3-x}\sqrt{3-(x+h)}}$

Rationalize the numerator by multiplying uniformly by the conjugate of the numerator, and simplify the result:

$f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x} - 4\sqrt{3-(x+h)}\right)\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x}\right)^2 - \left(4\sqrt{3-(x+h)}\right)^2}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16(3-x) - 16(3-(x+h))}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16h}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}$

Now divide this by <em>h</em> and take the limit as <em>h</em> approaches 0 :

$\dfrac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ \displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-x}\left(4\sqrt{3-x} + 4\sqrt{3-x}\right)} \\\\ \implies f'(x) = \dfrac{16}{4\left(\sqrt{3-x}\right)^3} = \boxed{\dfrac4{(3-x)^{3/2}}}$

3.1.1.

$f(x) = 4x^5 - \dfrac1{4x^2} + \sqrt[3]{x} - \pi^2 + 10e^3$

Differentiate one term at a time:

• power rule

$\left(4x^5\right)' = 4\left(x^5\right)' = 4\cdot5x^4 = 20x^4$

$\left(\dfrac1{4x^2}\right)' = \dfrac14\left(x^{-2}\right)' = \dfrac14\cdot-2x^{-3} = -\dfrac1{2x^3}$

$\left(\sqrt[3]{x}\right)' = \left(x^{1/3}\right)' = \dfrac13 x^{-2/3} = \dfrac1{3x^{2/3}}$

The last two terms are constant, so their derivatives are both zero.

So you end up with

$f'(x) = \boxed{20x^4 + \dfrac1{2x^3} + \dfrac1{3x^{2/3}}}$

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1 year ago

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Grabbing dem points up ✨YeAh✨

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2 years ago

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If the measure of cae is 95° what is the measure of cba

xxTIMURxx [149]

<span>CAE=95 GAE=90 CAG=95-90=5 ACG=5 CGA=180-(5+5)=170 CBA=12—170=85</span>

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2 years ago

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