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Alex777 [14]
3 years ago
9

A real estate firm charges 2% commission fee on sales. If Maha sold her house and was charged $13,400 for commission, what was t

he sale price of her house?
Mathematics
1 answer:
Vinil7 [7]3 years ago
5 0

Answer:

$670,000

Step-by-step explanation:

13,400 multiplied by 50 is 670,000

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In △ABC , m∠A=53°,m∠B=17°, and a=27. Find the perimeter of the triangle.
sdas [7]
Using the Law of Sines  (sinA/a=sinB/b=sinC/c) and the fact that all triangles have a sum of 180° for their angles.

The third angle is C is 180-53-17=110°

27/sin53=b/sin17=c/sin110

b=27sin17/sin53, c=27sin110/sin53

And the perimeter is a+b+c so

p=27+27sin17/sin53+27sin110/sin53 units

p≈68.65 units  (to nearest hundredth of a unit)
6 0
3 years ago
What is the solution to the linear equation 2/3 x - 1/2 = 1/3 + 5/6x
svp [43]

Answer:

Simplify and solve and get x=-5

Step-by-step explanation:

2/3x - 1/2= 1/3 + 5/6x

2/3x - 5/6x= 1/3 + 1/2

-1/6x= 5/6

x=-5

7 0
3 years ago
Simplify this expression.<br> 3^-3
Sphinxa [80]

Answer:

1/27 is the answer

Step-by-step explanation:

It is 1/(3^3) which is 1/27

4 0
3 years ago
Which parent function is represented by the graph?
Sergio [31]

Answer:

Linear

Step-by-step explanation:

pretty self explanatory

7 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}}}

8 0
2 years ago
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