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guapka [62]
3 years ago
12

Kayden was offered a job that paid a salary of $39, 500 in its first year. The salary

Mathematics
1 answer:
Allushta [10]3 years ago
6 0

Answer:

FV= $385,307.82

Step-by-step explanation:

Giving the following formula:

Annual salary (A)= $39,500

Growth rate (g)= 2% annual

Number of years (n)= 9 years

<u>To calculate the total amount earned over 9 years, we need to use the following formula:</u>

FV= {A*[(1+g)^n-1]}/g

A= annual deposit

FV= {39,500*[(1.02^9) - 1]} / 0.02

FV= $385,307.82

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Mademuasel [1]
The answer #6 is j. 100. He uses the treadmill for 40% so that means 60% of his exercise is not on the treadmill. 

40%= 40 mins 
60%= 60 mins
 total: 100%= 100 mins.
Hope this helps!
8 0
3 years ago
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Explain how you can use a quick picture to find 3 x 2.7
umka2103 [35]
You can make 3 tens blocks and then 3 tens blocks but only two should be filled than for the third one fill 7 only
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3 years ago
4−d&lt;4+d, solve for d<br><br> d&gt;0<br> d&gt;−4<br> d&gt;−8<br> d&gt;8
Harrizon [31]
4-2d<4
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4 0
3 years ago
A degree 4 polynomial P(x) with integer coefficients has zeros 5 i and 3, with 3 being a zero of multiplicity 2. Moreover, the c
Sergio039 [100]

Multiplicity is how many times the root repeats

roots r1 and r2 of a polynomial factor to

(x-r1)(x-r2)

so 3 multiplicty 2 means (x-3)^2 is in the factorization of that polynomial


also, for a polynomial with real coefients, if a+bi is a roots, a-bi is also a root

5i is a root, therefor -5i is a root as well


roots are

(x-5i)(x+5i)(x-3)^2

if we expand

x^4-6x^3+34x^2-150x+225

the polynomial is

f(x)=x^4-6x^3+34x^2-150x+225


6 0
3 years ago
The area of a right triangle is sixty square centimetres. Its base is one centimetre less than twice its height. If the base and
NemiM [27]

Answer:

5. None of the above

Step-by-step explanation:

We have this information:

A = 60 cm2

b = 2*h - 1

Let's find out b and h:

The equation of the are of a triangle is A = (b*h)/2, if we replace it with our information, we have

60 = ((2*h - 1) * h)/2\\ 120 = 2*h^2 - h\\ 0 = 2*h^2 - h - 120

Let's find h with the quadratic formula:

\fbox {Quadratic formula:}  \frac{-b\pm\sqrt{b^2-4ac}}{2a}

h = \frac{-(-1)\pm\sqrt{(-1)^2-4*2*(-120)}}{2*2} = \frac{1\pm\sqrt{961}}{4} = \frac{1\pm\ 31}{4}

h = 8 or h = -7.5

But h represents the height of the triangle, so it has to be a positive number, that's h = 8.

If we replace this in the equation we had for b, we have that b = 2*8 - 1 = 15.

Now we can calculate the hypotenuse with the Pythagorean equation  

\fbox {Pythagorean equation:} <em>The square of the length of the hypotenuse (the side opposite the right angle) of a right triangle is equal to the sum of the squares of the two legs (the two sides that meet at a right angle). </em>

The base and the height are our legs. We will use "H" for the hypotenuse

H^2 = b^2 + h^2 = 15^2 + 8^2 = 289\\ H = \sqrt {289} = 17

H = 17

If we decrease the base and the height by 2 centimeters, we have  

b' = 15 - 2 = 13 and h' = 8 - 2 = 6

With this, let's calculate the new hypotenuse:

H'^2 = b'^2 + h'^2 = 13^2 + 6^2 = 205\\ H' = \sqrt {205} \approx 14.3

H' \approx 14.3

So, the hypotenuse decreases H - H' \approx 17 - 14.3 = 2.7

3 0
3 years ago
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