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3 years ago

How do I solve this

Mathematics

1 answer:

xeze [42]3 years ago

8 0

Step-by-step explanation:

subtracting equation i from it got cropped

so the answer for x is 7 and y is -17

hope this helps. :)

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Suppose you want to accumulate $25,000 as down payment on a house and the best you can do is to put aside $200 a month. If you d

Sunny_sXe [5.5K]

Answer:

It will take 88.2 months to accumulate the amount

Step-by-step explanation:

Given;

Future value of money, FV = $25,000

investment per compound period, P = $200

interest rate, i = 0.75% x 12 = 9%

The number of monthly installments required to amount to FV is given by;

$n = \frac{ln(FVi +cp)-ln(cp)}{ln(c+i)-ln(c)}\\\\ n = \frac{ln(25000*0.09 +200*12)-ln(200*12)}{ln(12+0.09)-ln(12)}\\\\n = \frac{8.4446 -7.7832}{2.4924-2.4849}\\\\n = 88.2 \ months$

Therefore, it will take 88.2 months to accumulate the amount.

7 0

4 years ago

A survey was taken of students in math classes to find out how many hours per day students spend

Hoochie [10]

Answer:

1. Mean, because there are no outliers that affect the center

Step-by-step explanation:

Second period: 3,2,3,1,3, 4, 2, 4, 3, 1, 0, 2, 3, 1, 2

Sorted values : 0, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4

The mean = ΣX / n

n = sample size, n = 15

Mean = 34 / 15 = 2.666

The median = 1/2(n+1)th term.

1/2(16)th term = 8th term.

The 8th term = 2

The best measure of centre is the mean because the values for the second period has no outliers that might have affected the centre of the distribution.

Both interquartile range and standard deviation are measures of spread and not measures of centre.

6 0

3 years ago

What are the zeros of the function y=2x^2-3x-20

4vir4ik [10]

-2.5 and 4

You can get this by factoring to the equation (2x + 5)(x - 4)

4 0

3 years ago

Set Q contains 20 positive integer values. The smallest value in Set Q is a single digit value and the largest value in Set Q is

Lerok [7]

Answer: possible values of Range will be values that are >=91 or <=998

Step-by-step explanation:

Given that :

Set Q contains 20 positive integer values. The smallest value in Set Q is a single digit value and the largest value in Set Q is a three digit value.

Therefore,

given that the smallest value in set Q is a one digit number :

Then lower unit = 1, upper unit = 9( this represents the lowest and highest one digit number)

Also, the largest value in Set Q is a three digit value:

Then lower unit = 100, upper unit = 999 ( this represents the lowest and highest 3 digit numbers).

Therefore, the possible values of the range in SET Q:

The maximum possible range of the values in set Q = (Highest possible three digit value - lowest possible one digit) = (999 - 1) = 998

The least possible range of values in set Q = (lowest possible three digit value - highest possible one digit value) = (100 - 9) = 91

5 0

3 years ago

How do you solve his with working

AlexFokin [52]

Check the picture below.

a)

so the perimeter will include "part" of the circumference of the green circle, and it will include "part" of the red encircled section, plus the endpoints where the pathway ends.

the endpoints, are just 2 meters long, as you can see 2+15+2 is 19, or the radius of the "outer radius".

let's find the circumference of the green circle, and then subtract the arc of that sector that's not part of the perimeter.

and then let's get the circumference of the red encircled section, and also subtract the arc of that sector, and then we add the endpoints and that's the perimeter.

$\bf \begin{array}{cllll}
\textit{circumference of a circle}\\\\ 
2\pi r
\end{array}\qquad \qquad \qquad \qquad 
\begin{array}{cllll}
\textit{arc's length}\\\\
s=\cfrac{\theta r\pi }{180}
\end{array}\\\\
-------------------------------$

$\bf \stackrel{\stackrel{green~circle}{perimeter}}{2\pi(7.5) }~-~\stackrel{\stackrel{green~circle}{arc}}{\cfrac{(135)(7.5)\pi }{180}}~+
\stackrel{\stackrel{red~section}{perimeter}}{2\pi(9.5) }~-~\stackrel{\stackrel{red~section}{arc}}{\cfrac{(135)(9.5)\pi }{180}}+\stackrel{endpoints}{2+2}
\\\\\\
15\pi -\cfrac{45\pi }{8}+19\pi -\cfrac{57\pi }{8}+4\implies \cfrac{85\pi }{4}+4\quad \approx \quad 70.7588438888$

b)

we do about the same here as well, we get the full area of the red encircled area, and then subtract the sector with 135°, and then subtract the sector of the green circle that is 360° - 135°, or 225°, the part that wasn't included in the previous subtraction.

$\bf \begin{array}{cllll}
\textit{area of a circle}\\\\ 
\pi r^2
\end{array}\qquad \qquad \qquad \qquad 
\begin{array}{cllll}
\textit{area of a sector of a circle}\\\\
s=\cfrac{\theta r^2\pi }{360}
\end{array}\\\\
-------------------------------$

$\bf \stackrel{\stackrel{red~section}{area}}{\pi(9.5^2) }~-~\stackrel{\stackrel{red~section}{sector}}{\cfrac{(135)(9.5^2)\pi }{360}}-\stackrel{\stackrel{green~circle}{sector}}{\cfrac{(225)(7.5^2)\pi }{360}}
\\\\\\
90.25\pi -\cfrac{1083\pi }{32}-\cfrac{1125\pi }{32}\implies \cfrac{85\pi }{4}\quad \approx\quad 66.75884$

7 0

3 years ago