Let the number of months = x , and total budget = y
<span>An inequality to represent the situation:
For budget = $155
∴ 155 = 20 x +15
solve for x
20 x = 155 - 15 = 140
∴ x = 140/20 = 7
∴ T</span>he greatest number of months = 7 <span>months.</span>
Let's assign three blanks for each digit of the unknown number. But let's fill in the tens digit because it is already specified.
_ 4 _
The last digit should be even to make it even. The possible digits for this are 2, 4, 6, and 8. The first digit could be any digit from 1 to 9. Therefore, the possible answers are
142 242 342 442 542 642 742 842 942
144 244 344 444 544 644 744 844 944
146 246 346 446 546 646 746 846 946
148 248 348 448 548 648 748 848 948
Therefore, there are a total of 36 possible answers.
Answer:
43cm
Step-by-step explanation:
use the pythagorean theorem:
a² + b² = c²
25² + 35² = c²
625 + 1225 = c²
1850 = c²
√1850 = c
43 cm = c
your a, b and c values depend on where the values are on the triangle.
To solve the question we use the compound interest formula which is given by:
A=p(1+r)^(nt)
where:
A=future value
p=principle
r=rate
n=number of terms
t=time
thus plugging in the values in the formula we shall have:
A=835(1+0.04)^(4t)
simplifying this we get the sequence:
A=835(1.040)^(4t)
Thus the answer to the sequence will be:
A=835(1.040)^(4t)
![\bf \begin{cases} x=1\implies &x-1=0\\ x=1\implies &x-1=0\\ x=-\frac{1}{2}\implies 2x=-1\implies &2x+1=0\\ x=2+i\implies &x-2-i=0\\ x=2-i\implies &x-2+i=0 \end{cases} \\\\\\ (x-1)(x-1)(2x+1)(x-2-i)(x-2+i)=\stackrel{original~polynomial}{0} \\\\\\ (x-1)^2(2x+1)~\stackrel{\textit{difference of squares}}{[(x-2)-(i)][(x-2)+(i)]}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%0Ax%3D1%5Cimplies%20%26x-1%3D0%5C%5C%0Ax%3D1%5Cimplies%20%26x-1%3D0%5C%5C%0Ax%3D-%5Cfrac%7B1%7D%7B2%7D%5Cimplies%202x%3D-1%5Cimplies%20%262x%2B1%3D0%5C%5C%0Ax%3D2%2Bi%5Cimplies%20%26x-2-i%3D0%5C%5C%0Ax%3D2-i%5Cimplies%20%26x-2%2Bi%3D0%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0A%28x-1%29%28x-1%29%282x%2B1%29%28x-2-i%29%28x-2%2Bi%29%3D%5Cstackrel%7Boriginal~polynomial%7D%7B0%7D%0A%5C%5C%5C%5C%5C%5C%0A%28x-1%29%5E2%282x%2B1%29~%5Cstackrel%7B%5Ctextit%7Bdifference%20of%20squares%7D%7D%7B%5B%28x-2%29-%28i%29%5D%5B%28x-2%29%2B%28i%29%5D%7D)
![\bf (x^2-2x+1)(2x+1)~[(x-2)^2-(i)^2] \\\\\\ (x^2-2x+1)(2x+1)~[(x^2-4x+4)-(-1)] \\\\\\ (x^2-2x+1)(2x+1)~[(x^2-4x+4)+1] \\\\\\ (x^2-2x+1)(2x+1)~[x^2-4x+5] \\\\\\ (x^2-2x+1)(2x+1)(x^2-4x+5)](https://tex.z-dn.net/?f=%5Cbf%20%28x%5E2-2x%2B1%29%282x%2B1%29~%5B%28x-2%29%5E2-%28i%29%5E2%5D%0A%5C%5C%5C%5C%5C%5C%0A%28x%5E2-2x%2B1%29%282x%2B1%29~%5B%28x%5E2-4x%2B4%29-%28-1%29%5D%0A%5C%5C%5C%5C%5C%5C%0A%28x%5E2-2x%2B1%29%282x%2B1%29~%5B%28x%5E2-4x%2B4%29%2B1%5D%0A%5C%5C%5C%5C%5C%5C%0A%28x%5E2-2x%2B1%29%282x%2B1%29~%5Bx%5E2-4x%2B5%5D%0A%5C%5C%5C%5C%5C%5C%0A%28x%5E2-2x%2B1%29%282x%2B1%29%28x%5E2-4x%2B5%29)
of course, you can always use (x-1)(x-1)(2x+1)(x²-4x+5) as well.
