Answer:
(3x+4)(5x+7)
Step-by-step explanation:
15x^2
+41x+28
Factor the expression by grouping. First, the expression needs to be rewritten as 15x^2
+ax+bx+28. To find a and b, set up a system to be solved.
a+b=41
ab=15×28=420
Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 420.
1,420
2,210
3,140
4,105
5,84
6,70
7,60
10,42
12,35
14,30
15,28
20,21
Calculate the sum for each pair.
1+420=421
2+210=212
3+140=143
4+105=109
5+84=89
6+70=76
7+60=67
10+42=52
12+35=47
14+30=44
15+28=43
20+21=41
The solution is the pair that gives sum 41.
a=20
b=21
Rewrite 15x^2
+41x+28 as (15x^2
+20x)+(21x+28).
(15x^2
+20x)+(21x+28)
Factor out 5x in the first and 7 in the second group.
5x(3x+4)+7(3x+4)
Factor out common term 3x+4 by using distributive property.
(3x+4)(5x+7)
Answer:
1= 2,400 feet
2= 154 meters
3= 120 pints
4= 2. 056 centigrams
5= 16.1 C
6= 50F
7= -2.8C
1= multiply the length value by 3
2= divide the length value by 10
3= multiply the volume value by 8
4= divide the mass value by 10
5= Take the °F temperature and subtract 32. Multiply this number by 5. Divide this number by 9 to obtain your answer in °C.
6= multiply the temperature in degrees Celsius by 2, and then add 30 to get the (estimated) temperature in degrees Fahrenheit.
is a complex number that satisfies
![\begin{cases}r\cos x=-3\\[1ex]r\sin x=4\\[1ex]r=\sqrt{(-3)^2+4^2}\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7Dr%5Ccos%20x%3D-3%5C%5C%5B1ex%5Dr%5Csin%20x%3D4%5C%5C%5B1ex%5Dr%3D%5Csqrt%7B%28-3%29%5E2%2B4%5E2%7D%5Cend%7Bcases%7D)
The last equation immediately tells you that
.
So you have
![\begin{cases}\cos x=-\dfrac35\\[1ex]\sin x=\dfrac45\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%5Ccos%20x%3D-%5Cdfrac35%5C%5C%5B1ex%5D%5Csin%20x%3D%5Cdfrac45%5Cend%7Bcases%7D)
Dividing the second equation by the first, you end up with
Because the argument's cosine is negative and its sine is positive, you know that
. This is important to know because it's only the case that
whenever
. The inverse doesn't exist otherwise.
However, you can restrict the domain of the tangent function so that an inverse can be defined. By shifting the argument of tangent by
, we have
All this to say
So,
.
