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Ivahew [28]
3 years ago
14

Help Please Someone Asap

Mathematics
1 answer:
galben [10]3 years ago
5 0
33 over 1

and in 5 hours he would go to 1 mile and 2/3
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46 divided by 5 long divison
DIA [1.3K]
9.2 (b) is the answer hope this helped!
5 0
3 years ago
Suppose the equation ax^2+bx+c=0 has no real solution and a graph of the related function has a vertex that lies in the second q
Veseljchak [2.6K]
If it has no real solutions, that means the graph does not intersect the x axis

since we have ax^2+bx+c=0, the parabola opens either up or down
since the vertex is in the second quadrant (x is negative and y is positive in this reigon) and the graph does not cross the x axis, the parabola must open up
if the value of 'a' is positive, then the parabola opens up

so 'a' must be positive



if it is translated to the 4th quadrant, then the vertex is now below the x axis
it will now have 2 x intercepts because the vertex is in the 4th quadrant and look at a graph of a parabola opening up with vertex in 4th quadrant and seehow many time it crosses the x axis
7 0
3 years ago
Y=(2x-3)^5(2-x^4)^3 differentiate the function please
BARSIC [14]
\bf y=(2x-3)^5(2-x^4)^3\\\\
-------------------------------\\\\
\cfrac{dy}{dx}=\stackrel{\textit{product rule}}{[5(2x-3)^4\cdot 2(2-x^4)3]~~+~~[(2x-3)^5[3(2-x^4)^2(-4x^3)]]}
\\\\\\
\cfrac{dy}{dx}=10(2x-3)^4(2-x^4)^3~~-~~12x^3(2x-3)^5(2-x^4)^2
\\\\\\
\cfrac{dy}{dx}=\stackrel{\textit{common factor}}{2(2x-3)^4(2-x^4)^2}~[5(2-x^4)-6x^3(2x-3)]
\\\\\\
\cfrac{dy}{dx}=2(2x-3)^4(2-x^4)^2~[10-5x^4-12x^4+18x^3]
\\\\\\
\cfrac{dy}{dx}=2(2x-3)^4(2-x^4)^2(10-17x^4+18x^3)
5 0
3 years ago
Make non proportional equation for the tables
TEA [102]
Hahahahahhahahahahahhahhahahahahhahahahhahahahahahahhahahahahhahahahahahahahhahahahhahahahahhahahhahahahhahhahahhabbabaabbahahhahahahabahahhaabbahahahhahhababhahahhahabhahahhahahha thanks for the points !!!!!!
4 0
2 years ago
If a + bi is a zero of a function and the coefficients of the polynomial are real, what will be the other complex zero of the fu
lukranit [14]
A - bi would be the other complex zero.
When you take the sqrt when solving an equation you get a + and - solution.
8 0
3 years ago
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