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prisoha [69]
3 years ago
9

I don’t understand how to do this without plugging in numbers or using calculus. Please help. I’m studying for the SAT and I onl

y know how to solve with calculus.

Mathematics
1 answer:
Svetllana [295]3 years ago
8 0

Step-by-step explanation:

Recall that for a quadratic equation y = ax² + bx + c in the X-Y plane, the x-location of the vertex (i.e maximum or minimum point) is given by

x @vertex = -b/2a

in this case your quadratic equation is

h = 4 + 20t - 5t²   (rearranging in the form y = ax² + bx + c )

h = - 5t² + 20t + 4

hence a= -5, b = 20 and c = 4

applying the formula for vertex

t @ vertex = -b  /2a = -(20) / (2)(-5) = -20/-10 = 2

therefore t = 2

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What is the square root of PI (this is not a test question i just wanna know what it is)
adell [148]

Answer:

its about 1.77245385

Step-by-step explanation:

you need a calculator to calculate irrational square roots

8 0
3 years ago
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4. If y - 2x = 3 and y - x = 1, what are the values of x and y?
ioda

y - x = 1.

or, y= 1+x

y - 2x = 3.

=>1+x -2x= 3 (substituting the value of y)

=>-x= 2

=>x= -2

y = -2+1 = -1

Therefore, x = -2 and y = -1

3 0
3 years ago
3 - 9a =27 show work pls​
Sladkaya [172]

Hey there! :)

Answer:

a = -8/3

Step-by-step explanation:

Given:

3 - 9a = 27

Subtract 3 from both sides:

3 - 3 -9a = 27 -3

-9a = 24

Divide both sides by -9:

-9a/(-9) = 24/(-9)

a = -24/9

Simplify the fraction:

a = -8/3

8 0
3 years ago
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What are the roots of the equation x^4- 4x^3=6x^2- 12x. Using a graphing calculator and system of roots.
Vesna [10]

I don't have a graphing calculator and I don't know what the system of roots is.


But I can solve this anyway.


x^4 - 4x^3 - 6x^2 + 12 x = 0


x(x^3 - 4x^2 - 6x + 12) = 0


So x = 0 is a root and we have a cubic equation to solve.


In school the best way to solve cubic equations is to just try some small numbers; teachers seldom venture beyond 1, -1, 2 or -2 as roots. I guess a graphing calculator saves us the search.


Here we see x=-2 is a root; (-2)^3 - 4(-2)^2 - 6(-2) + 12 = -8 - 16 + 12 + 12 = 0.


So, continuing our factoring, x+2 is a factor, and we divide to get the other factor


        x^2 - 6x + 6

x+2 | x^3 - 4x^2 - 6x + 12

         x^3 + 2x^2

                 -6x^2 - 6x

                 - 6x^2 - 12 x

                               6x + 12

So fully factored we have


0 = x(x+2)( x^2 - 6x +6 )


There's a little shortcut I call the Shakespeare Quadratic Formula (2b or -2b) which works when the linear term is even:


x^2 - 2bx + c \textrm{ has zeros } x =b \pm \sqrt{b^2-c}


Here that means we have two additional roots,


x = 3 \pm \sqrt{3^2-6} = 3 \pm \sqrt{3}


Full set of roots:


x = 0 \textrm{ or } -2  \textrm{ or } 3 + \sqrt{3}  \textrm{ or } 3 - \sqrt{3}


7 0
4 years ago
Construct a rhombus whose diagonals are 6.2cm 8.4cm long​
Vladimir [108]

Answer:

you have to construct a rombus

6 0
3 years ago
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