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Daniel [21]
3 years ago
13

What is the x-coordinate of the vertex of the parabola whose equation is y = 3x2 + 9x?

Mathematics
2 answers:
Westkost [7]3 years ago
4 0
The vertex is (-1.5, -6.75). So the x-coordinate is -1.5. Hope this helps and if so please leave a good rating and thanks. I really appreciate it. Have a nice day!
lisabon 2012 [21]3 years ago
3 0

Answer:

-1.5

Step-by-step explanation:

We have been given an equation y=3x^2+9x. We are asked to find the x-coordinate of the vertex of the parabola for our given equation.

We will use formula \frac{-b}{2a} to finc the x-coordinate of vertex of given parabola.

Upon substituting our given values in above formula we will get,

\frac{-9}{2*3}

\frac{-3*3}{2*3}

\frac{-3}{2}

-1.5

Therefore, the x-coordinate of vertex of given parabola is -1.5.

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4 bags of cat food because food for pets are expensive

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Which expressions are equivalent to − 56 z + 28 −56z+28minus, 56, z, plus, 28? Choose 2 answers: Choose 2 answers: (Choice A) A
Dmitrij [34]

Answer:

(D)-2(28z-14) \: and\:\\(E) -7(8z-4)

Step-by-step explanation:

We are to determine which of these expressions are equivalent to: -56 z + 28

(A)\frac{1}{2}\cdot(-28z+14) \\(B)(-1.4z+0.7)40\\(C)(14-7z)\cdot(-4)\\(D)(8z-4)(-7)\\(E)-2(-28z-14)

-56 z + 28

  • If we factor out -2

-56 z + 28=-2(\frac{ -56 z}{-2} + \frac{28}{-2})=-2(28z-14)

  • If we factor out -7

-56 z + 28=-7(\frac{ -56 z}{-7} + \frac{28}{-7})=-7(8z-4)

Therefore, the two equivalent expressions are:

-2(28z-14) \: and\: -7(8z-4)

5 0
3 years ago
What is the correct order of the following numbers from least to greater?<br> 2/3, -0.6, 0.65, 4/5
ExtremeBDS [4]
We can either convert them to decimals or to fractions
if decimals
2/3=0.66666666
4/5=8/10=0.8
so the order is
-0.6, 0.65, 0.66, 0.8 or
-0.6, 0.65, 2/3, 4/5

if conver to fractions then
since -0.6 is negative, it is smallest, no need to convert
0.65=65 hundreths=65/100
2/3=66/99=prety close
4/5=8/10=80/100
order is
-0.6, 65/100, 66/99, 80/100 or
-0.6, 0.65, 2/3, 0.8
5 0
3 years ago
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Line segment 19 units long running from (x,0) ti (0, y) show the area of the triangle enclosed by the segment is largest when x=
Debora [2.8K]
The area of the triangle is

A = (xy)/2

Also,

sqrt(x^2 + y^2) = 19

We solve this for y.

x^2 + y^2 = 361

y^2 = 361 - x^2

y = sqrt(361 - x^2)

Now we substitute this expression for y in the area equation.

A = (1/2)(x)(sqrt(361 - x^2))

A = (1/2)(x)(361 - x^2)^(1/2)

We take the derivative of A with respect to x.

dA/dx = (1/2)[(x) * d/dx(361 - x^2)^(1/2) + (361 - x^2)^(1/2)]

dA/dx = (1/2)[(x) * (1/2)(361 - x^2)^(-1/2)(-2x) + (361 - x^2)^(1/2)]

dA/dx = (1/2)[(361 - x^2)^(-1/2)(-x^2) + (361 - x^2)^(1/2)]

dA/dx = (1/2)[(-x^2)/(361 - x^2)^(1/2) + (361 - x^2)/(361 - x^2)^(1/2)]

dA/dx = (1/2)[(-x^2 - x^2 + 361)/(361 - x^2)^(1/2)]

dA/dx = (-2x^2 + 361)/[2(361 - x^2)^(1/2)]

Now we set the derivative equal to zero.

(-2x^2 + 361)/[2(361 - x^2)^(1/2)] = 0

-2x^2 + 361 = 0

-2x^2 = -361

2x^2 = 361

x^2 = 361/2

x = 19/sqrt(2)

x^2 + y^2 = 361

(19/sqrt(2))^2 + y^2 = 361

361/2 + y^2 = 361

y^2 = 361/2

y = 19/sqrt(2)

We have maximum area at x = 19/sqrt(2) and y = 19/sqrt(2), or when x = y.
3 0
3 years ago
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