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ivann1987 [24]
3 years ago
11

A. 325/27 B. 95/9 C. 65/27 D. 35/9

Mathematics
1 answer:
Elis [28]3 years ago
4 0
The answer is C. (65/27)
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Find the x-intercepts of the parabola with vertex (1,1) and y-intercept (0,-3). Write your answer in this form (x of 1 , y of 1)
Vikki [24]
Remember that the vertex form of a parabola or quadratic equation is:

y=a(x-h)^2+k, where (h,k) is the "vertex" which is the maximum or minimum point of the parabola  (and a is half the acceleration of the of the function, but that is maybe too much :P)

In this case we are given that the vertex is (1,1) so we have:

y=a(x-1)^2+1, and then we are told that there is a point (0,-3) so we can say:

-3=a(0-1)^2+1

-3=a+1

-4=a so our complete equation in vertex form is:

y=-4(x-1)^2+1

Now you wish to know where the x-intercepts are.  x-intercepts are when the graph touches the x-axis, ie, when y=0 so

0=-4(x-1)^2+1  add 4(x-1)^2 to both sides

4(x-1)^2=1  divide both sides by 4

(x-1)^2=1/4  take the square root of both sides

x-1=±√(1/4)  which is equal to

x-1=±1/2  add 1 to both sides

x=1±1/2

So x=0.5 and 1.5, thus the x-intercept points are:

(0.5, 0) and (1.5, 0)  or if you like fractions:

(1/2, 0) and (3/2, 0) :P
4 0
3 years ago
Which of the following equations has only one solution? x2 + 4x + 4 = 0 x2 + x = 0 x2 - 1 = 0
Nady [450]
<span>you can solve all quadratic equations using this method:
---
x^2 + 4x + 4 = 0
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the above quadratic equation is in standard form, with a=1, b=4, and c=4
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to solve the quadratic equation, by using the quadratic formula, copy and paste this:
1 4 4
into this solver: https://sooeet.com/math/quadratic-equation-solver.php
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this quadratic has ONE real root (the x-intercept), which is:
x = -2
x = -2
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---
---
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x^2 + x = 0
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the above quadratic equation is in standard form, with a=1, b=1, and c=0</span>
4 0
3 years ago
Read 2 more answers
( 3 x 2 ) ^ 4 x 3 ^ -3 / 2 ^ 3 x 3 =<br><br> Simplify the exponential expression
Korvikt [17]

Answer:

2

Step-by-step explanation:

\frac{(3*2)^4*3^{-3}}{2^3*3}

we start simplifying by removing the parenthesis

Multiply the exponents inside the the parenthesis

3^4  * 2^4

\frac{3^4*2^4*3^{-3}}{2^3*3}

Now we apply exponential property

a^m * a^n = a^ (m+n)

3^4 * 3^-3 = 3^ (4-3) = 3^1

3 or 3^1  are same

\frac{3^1*2^4}{2^3*3^1}

3^1 at the top and bottom so we cancel it out

\frac{2^4}{2^3}

we apply log property . a^m / a^n = a^m-n

Now subtract the exponents

2^(4-3) = 2^1 = 2



4 0
3 years ago
I need help, if you know this plz help me. cuz i'm lost. Thankssss
Musya8 [376]

Answer:

Are there directions?

Step-by-step explanation:

6 0
3 years ago
If f(x) = 9x2 − x3, find f '(1) and use this value to find an equation of the tangent line to the curve y = 9x2 − x3 at the poin
yaroslaw [1]

Answer:

f(x)=9×2_x3=18 hope it help good luck

8 0
3 years ago
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