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SOVA2 [1]
3 years ago
10

Use the table to write a quadratic function in vertex form. Then rewrite the function in standard form.

Mathematics
1 answer:
Kisachek [45]3 years ago
8 0

Answer:

Step-by-step explanation:

The graph decreases at first, then changes direction at (2, 5).

y = a(x-2)^2 + 5

Plug in (1,11) and solve for a:

11 = a(1-2)^2 + 5

a = 6

Equation in vertex form:

y = 6(x-2)^2 + 5

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What is the solution to the equation
Dafna11 [192]
\sqrt{4t+5}=3-\sqrt{t+5}\\\\D:4t+5\geq0\ \wedge\ t+5\geq0\ \wedge\ \sqrt{t+5}\leq3\\\\t\geq-\dfrac{5}{4}\ \wedge\ t\geq-5\ \wedge\ t\leq6

therefore
D:x\in\left< -\dfrac{5}{4};\ 6\right>

\sqrt{4t+5}=3-\sqrt{t+5}\ \ \ |^2\\\\(\sqrt{4t+5})^2=(3-\sqrt{t+5})^2\ \ \ |use:(a-b)^2=a^2-2ab+b^2\\\\4t+5=3^2-2\cdot3\cdot\sqrt{t+5}+(\sqrt{t+5})^2\\\\4t+5=9-6\sqrt{t+5}+t+5\ \ \ \ |-t\\\\3t+5=14-6\sqrt{t+5}\ \ \ \ |-14\\\\3t-9=-6\sqrt{t+5}\ \ \ \ |change\ signs
9-3t=6\sqrt{t+5}\ \ \ \ |:3\\\\3-t=2\sqrt{t+5}\ \ \ \ |^2\\\\(3-t)^2=(2\sqrt{t+5})^2\\\\3^2-2\cdot3\cdot t+t^2=4(t+5)\\\\9-6t+t^2=4t+20\ \ \ |-4t-20\\\\t^2-10t-11=0\\\\t^2+t-11t-11=0\\\\t(t+1)-11(t+1)=0\\\\(t+1)(t-11)=0\iff t+1=0\ \vee\ t-11=0\\\\t=-1\in D\ \vee\ t=11\notin D
Answer: t = -1.


7 0
3 years ago
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Nat2105 [25]
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4 0
2 years ago
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Which of these is the algebraic expression for the verbal expression "ten times the difference of a number and twelve?"
fredd [130]
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5 0
3 years ago
What is the solution to |x + 2| &lt; 1?
Katyanochek1 [597]

Answer:

-3 <x <-1

Step-by-step explanation:

|x + 2| < 1

There are two solutions, one positive and one negative ( remember to flip the inequality for the negative)

x+2 <1    and  x+2 > -1

Subtract 2 from each side

x+2-2 < 1-2  and  x+2-2 > -1-2

x < -1      and       x >-3

-3 <x <-1

8 0
3 years ago
The quadratic function f(x)=-5x^2+5 does the parabola open up or down
yulyashka [42]
It opens down.
Since the beginning number -5x^2 is negative, it will open down.
5 0
3 years ago
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