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3 years ago

5

Find a, b and c for the function x^2 + 2x - 6 = 0

1 answer:

ss7ja [257]3 years ago

3 0

Remember y = ax^2 + bx + c

That means a = 1, b= 2 and c = -6

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Find the value of x in the triangle shown below.

ycow [4]

Answer:

30 Degrees for sure

6 0

3 years ago

Please help math thank you

Alina [70]

a: 30

b:40

c:78

d:36

e:20

f:96

5 0

4 years ago

Read 2 more answers

In Exercises 45–48, let f(x) = (x - 2)2 + 1. Match the<br> function with its graph

MA_775_DIABLO [31]

Answer:

45) The function corresponds to graph A

46) The function corresponds to graph C

47) The function corresponds to graph B

48) The function corresponds to graph D

Step-by-step explanation:

We know that the function f(x) is:

$f(x)=(x-2)^{2}+1$

45)

The function g(x) is given by:

$g(x)=f(x-1)$

using f(x) we can find f(x-1)

$g(x)=((x-1)-2)^{2}+1=(x-3)^{2}+1$

If we take the derivative and equal to zero we will find the minimum value of the parabolla (x,y) and then find the correct graph.

$g(x)'=2(x-3)$

$2(x-3)=0$

$x=3$

Puting it on g(x) we will get y value.

$y=g(3)=(3-3)^{2}+1$

$y=g(3)=1$

<u>Then, the minimum point of this function is (3,1) and it corresponds to (A)</u>

46)

Let's use the same method here.

$g(x)=f(x+2)$

$g(x)=((x+2)-2)^{2}+1$

$g(x)=(x)^{2}+1$

Let's find the first derivative and equal to zero to find x and y minimum value.

$g'(x)=2x$

$0=2x$

$x=0$

Evaluatinf g(x) at this value of x we have:

$g(0)=(x)^{2}+1$

$g(0)=1$

<u>Then, the minimum point of this function is (0,1) and it corresponds to (C)</u>

47)

Let's use the same method here.

$g(x)=f(x)+2$

$g(x)=(x-2)^{2}+1+2$

$g(x)=(x-2)^{2}+3$

Let's find the first derivative and equal to zero to find x and y minimum value.

$g'(x)=2(x-2)$

$0=2(x-2)$

$x=2$

Evaluatinf g(x) at this value of x we have:

$g(2)=(2-2)^{2}+3$

$g(2)=3$

<u>Then, the minimum point of this function is (2,3) and it corresponds to (B)</u>

48)

Let's use the same method here.

$g(x)=f(x)-3$

$g(x)=(x-2)^{2}+1-3$

$g(x)=(x-2)^{2}-2$

Let's find the first derivative and equal to zero to find x and y minimum value.

$g'(x)=2(x-2)$

$0=2(x-2)$

$x=2$

Evaluatinf g(x) at this value of x we have:

$g(2)=(2-2)^{2}-2$

$g(2)=-2$

<u>Then, the minimum point of this function is (2,-2) and it corresponds to (D)</u>

<u />

I hope it helps you!

<u />

8 0

3 years ago

Hi, can someone please help, and possibly explain the answer please.

mr_godi [17]

Answer:

$x=\frac{-(-2)\±\sqrt{(-2)^2-4(3)(0)} }{2(3)}$

Step-by-step explanation:

Quadratic formula: $x=\frac{-b\±\sqrt{b^2-4ac} }{2a}$ when the equation is $0=ax^2+bx+c$

The given equation is $1=-2x+3x^2+1$ . Let's first arrange this so its format looks like $y=ax^2+bx+c$ :

$1=-2x+3x^2+1$

$1=3x^2-2x+1$

Subtract 1 from both sides of the equation

$1-1=3x^2-2x+1-1\\0=3x^2-2x+0$

Now, we can easily identify 3 as a, -2 as b and 0 as c. Plug these into the quadratic formula:

$x=\frac{-b\±\sqrt{b^2-4ac} }{2a}\\x=\frac{-(-2)\±\sqrt{(-2)^2-4(3)(0)} }{2(3)}$

I hope this helps!

8 0

3 years ago

If 10 miles is about 16 kilometers and the distance between two towns is 45 miles ,use a ratio table to find the distance betwee

diamong [38]

10m : 16 km
x 4.5
45m : 72km.

8 0

3 years ago

Read 2 more answers