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Morgarella [4.7K]
3 years ago
14

Nick is solving the equation 3x^2=20-7x with the quadratic formula

Mathematics
1 answer:
krek1111 [17]3 years ago
6 0

Answer:

C. a = 3

b = 7

c = -20

Step-by-step explanation:

x = -b+/- (sq) b^2 - 4ac.

Well the way to plug the numbers in is by using this equation: Ax^2 + Bx + C = 0

S if you use your equation : 3x^2 = 20 - 7x

First you have to move everything to one side.

So it'll look like this:

3x^2 = 20 - 7x

-20      -20

3x^2 - 20 = -7x

+7x  +7x

3x^2 - 20 + 7x = 0

don't forget that... Ax^2 +Bx+ C= 0

3 = A

7 = B

-20 = C

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0.7km in miles<br> Please answer
iVinArrow [24]

Answer:

0.43496 miles

Step-by-step explanation:

To convert from km to miles you can divide the km by 1.609 and that should give you an aproximate value for miles.

7 0
3 years ago
Which function has a vertex on the y-axis? f(x) = (x – 2)2 f(x) = x(x + 2) f(x) = (x – 2)(x + 2) f(x) = (x + 1)(x – 2)
strojnjashka [21]

we know that

If the vertex is on the y-axis, then the x-coordinate of the vertex is equal to zero

we are going to verify the vertex of each one of the functions to determine the solution

Remember that

The equation in vertex form of a vertical parabola is equal to

y=a(x-h)^{2} +k

where

(h,k) is the vertex

if a>0 -------> the parabola open upward (vertex is a minimum)

if a -------> the parabola open downward (vertex is a maximun)

<u>case A)</u> f(x)=(x-2)^{2}

This is a vertical parabola open upward

the vertex is the point (2,0)

therefore

The function f(x)=(x-2)^{2}  does not have a vertex on the y-axis

<u>case B)</u> f(x)=x(x+2)

f(x)=x(x+2)=x^{2}+2x

convert to vertex form

Complete the square. Remember to balance the equation by adding the same constants to each side.

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

Rewrite as perfect squares

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

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

the vertex is the point (-1,-1)

therefore

The function f(x)=x(x+2) does not have a vertex on the y-axis

<u>case C)</u> f(x)=(x-2)(x+2)

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

f(x)=x^{2}-4

the vertex is the point (0,-4)

The x-coordinate of the vertex is equal to zero

therefore

The function f(x)=(x-2)(x+2) has a vertex on the y-axis

<u>case D)</u> f(x)=(x+1)(x-2)

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

convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

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

Complete the square. Remember to balance the equation by adding the same constants to each side.

f(x)+2+0.25= x^{2} -x+0.25

f(x)+2.25= x^{2} -x+0.25

Rewrite as perfect squares

f(x)+2.25= (x-0.50)^{2}

f(x)=(x-0.50)^{2}-2.25

the vertex is the point (0.5,-2.25)

therefore

The function f(x)=(x+1)(x-2) does not have a vertex on the y-axis

<u>the answer is</u>

f(x)=(x-2)(x+2)

6 0
3 years ago
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Answer:

A.

Step-by-step explanation:

Not reasonable, Why did you put that as your answer?

The whole numbers add up to more then 4 cups.

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soldier1979 [14.2K]
<h2>i cant answer it without text sorry. i just cant see the question :(</h2>
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Please help I’m having trouble with this question!
mezya [45]

ok thanks for letting me know you need help

4 0
3 years ago
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