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Lisa [10]
3 years ago
14

When the following quadratic equation is written in standard form, what is the value of "c"?

Mathematics
1 answer:
Nikolay [14]3 years ago
7 0

Answer:

Then value is 2

Step-by-step explanation:

standard of quadratic is Ax^{2}+Bx+C=0

and 2 is where C is.

sorry for the bad  explanation but, trust me I did this type of question in class yesterday

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ehidna [41]

Answer:

We conclude that we must add 4² or 16 to complete the square.

Hence, option C i.e. 16 is the correct answer.

Step-by-step explanation:

We know that the perfect square formula is

<em>(a + b)² = a² + b² + 2ab</em>

Given the equation

x^2+8x

Let us add 4² or 16 in the equation

x^2+8x+4^2

x^2+\:2\left(1\right)\left(4\right)x+4^2

as <em>(a + b)² = a² + b² + 2ab, </em>so

<em />x^2+8x+4^2=\left(x+4\right)^2<em />

<em />

Therefore, we conclude that we must add 4² or 16 to complete the square.

Hence, option C i.e. 16 is the correct answer.

4 0
3 years ago
Read 2 more answers
A single number between 0 and 9 occurring either alone or in a larger number is called a what
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Answer:

A chronilagical corientation

Step-by-step explanation:

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6 0
3 years ago
The sum of 5 times a larger number and twice a smaller is 5. The difference of 4 times the larger and the smaller number is 4. F
Gemiola [76]

Answer:

1 and 0

Step-by-step explanation:

Let's call the smaller number x and the larger number y.

This means 5y+2x=5.

Also, 4y-x=4.

This is a system of equations. There are two main ways two solve this.

Method 1: Substitution

5y+2x=5

4y-x=4

Solve for one of the variables, then substitute it back into one of the equations to solve for the other variable.

I will take 4y-x=4 and solve for x.

4y=4+x

x=4y-4

Then substitute for x in the other equation and solve for y.

5y+2x=5

5y+2(4y-4)=5

5y+8y-8=5

13y=13

y=1

Now we can go back to our equation for x and substitute for y.

x=4y-4

x=4(1)-4=0

y, the larger number, is 1 and

x, the smaller number, is 0

Method 2: Elimination

5y+2x=5

4y-x=4

Change the equations so that the coefficient of one of the variables in an equation is opposite the one in the other equation. Then add the equations and solve for each variable.

If I multiply both sides of the second equation, 4y-x=4, by 2, the coefficient of x will become -2. In the first equation, the coefficient is 2, so they will become opposites.

4y-x=4

becomes

8y-2x=8

Now we have 5y+2x=5 and 8y-2x=8.

Add the sides of both equations to cancel the x values.

5y+2x+8y-2x=5+8

13y=13

y=1

like last time, we can substitute for y in any of the original equations to get x.

5y+2x=5

5(1)+2x=5

5+2x=5

2x=0

x=0

y, the larger number, is 1 and

x, the smaller number, is 0

6 0
3 years ago
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