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Harrizon [31]
3 years ago
7

What number must be added to the expression below to complete the square? x2+x

Mathematics
1 answer:
iren2701 [21]3 years ago
4 0

Answer:

answer is 1/4

Step-by-step explanation:

proof is (x-1/2)²= x²-2*1/2*x+1/4= x²-x+1/4

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Which expression is equivalent to 6x+7-12*2-(3 to the power 2 +3)-x
kow [346]

Step-by-step explanation:

Questions about equivalent expressions usually feature both simple expressions and complex expressions. To check which complex expression is equivalent to the simple expression:

Distribute any coefficients: a(bx\pm c)=abx\pm aca(bx±c)=abx±aca, left parenthesis, b, x, plus minus, c, right parenthesis, equals, a, b, x, plus minus, a, c.

Combine any like terms on each side of the equation: xxx-terms with xxx-terms and constants with constants.

Arrange the terms in the same order, usually xxx-term before constants.

If all of the terms in the two expressions are identical, then the two expressions are equivalent.

Example

How do we solve for unknown coefficients?

Some questions will present us with an equation with algebraic expressions on both sides. On one side, there will be an unknown coeffient, and the question will ask us to find its value.

For the equation to be true for all values of the variable, the two expressions on each side of the equation must be equivalent. For example, if ax+b=cx+dax+b=cx+da, x, plus, b, equals, c, x, plus, d for all values of xxx, then:

aaa must equal ccc.

bbb must equal ddd.

To find the value of unknown coefficients:

Distribute any coefficients on each side of the equation.

Combine any like terms on each side of the equation.

Set the coefficients on each side of the equation equal to each other.

Solve for the unknown coefficient.

Example

How do we rearrange formulas?

Formulas are equations that contain 222 or more variables; they describe relationships and help us solve problems in geometry, physics, etc.

Since a formula contains multiple variables, sometimes we're interested in writing a specific variable in terms of the others. For example, the formula for the area, AAA, for a rectangle with length lll and width www is A=lwA=lwA, equals, l, w. It's easy to calculate AAA using the formula if we know lll and www. However, if we know AAA and www and want to calculate lll, the formula that best helps us with that is an equation in which lll is in terms of AAA and www, or l=\dfrac{A}{w}l=

w

A

l, equals, start fraction, A, divided by, w, end fraction.

Just as we can add, subtract, multiply, and divide constants, we can do so with variables. To isolate a specific variable, perform the same operations on both sides of the equation until the variable is isolated. The new equation is equivalent to the original equation.

7 0
3 years ago
Read 2 more answers
How many solutions exist for the given equation?
andrey2020 [161]

Answer:

1

Step-by-step explanation:

as the maximum degree is 1

3 0
2 years ago
100 Freebie points!!! YW Nerds
blondinia [14]
Thanks for the points.
8 0
3 years ago
Read 2 more answers
On Monday Sophie ran 4 5/6 laps of a running track and on Tuesday she ran 5 3/4
MatroZZZ [7]

Answer:

10  7/12

hope this is right

Step-by-step explanation:

4  5/6 = 29/6 as an improper fraction

5  3/4 = 23/4 as an improper fraction

you times the two denominators together = 24

you times 29 by 4 = 116

you times 23 by 6 = 138

you add 116 and 138 = 254

you put this over 24 =  254/24

simplify this = 127/12

you make this a mixed number = 10 7/12

the answer is 10 7/12

3 0
3 years ago
Please answer correctly !!!!!!!!! Will<br> Mark brainliest !!!!!!!!!!!!!
il63 [147K]

Answer:

x=-\frac{-20+\sqrt{-20w+3600}}{10},\:x=-\frac{-20-\sqrt{-20w+3600}}{10}

Step-by-step explanation:

w=-5\left(x-8\right)\left(x+4\right)\\\mathrm{Expand\:}-5\left(x-8\right)\left(x+4\right):\quad -5x^2+20x+160\\w=-5x^2+20x+160\\Switch\:sides\\-5x^2+20x+160=w\\\mathrm{Subtract\:}w\mathrm{\:from\:both\:sides}\\-5x^2+20x+160-w=w-w\\Simplify\\-5x^2+20x+160-w=0\\Solve\:with\:the\:quadratic\:formula\\\mathrm{Quadratic\:Equation\:Formula:}\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:}\quad a=-5,\:b=20,\:c=160-w:\quad x_{1,\:2}=\frac{-20\pm \sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}\\x=\frac{-20+\sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}:\quad -\frac{-20+\sqrt{-20w+3600}}{10}\\x=\frac{-20-\sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}:\quad -\frac{-20-\sqrt{-20w+3600}}{10}\\The\:solutions\:to\:the\:quadratic\:equation\:are\\x=-\frac{-20+\sqrt{-20w+3600}}{10},\:x=-\frac{-20-\sqrt{-20w+3600}}{10}

6 0
3 years ago
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