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dmitriy555 [2]
3 years ago
6

Ebony is solving a quadratic equation. she wants to find the value of x by taking the square root of both sides of the equation.

which equation allows her to do this?
a. x^2 - 8x + 64 = 32

b. x^2 - 144x + 12 = 13

c. x^2 - 6x - 9 = 15

d. x^2 - 4x + 4 = 36
Mathematics
1 answer:
Fantom [35]3 years ago
3 0

Answer:

D

Step-by-step explanation:

x^2 - 4x + 4 = 36

Take the square root on both sides.

x - 2x + 2 = 6

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Find the product.<br> (2x + 3)(2x + 3)
solniwko [45]

Answer:

=4x2+12x+9

Step-by-step explanation:

(2x+3)(2x+3)

(happy to help)

5 0
4 years ago
The answer of the problem
k0ka [10]

Answer:

x=16 and x=-3

Step-by-step explanation:

A <em>second-degree equation</em> or <em>quadratic equation of a variable</em> is an equation that it has the general expression:

ax^{2} +bx+c=0 with a\neq 0

Where x is the variable, and a, b and c constants; a is the quadratic coefficient (other than 0), b the linear coefficient and c is the independent term. This polynomial can be interpreted by means of the graph of a quadratic function, that is, by a parabola. This graphical representation is useful, because the abscissas of the intersections or point of tangency of this graph, in the case of existing, with the X axis are the real roots of the equation. If the parabola does not cut the X axis the roots are complex numbers, they correspond to a negative discriminant.

Second degree equation solutions

For a quadratic equation with real or complex coefficients there are always two solutions, not necessarily different, called roots, which can be real or complex (if the coefficients are real and there are two non-real solutions, then they must be complex conjugates). General formula for obtaining roots:

x=\frac{-b±\sqrt{b^{2} -4ac} }{2a}

The discriminant serves to analyze the nature of the roots that can be real or complex.

Δ=b^{2} -4ac

Solving the problem of the answer.

x^{2} -13x-48=0 with a = 1, b = -13, and c = -48

Substituting the values in the general formula for a quadratic equation.

x=\frac{-(-13)±\sqrt{(-13)^{2} -4(1)(-48)} }{2(1)}

x=\frac{13±\sqrt{169+192} }{2}

x=\frac{13±\sqrt{361} }{2}

Then, we obtain the roots:

x=\frac{13+\sqrt{361} }{2} and x=\frac{13-\sqrt{361} }{2}

Solving the roots:

x=\frac{13+\sqrt{361} }{2}\\x=\frac{13+19}{2}\\x=\frac{32}{2}\\x=16

x=\frac{13-\sqrt{361} }{2}\\x=\frac{13-19}{2}\\x=\frac{-6}{2}\\x=-3

4 0
3 years ago
Ed’s rental charges a flat fee of $5 plus $2 per hour to rent a pair of skates.
zavuch27 [327]
Flat doesn't change
2 changes with x
y=2x+5
yint is 5

to graph
start at center and go 5 up, that is y intercept
go 1 right and 2 up, that is another point

y=2(2)+5
y=4+5
y=9
cost $9 to rent for 2 hours
5 0
3 years ago
Help me plz asap (another post but better photo)
Katena32 [7]

You could survey people about things you like or are interested in. Such as "What's your favorite food?", 'What's your favorite sport?" and get 10 data values. Put it in a graph or a chart.
6 0
4 years ago
Evaluate 7+q(r−8)2 when q = 3 and r = 10.
liberstina [14]

<u>Answer:</u>

  • The solution of the expression is 19.

<u>Step-by-step explanation:</u>

<u>Given:</u>

  • q = 3
  • r = 10

<u>Work:</u>

  • => 7 + q(r − 8)2
  • => 7 + 3(10 - 8)2
  • => 7 + 6(10 - 8)
  • => 7 + 6(2)
  • => 7 + 12
  • => 19

Hence, <u>the solution of the expression is 19.</u>

Hoped this helped.

BrainiacUser1357

8 0
3 years ago
Read 2 more answers
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