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Zina [86]
3 years ago
9

This is the equation and we need to solve for y 2x+y=7

Mathematics
2 answers:
vitfil [10]3 years ago
8 0

Answer:

y= -2x+7

Step-by-step explanation:

We need to keep y on one side of the equation.

2x+y= 7

=2x     -2x

y= -2x+7

Hope this helps!

marysya [2.9K]3 years ago
7 0

Answer: y = -2x + 7

Step-by-step explanation:

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Solve the equation for all real solutions in simplest form.<br> 3x2 - 12x + 11 = x
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Answer:

(13 + or - (sign) square root(37)) / 6

Step-by-step explanation:

First search up the quadratic formula which is

x = (- b + or - square root(b^2 - 4ac)) / 2a

and to find a, b, and c you need this formula

ax^2 + bx + c = 0 (btw this is the safest solution to finding your answer with a number in-front of a quadratic equation.)

Math:

a = 3, b = -13, c = 11

b = -13, because of the x on the other end of the equation as the equation above "ax^2 + bx + c = 0" you need the zero to proceed to this next step.

Side Note: You only take the number and not the variable, variables are x, y, w, z, or etc.

(- (- 13) + or - square root((-13)^2 - 4(3)(11) ) ) / 2 * 3

13 + or - square root(37) / 6

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Which of the following measuring instruments is used to test the quality of wire insulation?
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Identify the vertex, axis of symmetry, minimum or maximum, domain, and range of the function f(
alekssr [168]

Identify the vertex, axis of symmetry, minimum or maximum, domain, and range of the function ()=−(+)^−

<em><u>Answer:</u></em>

vertex = (-4, -5)

Axis of symmetry = -4

use the (-4, -5) to find the minimum value

Domain = ( - \infty, \infty ) , [ x | x\ is\ real ]\\\\Range = [ -5, \infty ), y\geq -5

<em><u>Solution:</u></em>

Given function is:

f(x) = (x+4)^2 - 5

The equation in vertex form is given as:

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

Where, (h, k) is constant

On comparing give function with vertex form,

h = -4

k = -5

Vertex is (-4 , -5)

Axis of symmetry : x co-ordinate of vertex

Thus, axis of symmetry = -4

The coefficient of x^2 is positive in given function.

Thus the vertex point will be a minimum

Minimum\ value = f(\frac{-b}{a})

f(x) = x^2 + 8x + 16 - 5\\\\f(x) = x^2 + 8x + 11

f(x) = ax^2+bx+c

On comparing,

a = 1

b = 8

x = \frac{-b}{2a} = \frac{-8}{2 \times 1} = -4

f(-4) = (-4)^2 + 8(-4) + 11 = 16 - 32 + 11 = -5

Thus, use the (-4, -5) to find the minimum value

Domain and range

f(x) = (x+4)^2 - 5

The domain is the input values shown on the x-axis

The range is the set of possible output values f(x)

Therefore,

Domain = ( - \infty, \infty ) , [ x | x\ is\ real ]\\\\Range = [ -5, \infty ), y\geq -5

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