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AnnZ [28]
3 years ago
12

How many times does the graph of 4x = 32 - x2 cross the x-axis ?

Mathematics
2 answers:
Alisiya [41]3 years ago
8 0

Answer:

Step-by-step explanation:

● 4x = 32 - x^2

Add x^2 to both sides

● 4x +x^2 = 32 -x^2+x^2

● x^2 +4x = 32

Substract 32 from both sides

● x^2+4x -32 =32-32

● x^2 +4x-32=0

That is a quadratic equation

We will use the determinant method

The determinant is b^2-4ac

● a = 1

● b = 4

● c = 32

● b^2-4ac = 4^2 -4×1×(-32) = 144

144 is positive so the parabola will cross the x-axis two times

● x = (-b -/+ √(b^2-4ac))/2a

● x = (-4 +/- 12) /2

● x = 8/2 or x= -16/2

● x = 4 or x =-8

The solutions are 4 and -8

Molodets [167]3 years ago
5 0

Answer:

x=4 orx=- 8

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

4

x

=

−

x

2

+

32

Step 2: Subtract -x^2+32 from both sides.

4

x

−

(

−

x

2

+

32

)

=

−

x

2

+

32

−

(

−

x

2

+

32

)

x

2

+

4

x

−

32

=

0

Step 3: Factor left side of equation.

(

x

−

4

)

(

x

+

8

)

=

0

Step 4: Set factors equal to 0.

x

−

4

=

0

or

x

+

8

=

0

x

=

4

or

x

=

−

8

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Answer:

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Step-by-step explanation:

A = 30° and r = 12

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\sin 30 = \frac{12}{12 + x}

\frac{1}{2} = \frac{12}{12 + x}

1(12 + x) = 2(12)

12 + x = 24

x = 24 - 12

x = 12

.

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IrinaK [193]

Answer: (-2,3), r=5\ \text{units}

Step-by-step explanation:

Given

the equation is x^2+y^2+4x-6y-12=0

completing the square

\Rightarrow (x^2+2\times 2\times x+4-4)+(y^2-2\times3\times y+9-9)-12=0\\\Rightarrow (x^2+2\times 2\times x+4)+(y^2-2\times3\times y+9)-12-4-9=0\\\Rightarrow (x+2)^2+(y-3)^2=25\\\Rightarrow (x+2)^2+(y-3)^2=5^2\\\Rightarrow (x-(-2))^2+(y-3)^2=5^2

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