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Amanda [17]
4 years ago
10

Find the zero of y=x^2+6x-4 by completing the square

Mathematics
1 answer:
REY [17]4 years ago
3 0
X^2 + ( 6/2 )^2 - ( 6/2 )^2 - 4 = 0
( x + 3 )^2 - 3 - 4 = 0
( x + 3 )^2 - 7 = 0
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Solve the inequality.
Olegator [25]

Answer:

x = 14

x + 23 < 37 ( move constant to right-hand side and change its sign)

x<37-23 ( subtract numbers)

x < 14

3 0
3 years ago
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(3x2 + 10x + 3) by (x + 3)​
Nata [24]
Explanation: 3 x 2 + 10 x + 3 We can Split the Middle Term of this expression to factorise it. In this technique, if we have to factorise an expression like a x 2 + b x + c , we need to think of 2 numbers such that: N 1 ⋅ N 2 = a ⋅ c = 3 ⋅ 3 = 9 and, N 1 + N 2 = b = 10 After trying out a few numbers we get: N 1 = 9 and N 2 = 1 9 ⋅ 1 = 9 , and 9 + ( 1 ) = 10 3 x 2 + 10 x + 3 = 3 x 2 + 9 x + 1 x + 3 = 3 x ( x + 3 ) + 1 ( x + 3 ) ( 3 x + 1 ) ( x + 3 ) is the factorised form for the expression.

c
, we need to think of 2 numbers such that:
N
1
⋅
N
2
=
a
⋅
c
=
3
⋅
3
=
9

and,
N
1
+
N
2
=
b
=
10
After trying out a few numbers we get:
N
1
=
9
and
N
2
=
1

9
⋅
1
=
9
, and
9
+
(
1
)
=
10
3
x
2
+
10
x
+
3
=
3
x
2
+
9
x
+
1
x
+
3

=
3
x
(
x
+
3
)
+
1
(
x
+
3
)
(
3
x
+
1
)
(
x
+
3
)
is the factorised form for the expression.
6 0
3 years ago
How would a vertical line test that a circle is not function?
AleksAgata [21]

The vertical line test is a visual way to determine if a curve is a graph of a function or not. A function can only have one output, y, for each unique input, x. If the vertical line you drew intersects the graph more than once for any value of x then the graph is not the graph of a function

5 0
3 years ago
WHAT IS THE EXACT VALUE OF PI ?
alekssr [168]

Answer:

3.14159

Step-by-step explanation:

3 0
2 years ago
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You are given the parametric equations x=2cos(θ),y=sin(2θ). (a) List all of the points (x,y) where the tangent line is horizonta
vladimir1956 [14]

Answer:

The solutions listed from the smallest to the greatest are:

x:  -\sqrt{2}   -\sqrt{2}  \sqrt{2}  \sqrt{2}

y:      -1         1     -1     1

Step-by-step explanation:

The slope of the tangent line at a point of the curve is:

m = \frac{\frac{dy}{dt} }{\frac{dx}{dt} }

m = -\frac{\cos 2\theta}{\sin \theta}

The tangent line is horizontal when m = 0. Then:

\cos 2\theta = 0

2\theta = \cos^{-1}0

\theta = \frac{1}{2}\cdot \cos^{-1} 0

\theta = \frac{1}{2}\cdot \left(\frac{\pi}{2}+i\cdot \pi \right), for all i \in \mathbb{N}_{O}

\theta = \frac{\pi}{4} + i\cdot \frac{\pi}{2}, for all i \in \mathbb{N}_{O}

The first four solutions are:

x:   \sqrt{2}   -\sqrt{2}  -\sqrt{2}  \sqrt{2}

y:     1        -1        1     -1

The solutions listed from the smallest to the greatest are:

x:  -\sqrt{2}   -\sqrt{2}  \sqrt{2}  \sqrt{2}

y:      -1         1     -1     1

6 0
3 years ago
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