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Roman55 [17]
4 years ago
9

If ||M and M6 = 4X - 15 and m7 = x + 30 then m 6 =​

Mathematics
2 answers:
TEA [102]4 years ago
8 0

Answer:

<h2>15</h2>

Step-by-step explanation:

∠6 and ∠7 are Alternate Interior Angles.

The lines l and m are parallel, therefore ∠6 and ∠7 are congruent.

m∠6 = m∠7 → 4x - 15 = x + 30        <em>add 15 to both sides</em>

4x = x + 45          <em>subract x from both sides</em>

3x = 45            <em>divide both sides by 3</em>

x = 15

PilotLPTM [1.2K]4 years ago
8 0

Answer:

\angle 6 =45^{\circ}

Step-by-step explanation:

We are given that l || m

\angle 6 = 4x-15

\angle 7 = x+30

Since l || m

So, \angle 6 = \angle 7 (Alternate interior angles are equal )

So, 4x-15 =  x+30

3x= 45

x= 15

\angle 6 = 4x-15=4(15)-15 = 45^{\circ}

So, \angle 6 =45^{\circ}

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Which of the following statements are true of the graph of the function f(x) = (x+5)(x-3)
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Goshia [24]

Answer:

a) The horizontal asymptote is y = 0

The y-intercept is (0, 9)

b) The horizontal asymptote is y = 0

The y-intercept is (0, 5)

c) The horizontal asymptote is y = 3

The y-intercept is (0, 4)

d) The horizontal asymptote is y = 3

The y-intercept is (0, 4)

e) The horizontal asymptote is y = -1

The y-intercept is (0, 7)

The x-intercept is (-3, 0)

f) The asymptote is y = 2

The y-intercept is (0, 6)

Step-by-step explanation:

a) f(x) = 3^{x + 2}

The asymptote is given as x → -∞, f(x) = 3^{x + 2} → 0

∴ The horizontal asymptote is f(x) = y = 0

The y-intercept is given when x = 0, we get;

f(x) = 3^{0 + 2} = 9

The y-intercept is f(x) = (0, 9)

b) f(x) = 5^{1  - x}

The asymptote is fx) = 0 as x → ∞

The asymptote is y = 0

Similar to question (1) above, the y-intercept is f(x) = 5^{1  - 0} = 5

The y-intercept is (0, 5)

c) f(x) = 3ˣ + 3

The asymptote is 3ˣ → 0 and f(x) → 3 as x → ∞

The asymptote is y = 3

The y-intercept is f(x) = 3⁰ + 3= 4

The y-intercept is (0, 4)

d) f(x) = 6⁻ˣ + 3

The asymptote is 6⁻ˣ → 0 and f(x) → 3 as x → ∞

The horizontal asymptote is y = 3

The y-intercept is f(x) = 6⁻⁰ + 3 = 4

The y-intercept is (0, 4)

e) f(x) = 2^{x + 3} - 1

The asymptote is 2^{x + 3}  → 0 and f(x) → -1 as x → -∞

The horizontal asymptote is y = -1

The y-intercept is f(x) =  2^{0 + 3} - 1 = 7

The y-intercept is (0, 7)

When f(x) = 0, 2^{x + 3} - 1 = 0

2^{x + 3} = 1

x + 3 = 0, x = -3

The x-intercept is (-3, 0)

f) f(x) = \left (\dfrac{1}{2} \right)^{x - 2} + 2

The asymptote is \left (\dfrac{1}{2} \right)^{x - 2} → 0 and f(x) → 2 as x → ∞

The asymptote is y = 2

The y-intercept is f(x) = f(0) = \left (\dfrac{1}{2} \right)^{0 - 2} + 2 = 6

The y-intercept is (0, 6)

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