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4 years ago

Which equation represents a line that passes through (-2, 4) and has a slope 71

Mathematics

1 answer:

12345 [234]4 years ago

7 0

Step-by-step explanation:

The point-slope of an equation of a line:

$y-y_1=m(x-x_1)$

m - slope

We have the slope m = 71 and the point (-2, 4).

Substitute:

$y-4=71(x-(-2))\\\\\bold{y-4=71(x+2)}$

The slope-intercept form of an equation of a line:

$y=mx+b$

Convert:

$y-4=71(x+2)$ use the distributive property

$y-4=71x+142$ add 4 to both sides

$\bold{y=71x+146}$

The standard form of an equation of a line:

$Ax+By=C$

Convert:

$y=71x+146$ subtract 71x from both sides

$-71x+y=146$ change the signs

$\bold{71x-y=-146}$

The general form of an equation of a line:

$Ax+By+C=0$

Convert:

$71x-y=-146$ add 146 to both sides

$\bold{71x-y+146=0}$

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Points E, F, and D are on circle C, and angle G measures 60°. The measure of arc EF equals the measure of arc FD.

galben [10]

Answer:

∠EFD ≅ ∠EGD ⇒ A

Arc ED ≅ arc FD ⇒ C

m arc FD = 120° ⇒ E

Step-by-step explanation:

Let us revise some facts

Equal chords subtended equal arcs
The measure of an inscribed angle is one-half the measure of the central angle which subtended by the same arc
The measure of a central angle is equal to the measure of its subtended arc
If one angle of an isosceles triangle measure 60° then the triangle is equilateral
The sum of the measures of the interior angles of any quadrilateral is 360°

In the quadrilateral CDGE

∵ m∠G = 60°

∵ m∠GDC = m∠GEC = 90°

- By using the 5th rule above

∴ m∠G + m∠GDC + m∠DCE + m∠GEC = 360°

∴ 60 + 90 + m∠DCE + 90 = 360

∴ 240 + m∠DCE = 360

- Subtract 240 from both sides

∵ m∠DCE = 120°

In circle C

∵ ∠DCE is a central angle subtended by arc DE

∵ ∠DFE is an inscribed angle subtended by arc DE

- By using the 2nd rule above

∴ m∠DFE = $\frac{1}{2}$ m∠∠DCE

∵ m∠DCE = 120°

∴ m∠DFE = $\frac{1}{2}$ (120)

∴ m∠DFE = 60°

- That means ∠EFD ≅ ∠EGD because their measure is 60°

∴ ∠EFD ≅ ∠EGD

In Δ EFD

∵ EF = FD

∵ m∠DFE = 60°

- By using the 4th rule above

∴ Δ EFD is an equilateral triangle

∴ ED = FD = FE

In circle C

∵ Side ED subtended by arc ED

∵ Side FD subtended by FD

∵ Side ED ≅ side FD ⇒ proved

- By using the 1st rule above

∴ Arc ED ≅ arc FD

∵ m∠ECD = 120°

∵ ∠ECD is a central angle subtended by arc ED

- By using the 3rd rule above

∴ m∠ECD = m arc ED

∴ m of arc ED = 120°

∵ Arc ED ≅ arc FD

∴ m arc ED = m arc FD

∴ m arc FD = 120°

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Let f(x)=5x3−60x+5 input the interval(s) on which f is increasing. (-inf,-2)u(2,inf) input the interval(s) on which f is decreas

o-na [289]

Answers:

(a) f is increasing at $(-\infty,-2) \cup (2,\infty)$ .

(b) f is decreasing at $(-2,2)$ .

(c) f is concave up at $(2, \infty)$

(d) f is concave down at $(-\infty, 2)$

Explanations:

(a) f is increasing when the derivative is positive. So, we find values of x such that the derivative is positive. Note that

$f'(x) = 15x^2 - 60
$

So,

$
f'(x) \ \textgreater \ 0
\\
\\ \Leftrightarrow 15x^2 - 60 \ \textgreater \ 0
\\
\\ \Leftrightarrow 15(x - 2)(x + 2) \ \textgreater \ 0
\\
\\ \Leftrightarrow \boxed{(x - 2)(x + 2) \ \textgreater \ 0} \text{ (1)}$

The zeroes of (x - 2)(x + 2) are 2 and -2. So we can obtain sign of (x - 2)(x + 2) by considering the following possible values of x:

-->> x < -2
-->> -2 < x < 2
--->> x > 2

If x < -2, then (x - 2) and (x + 2) are both negative. Thus, (x - 2)(x + 2) > 0.

If -2 < x < 2, then x + 2 is positive but x - 2 is negative. So, (x - 2)(x + 2) < 0.
If x > 2, then (x - 2) and (x + 2) are both positive. Thus, (x - 2)(x + 2) > 0.

So, (x - 2)(x + 2) is positive when x < -2 or x > 2. Since

$f'(x) \ \textgreater \ 0 \Leftrightarrow (x - 2)(x + 2) \ \textgreater \ 0$

Thus, f'(x) > 0 only when x < -2 or x > 2. Hence f is increasing at $(-\infty,-2) \cup (2,\infty)$ .

(b) f is decreasing only when the derivative of f is negative. Since

$f'(x) = 15x^2 - 60$

Using the similar computation in (a),

$f'(x) \ \textless \ \ 0 \\ \\ \Leftrightarrow 15x^2 - 60 \ \textless \ 0 \\ \\ \Leftrightarrow 15(x - 2)(x + 2) \ \ \textless \ 0 \\ \\ \Leftrightarrow \boxed{(x - 2)(x + 2) \ \textless \ 0} \text{ (2)}$

Based on the computation in (a), (x - 2)(x + 2) < 0 only when -2 < x < 2.

Thus, f'(x) < 0 if and only if -2 < x < 2. Hence f is decreasing at (-2, 2)

(c) f is concave up if and only if the second derivative of f is positive. Note that

$f''(x) = 30x - 60$

Since,

$f''(x) \ \textgreater \ 0
\\
\\ \Leftrightarrow 30x - 60 \ \textgreater \ 0
\\
\\ \Leftrightarrow 30(x - 2) \ \textgreater \ 0
\\
\\ \Leftrightarrow x - 2 \ \textgreater \ 0
\\
\\ \Leftrightarrow \boxed{x \ \textgreater \ 2}$

Therefore, f is concave up at $(2, \infty)$ .

(d) Note that f is concave down if and only if the second derivative of f is negative. Since,

$f''(x) = 30x - 60$

Using the similar computation in (c),

$f''(x) \ \textless \ 0 
\\ \\ \Leftrightarrow 30x - 60 \ \textless \ 0 
\\ \\ \Leftrightarrow 30(x - 2) \ \textless \ 0 
\\ \\ \Leftrightarrow x - 2 \ \textless \ 0 
\\ \\ \Leftrightarrow \boxed{x \ \textless \ 2}$

Therefore, f is concave down at $(-\infty, 2)$ .

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