Given f(x)=3x-1 and g(x)=2x-3, for which value of x does g(x)=f(2)?

GalinKa [24]

Part A. What is the slope of a line that is perpendicular to a line whose equation is −2y=3x+7?

Rewrite the equation −2y=3x+7 in the form $y=-\dfrac{3}{2}x-\dfrac{7}{2}.$ Here the slope of the given line is $m_1=-\dfrac{3}{2}.$ If $m_2$ is the slope of perpendicular line, then

$m_1\cdot m_2=-1,\\ \\m_2=-\dfrac{1}{m_1}=\dfrac{2}{3}.$

Answer 1: $\dfrac{2}{3}$

Part B. The slope of the line y=−2x+3 is -2. Since $-\dfrac{3}{2}\neq -2\quad \text{and}\quad \dfrac{2}{3}\neq -2,$ then lines from part A are not parallel to line a.

Since $-2\cdot \left(-\dfrac{3}{2}\right)=3\neq -1\quad \text{and}\quad -2\cdot \dfrac{2}{3}=-\dfrac{4}{3}\neq -1,$ both lines are not perpendicular to line a.

Answer 2: Neither parallel nor perpendicular to line a

Part C. The line parallel to the line 2x+5y=10 has the equation 2x+5y=b. This line passes through the point (5,-4), then

2·5+5·(-4)=b,

10-20=b,

b=-10.

Answer 3: 2x+5y=-10.

Part D. The slope of the line $y=\dfrac{x}{4}+5$ is $\dfrac{1}{4}.$ Then the slope of perpendicular line is -4 and the equation of the perpendicular line is y=-4x+b. This line passes through the point (2,7), then

7=-4·2+b,

b=7+8,

b=15.

Answer 4: y=-4x+15.

Part E. Consider vectors $\vec{p}_1=(-c-0,0-(-d))=(-c,d)\quad \text{and}\quad \vec{p}_2=(0-b,a-0)=(-b,a).$ These vectors are collinear, then

$\dfrac{-c}{-b}=\dfrac{d}{a},\quad \text{or}\quad -\dfrac{a}{b}=-\dfrac{d}{c}.$

Answer 5: $-\dfrac{a}{b}=-\dfrac{d}{c}.$

5 0

3 years ago

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What is the answer to 1.1x + 1.2x - 5.4 = -10

lord [1]

1.1x+1.2x-5.4=-10

11/10x+12/10x-54/10=10

11x/2.5+2^2-1*3/5x-3^3/5=-10

(11x)+2(2*3x)+2(-3^3)/2*5=-10

11x+2(6x)+2(-27)/2*5=-10

11x+2*6x-2*27/2*5=-10

11x+12x-54/2*5=-10

23x-54/2*5=-10

23x-54=-100

23x=-46

23x/23=-46/23

x=-2*23/23

x=-2

8 0

3 years ago

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Solve with solution...no spam answers pls

Agata [3.3K]

Let's analyze Hannah's work, step-by-step, to see if she made any mistakes.

In Step 1, Hannah wrote $\dfrac{d}{dx} (-3+8x)$ as the sum of two separate derivatives $\dfrac{d}{dx}(-3)+ \dfrac{d}{dx} (8x)$ using the sum rule.

This step is perfectly fine.

In Step 2, $\dfrac{d}{dx}(8x)$ was kept as it is, and $\dfrac{d}{dx}(-3)$ was rewritten as $0$ using the constant rule.Indeed, according to the constant rule, the derivative of a constant number is equal to zero.

This step is perfectly fine.

In Step 3, $\dfrac{d}{dx} (8x)$ was rewritten as $\dfrac{d}{dx}(8) \dfrac{d}{dx}(x)$ supposedly using the constant multiple rule.

The problem is that according to the constant multiple rule, $\dfrac{d}{dx}(8x)
$ should be rewritten as $8 \dfrac{d}{dx}(x)$ and not as $\dfrac{d}{dx}(8)\dfrac{d}{dx}(x)$ .

Therefore, Hannah made a mistake in this step.

6 0

3 years ago

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6x+9y=15 -12x-18y=-30 Solve system by elimination Giving 10,pts

larisa [96]

Answer:

dssfgffdddfggrrghygfdedfhyjjujtg

7 0

2 years ago

B) Express 0.6363......as a rational number in its lowest term.

bulgar [2K]

Answer: