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4 years ago

Which line is perpendicular to a line that has a slope of 1/2 ?

Mathematics

2 answers:

Lunna [17]4 years ago

8 0

The answer is HJ i took the test and got it right :)

lesya692 [45]4 years ago

4 0

Answer:

Option D. line HJ

Step-by-step explanation:

Slope of line AB which passes through points A(-4, -2) and B(4, 2)

$m1=\frac{2+2}{4+4}=\frac{4}{8}=1/2$

m1 = 1/2

Slope of line CD passing through C(-4, 0) and D(4, -4)

$m2=\frac{0+4}{-4-4}=\frac{4}{(-8)}=-\frac{1}{2}$

m2 = -1/2

Slope of the line HJ passing through H(-1, 3) and J(1, -1)

$m3 =\frac{-1-3}{1+1}=\frac{-4}{2}=(-2)$

m3 = (-2)

As we know that two line will be perpendicular if their m×m' = -1

Here we see that Line AB is having slope m1 = 1/2 and slope of HJ is m3 = (-2)

m1×m3 = (1/2)×(-2) = -1

Which clearly tells us that line HJ is perpendicular to line AB.

Therefore Option D. is the correct option.

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3 years ago

Read 2 more answers

Consider the equation below. (If you need to use -[infinity] or [infinity], enter -INFINITY or INFINITY.)f(x) = 2x3 + 3x2 − 180x

soldier1979 [14.2K]

Answer:

(a) The function is increasing $\left(-\infty, -6\right) \cup \left(5, \infty\right)$ and decreasing $\left(-6, 5\right)$

(b) The local minimum is x = 5 and the maximum is x = -6

(c) The inflection point is $x = -\frac{1}{2}$

(d) The function is concave upward on $\left(- \frac{1}{2}, \infty\right)$ and concave downward on $\left(-\infty, - \frac{1}{2}\right)$

Step-by-step explanation:

(a) To find the intervals where $f(x) = 2x^3 + 3x^2 -180x$ is increasing or decreasing you must:

1. Differentiate the function

$\frac{d}{dx}f(x) =\frac{d}{dx}(2x^3 + 3x^2 -180x) \\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\f'(x)=\frac{d}{dx}\left(2x^3\right)+\frac{d}{dx}\left(3x^2\right)-\frac{d}{dx}\left(180x\right)\\\\f'(x) =6x^2+6x-180$

2. Now we want to find the intervals where f'(x) is positive or negative. This is done using critical points, which are the points where f'(x) is either 0 or undefined.

$f'(x) =6x^2+6x-180 =0\\\\6x^2+6x-180 = 6\left(x-5\right)\left(x+6\right)=0\\\\x=5,\:x=-6$

These points divide the number line into three intervals:

$(-\infty,-6)$ , $(-6,5)$ , and $(5, \infty)$

Evaluate f'(x) at each interval to see if it's positive or negative on that interval.

$\left\begin{array}{cccc}Interval&x-value&f'(x)&Verdict\\(-\infty,-6)&-7&72&Increasing\\(-6,5)&0&-180&Decreasing\\(5, \infty)&6&72&Increasing\end{array}\right$

Therefore f(x) is increasing $\left(-\infty, -6\right) \cup \left(5, \infty\right)$ and decreasing $\left(-6, 5\right)$

(b) Now that we know the intervals where f(x) increases or decreases, we can find its extremum points. An extremum point would be a point where f(x) is defined and f'(x) changes signs.

We know that:

f(x) increases before x = -6, decreases after it, and is defined at x = -6. So f(x) has a relative maximum point at x = -6.

f(x) decreases before x = 5, increases after it, and is defined at x = 5. So f(x) has a relative minimum point at x = 5.

(c)-(d) An Inflection Point is where a curve changes from Concave upward to Concave downward (or vice versa).

Concave upward is when the slope increases and concave downward is when the slope decreases.

To find the inflection points of f(x), we need to use the f''(x)

$f''(x)=\frac{d}{dx}\left(6x^2+6x-180\right)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\f''(x)=\frac{d}{dx}\left(6x^2\right)+\frac{d}{dx}\left(6x\right)-\frac{d}{dx}\left(180\right)\\\\f''(x) =12x+6$

We set f''(x) = 0

$f''(x) =12x+6 =0\\\\x=-\frac{1}{2}$

Analyzing concavity, we get

$\left\begin{array}{cccc}Interval&x-value&f''(x)\\(-\infty,-1/2)&-2&-18\\(-1/2,\infty)&0&6\\\end{array}\right$

The function is concave upward on $(-1/2,\infty)$ because the f''(x) > 0 and concave downward on $(-\infty,-1/2)$ because the f''(x) < 0.

f(x) is concave down before $x = -\frac{1}{2}$ , concave up after it. So f(x) has an inflection point at $x = -\frac{1}{2}$ .

7 0

3 years ago

PLS HELP WILL MARK BRAINLIEST!!!!!!!! answer both questions (they are two separate questions)

Angelina_Jolie [31]

Answer A is correct for both questions.

Step-by-step explanation:

The area of the rectangle is lengthxwidth. To find the length, we can divide the area by the width.

$\frac{(4x^2-43x+63)}{(4x-7)}$ is the equation.

We need to simplify it (or just divide). Since the coefficient of the area (on the x^2) is the same as that on the width, we know that that same coefficient on the length is 1.

This gets us to the basic frame (1x+/-y).

To find the value of y, we need to pay attention to the "-43x+63" and "-7" aspects of the area and width, respectively. To get "63," the "-7" was multiplied by the y — by dividing 63 by -7, we know that the value of y is -9 (the numbers both have to be negative to multiply to a positive number).

We are left with the length (x-9). Put together, this means (x-9)(4x-7) is the area. Multiplying, that makes (4x^2-7x-36x+63), or (4x^2-43x+63). Since this is the area given to us, we know our answer is correct. For this question, the answer is A.

*****

Divide $9x^4-2-6x-x^2$ by $3x-1$ . First, put the first equation in order by exponents. We get $\frac{9x^4-x^2-6x-2}{3x-1}$ . Since the exponent on the upper equation goes up to x^4, and we are dividing by a simple x, we know that the first exponent in our answer will be x^3. Since our coefficient needs to have a product of 9 when multiplied by 3, it is 3. The first part of our answer is $(3x^3)$ . Since there is no exponent of 3 for x in the upper equation, and since the "x^2" that has to be the next term in the equation due to it being present in each answer choice, we know that it as to be a "+x^2" — the "3x^3 that result from the "-1" (in the lower equation) being multiplied by 3x^3 have to be cancelled out by a "-3x^3", and if the sign for the term "x^2" is negative, we end up with tw0 "3x^3" that add up to "6x^3" instead of cancelling each other out.

Now, we have $(3x^3-x^2)$ . We can immediately rule out C. Moving on.

We now have $(3x-1)(3x^3+x^2)=9x^4-x^2$ . We can rule out answer choice B because it is incomplete - we are missing the second part of the upper equation, "-6x-2." Both of the remaining answers include "-2" as the next term, whether with an x or without.

Honestly, I haven't done algebra in a few years — while I know there's a way to deduce the rest of the equation, let's solve the equation using the two remaining answer choices and see which one is correct.

A: $(3x-1)(3x^3+x^2-2-\frac{4}{3x-1} )$

$=9x^4+3x^3-6x -3x^3-x^2+2-4$ (FOIL) (3x-1 times $-\frac{4}{3x-1}$ = -4)

$=9x^4+3x^3-3x^3-x^2-6x+2-4$ (in order of exponents)

$=9x^4-x^2-6x+2-4$ (simplify)

$=9x^4-x^2-6x-2$ (simplify)

D: $(3x-1)(3x^3+x^2-2x-\frac{4}{3x-1} )$

$=9x^4+3x^3-6x^2-\frac{4}{1} -3x^3-x^2+3x+\frac{4}{1}$ (FOIL)

$=9x^4+3x^3-3x^3-7x^2-4+4$ (in order of exponents)

$=9x^4-7x^2$ (simplify)

A is exactly the long fraction we started with ( $9x^4-2-6x-x^2$ ), just in a different order! This means that answer A, when multiplied by (3x-1), equals the same thing which, if divided by (3x-1), yields answer A. Because of the rules of multiplication/division (xy=z, z/x=y, z/y=z), this means that we have the proper set of numbers. Answer A is correct.

I hope this helps!!!!! Let me know if there's anything else I can help with :)

5 0

2 years ago