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3 years ago

Which of the following is the equation of a line perpendicular to the line y=-10×+1 passing through the point (5,7)

1 answer:

3 0

Y = -10x + 1...slope here is -10. A perpendicular line will have a negative reciprocal slope. To find the negative reciprocal of a number, flip the number and change the sign. So our perpendicular line will need a slope of 1/10.

y = mx + b
slope(m) = 1/10
(5,7)....x = 5 and y = 7
now sub and find b, the y int
7 = 1/10(5) + b
7 = 1/2 + b
7 - 1/2 = b
14/2 - 1/2 = b
13/2 = b

so ur perpendicular line going thru (5,7) is y = 1/10x + 13/2

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A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y=2−x2. What are the dimensions of su

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Answer:

Step-by-step explanation:

Given that a rectangle is inscribed with its base on the x-axis and its upper corners on the parabola

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We can find that the rectangle also will be symmetrical about y axis.

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Then other two vertices would be (p,2-p^2) (-p,2-p^2) because the vertices lie on the parabola and satisfy the parabola equation

Now width = $2-p^2$

Area = l*w = $2(2p-p^3)$

Use derivative test

I derivative = $2(2-3p^2)$

II derivative = $-12p$

Equate I derivative to 0 and consider positive value only since we want maximum

p = $\sqrt{\frac{2}{3} }$

Thus width= $\sqrt{\frac{2}{3} }$

Length = $2\sqrt{\frac{2}{3} }$

Width = $2-2/3 = 4/3$

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3 years ago

M is the midpoint of AB. If A is at (5,7) and M is at (-3,-4), find the coordinates of B.

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3 years ago

If x>2, then x^2-x-6/x^2-4=

Viefleur [7K]

Consider the expression $\frac{x^{2} -x-6}{ x^{2} -4}$

To factorize the expression in the denominator we use difference of squares: $x^{2} -4=x^{2} - 2^{2} =(x-2)(x+2)$

To factorize $x^{2} -x-6$ we use the following method:

$x^{2} -x-6=(x-a)(x-b)$

where a, b are 2 numbers such that a+b= -1, the coefficient of x,

and a*b= -6, the constant.

such 2 numbers can be easily checked to be -3 and 2

(-3*2=6, -3+2=-1)

So $x^{2} -x-6=(x-a)(x-b)=(x+3)(x-2)$

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

$\frac{x+3}{x+2}= \frac{x+2+1}{x+2}= \frac{x+2}{x+2}+ \frac{1}{x+2}=1+ \frac{1}{x+2}$

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$\frac{1}{x+2}\ \textless \ \frac{1}{2+2}= \frac{1}{4}$

thus

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$1+ \frac{1}{x+2}\ \textless \ 1+ \frac{1}{4}= \frac{5}{4}$

Answer:

for x>2

$\frac{x^{2} -x-6}{ x^{2} -4} = \frac{x+3}{x+2} \ \textless \ \frac{5}{4}$ , (but the expression is never 0)

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3 years ago

Plssssssssssssssssssssss

daser333 [38]

Answer:

g = 118 degrees

h = 62 degrees

k = 98 degrees

m = 82 degrees

Step-by-step explanation:

180-118=62

180-98=82

7 0

2 years ago

Read 2 more answers