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3 years ago

PLEASE HELP ASAP SLOPES OF PARALLEL AND PERPENDICULAR LINES

Mathematics

2 answers:

Rainbow [258]3 years ago

6 0

The slopes of parallel lines are the same just the y-intercepts are different.

The slopes of perpendicular lines are negative reciprocals of each other.

1. -15y = -5x + 30

y = 1/3x + 2

The slope is 1/3. So, A.

2. y = 2/3x - 5

The slope is 2/3

3. x = 2 is a vertical line at x of 2

y = -5 is a horizontal line at y of -5

The two lines are perpendicular to each other. So, B.

4. y = -5/2x +6

The slope of the perpendicular line is 2/5. So, C.

5. y = -3x + 4

The slope of the perpendicular line is 1/3.

Hope this helps! Mark as Brainliest please! :)))

Kamila [148]3 years ago

3 0

1) Parallel lines have equal slopes.

We need to find the slope of the given line by solving for y.

5x - 15y = 30

-15y = -5x + 30

y = (1/3)x - 2

The slope of the given line is 1/3, so the slope of the parallel line is also 1/3.

2) Line 1 has slope 2/3.

Line 2 has slope 2/3.

3) Line 1 is a vertical line through x = 2. Line 2 is a horizontal line through y = -5. The lines are perpendicular.

4) The given line can is solved for y, so we can see its slope is -5/2. The slopes of perpendicular lines are negative reciprocals. The perpendicular line has slope 2/5. We have point (5, 0) which is (x1, y1) in the slope-point formula.

y - 0 = (2/5)(x - 5)

y = 2/5 x - 2

5) The given equation has slope -3. To find the negative reciprocal, write the slope as a fraction, flip it, and change the sign.

Slope of perpendicular line is 1/3.

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Let T be the plane-2x-2y+z =-13. Find the shortest distance d from the point Po=(-5,-5,-3) to T, and the point Q in T that is cl

GaryK [48]

Answer:

d=10u

Q(5/3,5/3,-19/3)

Step-by-step explanation:

The shortest distance between the plane and Po is also the distance between Po and Q. To find that distance and the point Q you need the perpendicular line x to the plane that intersects Po, this line will have the direction of the normal of the plane $n=(-2,-2,1)$ , then r will have the next parametric equations:

$x=-5-2\lambda\\y=-5-2\lambda\\z=-3+\lambda$

To find Q, the intersection between r and the plane T, substitute the parametric equations of r in T

$-2x-2y+z =-13\\-2(-5-2\lambda)-2(-5-2\lambda)+(-3+\lambda) =-13\\10+4\lambda+10+4\lambda-3+\lambda=-13\\9\lambda+17=-13\\9\lambda=-13-17\\\lambda=-30/9=-10/3$

Substitute the value of $\lambda$ in the parametric equations:

$x=-5-2(-10/3)=-5+20/3=5/3\\y=-5-2(-10/3)=5/3\\z=-3+(-10/3)=-19/3\\$

Those values are the coordinates of Q

Q(5/3,5/3,-19/3)

The distance from Po to the plane

$d=\left| {\to} \atop {PoQ}} \right|=\sqrt{(\frac{5}{3}-(-5))^2+(\frac{5}{3}-(-5))^2+(\frac{-19}{3}-(-3))^2} \\d=\sqrt{(\frac{5}{3}+5))^2+(\frac{5}{3}+5)^2+(\frac{-19}{3}+3)^2} \\d=\sqrt{(\frac{20}{3})^2+(\frac{20}{3})^2+(\frac{-10}{3})^2}\\d=\sqrt{\frac{400}{9}+\frac{400}{9}+\frac{100}{9}}\\d=\sqrt{\frac{900}{9}}=\sqrt{100}\\d=10u$

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3 years ago

A fence must be built to enclose a rectangular area of 45 comma 000 ftsquared. Fencing material costs $ 3 per foot for the two s

Tatiana [17]

Answer:

150 feet by 300 feet.

Step-by-step explanation:

The fence is to enclose a rectangular area of 45,000 ft squared.

If the dimensions of the rectangle are x and y

Area of a rectangle = xy

xy=45000
$x=\frac{45000}{y}$

Perimeter of the Rectangle =2x+2y

Fencing material costs $ 3 per foot for the two sides facing north and south and $6 per foot for the other two sides.

Cost of Fencing, C=$(6*2x+3*2y)=$(12x+6y)

Substitute $x=\frac{45000}{y}$ into the Cost to get C(y)

C=12x+6y

$C(y)=12(\frac{45000}{y})+6y\\C(y)=\frac{540000+6y^2}{y}$

The value at which the cost is least expensive is at the minimum point of C(y), when the derivative is zero.

$C^{'}(y)=\dfrac{6y^2-540000}{y^2}$

$\dfrac{6y^2-540000}{y^2}=0\\6y^2-540000=0\\6y^2=540000\\y^2=\frac{540000}{6} =90000\\y=\sqrt{90000}=300$

Recall,

$x=\frac{45000}{y}=\frac{45000}{300}=150$

Since x=150, y=300

The dimensions that will be least expensive to build is 150 feet by 300 feet.

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3 years ago

PLEASE HELP WITH THIS ONE AS WELL! I’m struggling with Transcendental Functions!

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$h(x)=20e^{kx}$

a. If $h(1)=20$ , then $20=20e^k\implies e^k=1\implies k=0$ .

b. If $h(1)=40$ , then $40=20e^k\implies e^k=2\implies k=\ln2$ , so that $2k+1=2\ln2+1$ .

c. If $h(1)=10$ , then $10=20e^k\implies e^k=\dfrac12\implies k=-\ln2$ , so that $k-3=-\ln2-3$ .

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3 years ago

If John has pairs of red, orange, yellow, blue and green socks, how many can he wear them in over 5 days, repetition is allowed

iren [92.7K]

Answer:

Total ways = 5×5×5×5×5

Total ways = 5^5

Total ways = 3125

Therefore, the correct option is C) 5^5

Step-by-step explanation:

John has pairs of red, orange, yellow, blue and green socks.

Which means that John has 5 different colors pairs of socks.

We are asked to find out in how many ways can he wear them over 5 days.

1st Day:

On the first day John has 5 ways to choose from.

2nd Day:

On the second day John has 5×5 ways to choose from.

(Since repetition is allowed)

3rd Day:

On the third day John has 5×5×5 ways to choose from.

4th Day:

On the fourth day John has 5×5×5×5 ways to choose from.

5th Day:

On the fifth day John has 5×5×5×5×5 ways to choose from.

Total ways = 5×5×5×5×5

Total ways = 5^5

Total ways = 3125

Therefore, the correct option is C) 5^5

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3 years ago