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2 years ago

Luis walked one-eighth of the distance to school. What part of distance to school does he still need to walk?

2 answers:

6 0

<span>Total Distance = x
Luis walked one-eighth of x = x(1/8) = x/8
He still need to walk = Total Distance - Distance covered
He still need to walk = x-x/8
He still need to walk = (8x-x)/8

He still need to walk = 7x/8
He still need to walk = x(7/8)
He still need to walk seven-eighth of the total distance.

</span>

algol [13]2 years ago

6 0

7/8 you subtract 1/8 from 8/8

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Segment XY represents the path of an airplane that passes through the coordinates (2, 1) and (4, 5). What is the slope of a line

Karo-lina-s [1.5K]

Answer:

Step-by-step explanation:

Firstly, we will need to calculate the slope of the line formed by joining the coordinates.

The slope of that line would be : m = (y2 - y1)/(x2 - x1)

= (5 - 1)/(4 - 2) = 4/2 = 2

The slope of the line joining the coordinates is 2. Since the other line is parallel to this, it simply means the slope of the second line is 2. This is because of two lines are parallel, they will have the same slope.

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2 years ago

Put 3/4 7/8 3/5 in order from least to greatest

andrew-mc [135]

I am not sure of the correct answer but cross multiply to find which has the bigger number and smallest to put them in order.

4 0

2 years ago

Read 2 more answers

a circle has a center (3,5) and a diameter AB. The coordinates of A are (-4,6). what are the coordinates of B?

Kruka [31]

Find the length of the radius.

$r= \sqrt{(3-(-4))^2+(5-6)^2}
\\r = \sqrt{7^2+(-1)^2}
\\r= \sqrt{50} 
\\ r=5 \sqrt{2}$

Find the length of the diameter.

d = 2r = 2 × 5√2 = 10√2

Point B must lie on a line AC, where C is a center of a circle.
Find equation of line AC.
A(–4, 6), C(3, 5)

$y-y_1= \frac{y_2-y_1}{x_2-x_1}(x-x_1)
\\y-6= \frac{5-6}{3-(-4)} (x-(-4))
\\y-6=- \frac{1}{7} (x+4)
\\7y-42=-x-4
\\x+7y-38=0$

The distance from B(x, y) to C(3, 5) is 5√2.
$\sqrt{(x-3)^2+(y-5)^2}=5 \sqrt{2} 
 \\(x-3)^2+(y-5)^2=(5 \sqrt{2})^2 
\\(x-3)^2+(y-5)^2=50$

Solve system of equations.
$x+7y-38=0
\\(x-3)^2+(y-5)^2=50
\\
\\x=38-7y
\\(38-7y-3)^2+(y-5)^2=50
\\(35-7y)^2+(y-5)^2=50
\\1225-490y+49y^2+y^2-10y+25-50=0
\\50y^2-500y+1200=0
\\y^2-10y+24=0
\\y^2-6y-4y+24=0
\\y(y-6)-4(y-6)=0 \\(y-6)(y-4)=0 \\y_1=6,y_2=4
$

$x_1=38-7y_1=38-7 \times 6 = 38-42=-4
\\x_2=38-7y_2=38-7 \times 4 = 38-28=10
$

Point B could have coordinates
$\\
\\(x_1,y_1)=(-4,6),(x_2,y_2)=(10,4)
$

But (–4, 6) are the coordinates of point A.
Therefore, point B has coordinates (10,4).

8 0

3 years ago

Find the equation of a line parallel to −5x+y=2 that contains the point (5,−2). Write the equation in slope-intercept form.

mario62 [17]

Answer:

L2: y-0 = 5/2(x-5)

y = 5/2x-25/2

Step-by-step explanation:

Parallel lines have same slopes.

Line 1, L1: 5x-2y=20 is in standard form Ax+By=C therefore slope m1= -A/B = -5/-2 = 5/2 or you can solve it for y so you will have the equation in slope-intercept form.

5x-2y = 20

-2y = -5x+20

y = (-5/-2)x+20/(-2)

y = (5/2)x-10 hence m1=5/2 and y-intercept is -10

Line 2 , L2: y-y1 = m (x-x1), m=m2=m1=5/2

Point p(5,0) or p(x1,y1) therefore x1=5 , y1=0 and m=5/2

L2: y-0 = 5/2(x-5)

y = 5/2x-25/2

5 0

2 years ago

Need help please<br>On all three

Pepsi [2]

Answer:

1. proportional relationship

2. constant of proportionality

3. unit rate.

I'm not exactly sure on these answers, so please double check them for me to make sure that they are correct.

Step-by-step explanation:

5 0

2 years ago