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andreyandreev [35.5K]
3 years ago
6

Which pair of points have a slope of -3/5?

Mathematics
1 answer:
Allushta [10]3 years ago
5 0

Answer:

No of these answers has a slope of  \frac{-3}{5} . My guess is B becasue its close to \frac{-3}{5} unless the 5 is supposed to be negative

Step-by-step explanation:

You have to use the slope formula \frac{y_{2} - y_{1} }{x_{2} - x_{1} }

(x_{1},y_{1}), (x_{2},y_{2})

(4, 0), (-2, 10)   \frac{10 - 0 }{-2 - 4 }= \frac{10}{6} you have to simplfy, the answer will be \frac{5}{3}

(10, -2), (0, 4)  \frac{4 - -2 }{0- 10 }= \frac{6}{-10} you have to simplfy, the answer will be \frac{3}{-5}

 (4, 2), (10, 4)  \frac{4 - 2 }{10 - 4 }= \frac{2}{6} you have to simplfy, the answer will be \frac{1}{3}

 (-4, -10),(0, -2) \frac{{-2- -10} }{0 - -4 }= \frac{8}{4} you have to simplfy, the answer will be \frac{2}{1} =2

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Step-by-step explanation:

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4 0
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It is 4 1/2 miles to the end of the trail. If Sophia bikes to the end and back in 60 minutes, what is her average speed in miles
PtichkaEL [24]
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4 0
3 years ago
Solve by taking the square root of both sides 3(x-1)^2-162=0
Natali5045456 [20]
We know that: <span>3(x-1)^2-162=0
or (x-1)</span>²= 162:3
and (x-1)²= 54
we <span>take the square root of both sides 
* x-1=</span>√54= 3√6 or x=1+3√6
* x-1= -<span>√54= -3√6 or x=1-3√6
This equation has 2 solutions</span>
7 0
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Read 2 more answers
PLEASE HELP!!!!!!!dgbdgdbhdndcn
bogdanovich [222]
Problem 1)

AC is only perpendicular to EF if angle ADE is 90 degrees

(angle ADE) + (angle DAE) + (angle AED) = 180
(angle ADE) + (44) + (48) = 180
(angle ADE) + 92 = 180
(angle ADE) + 92 - 92 = 180 - 92
angle ADE  = 88

Since angle ADE is actually 88 degrees, we do NOT have a right angle so we do NOT have a right triangle

Triangle AED is acute (all 3 angles are less than 90 degrees)

So because angle ADE is NOT 90 degrees, this means AC is NOT perpendicular to EF

-------------------------------------------------------------

Problem 2)

a) The center is (2,-3) 

The center is (h,k) and we can see that h = 2 and k = -3. It might help to write (x-2)^2+(y+3)^2 = 9 into (x-2)^2+(y-(-3))^2 = 3^3 then compare it to (x-h)^2 + (y-k)^2 = r^2

---------------------

b) The radius is 3 and the diameter is 6

From part a), we have (x-2)^2+(y-(-3))^2 = 3^3 matching (x-h)^2 + (y-k)^2 = r^2

where
h = 2
k = -3
r = 3

so, radius = r = 3
diameter = d = 2*r = 2*3 = 6

---------------------

c) The graph is shown in the image attachment. It is a circle with center point C = (2,-3) and radius r = 3.

Some points on the circle are

A = (2, 0)
B = (5, -3)
D = (2, -6)
E = (-1, -3)

Note how the distance from the center C to some point on the circle, say point B, is 3 units. In other words segment BC = 3.

6 0
3 years ago
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