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prisoha [69]
3 years ago
7

Which equation in point-slope form contains the points (0, 5) and (5, 8)?

Mathematics
1 answer:
Radda [10]3 years ago
4 0

Answer:

y-5=3/5(x-0)

or

y-8=3/5(x-5)

Step-by-step explanation:

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The only shape you can make is a parallelogram. If both pairs of opposite angles of a quadrilateral are congruent, then it's a parallelogram (converse of a property). If the diagonals of a quadrilateral bisect each other, then it's a parallelogram (converse of a property).

Step-by-step explanation:

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Answer:

The triangle is a  obtuse triangle

Step-by-step explanation:


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3 years ago
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3 years ago
Circle P is described by the equation (x−1)2+(y+6)2=9 and circle Q is described by the equation (x+4)2+(y+14)2=4. Select from th
xeze [42]

Answer:

(a) Circle Q is 9.4 units to the center of circle P

(b) Circle Q has a smaller radius

Step-by-step explanation:

Given

P:(x - 1)^2 + (y + 6)^2 = 9

Q:(x + 4)^2 + (y + 14)^2 = 4

Solving (a): The distance between both

The equation of a circle is:

(x - h)^2 + (y - k)^2 = r^2

Where

Center: (h,k)

Radius:r

P and Q can be rewritten as:

P:(x - 1)^2 + (y + 6)^2 = 3^2

Q:(x + 4)^2 + (y + 14)^2 = 2^2

So, for P:

Center = (1,-6)

r = 3

For Q:

Center = (-4,-14)

r = 2

The distance between them is:

d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2

Where:

Center = (1,-6) --- (x_1,y_1)

Center = (-4,-14) --- (x_2,y_2)

So:

d = \sqrt{(1 - -4)^2 + (-6 - -14)^2

d = \sqrt{(5)^2 + (8)^2

d = \sqrt{25 + 64

d = \sqrt{89

d = 9.4

Solving (b): The radius;

In (a), we have:

r = 3 --- circle P

r = 2 --- circle Q

By comparison

2 < 3

<em>Hence, circle Q has a smaller radius</em>

4 0
2 years ago
What is a quotient of 3/10÷8/5 out of these forms 1. 3/16 2. 4/9 3. 15/80 4. 24/50
Zanzabum

Answer:

option 1

Step-by-step explanation:

\frac{3}{10} ÷ \frac{8}{5}

• leave the first fraction

• change division to multiplication

•  turn the second fraction 'upside down'

= \frac{3}{10} × \frac{5}{8} ( cancel 5 and 10 by 5 )

= \frac{3}{2} × \frac{1}{8}

= \frac{3}{16}

8 0
2 years ago
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