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Komok [63]
3 years ago
11

The forecast calls for a 10% chance of rain on Monday. It’s _______ that it will rain on Monday.

Mathematics
1 answer:
nydimaria [60]3 years ago
5 0
If the forecast calls for 10%, it’s very unlikely that it’ll rain on monday
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The equation for the circle below is x^2+y^2=64 what is the length of the circle’s radius
FromTheMoon [43]

Answer:

<h2>r = 8</h2>

Step-by-step explanation:

The standard form of an equation of a circle:

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

(h, k) - center

r - radius

We have the equation:

x^2+y^2=64\\\\(x-0)^2+(y-0)^2=8^2

Therefore the center is (0, 0) and the radius r = 8

3 0
3 years ago
Which equation would shift the parabola down 3 units? A) y = x2 B) y = x2 - 3 C) y = x2 + 3 D) y = (x + 3)2
mrs_skeptik [129]

Answer

x^2-3

Step-by-step explanation:


5 0
3 years ago
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Given: ABCD is a ∥-gram, BF ⊥ CD , BE ⊥ AD, Prove: △ABE∼△CBF
drek231 [11]

Answer:

Hence proved △ABE∼△CBF.

Step-by-step explanation:

Given,

ABCD is a parallelogram.

BF ⊥ CD    and

BE ⊥ AD

To Prove : △ABE∼△CBF

We have drawn the diagram for your reference.

Proof:

Since ABCD is a parallelogram,

So according to the property of parallelogram opposite angles are equal in measure.

\therefore m\angle A = m\angle B ⇒1

And given that BF ⊥ CD and BE ⊥ AD.

So we can say that;

m\angle F=m\angle E=90\° ⇒2

Now In △ABE and △CBF

∠A = ∠C   (from 1)

∠E = ∠F    (from 2)

So by A.A. similarity postulate;

△ABE∼△CBF

7 0
3 years ago
The center of a hyperbola is located at the origin. One focus is located at (−50, 0) and its associated directrix is represented
leva [86]

The equation of the hyperbola is : \frac{x^{2}}{48^2}  - \frac{y^{2}}{14^2}  = 1

The center of a hyperbola is located at the origin that means at (0, 0) and one of the focus is at (-50, 0)

As both center and the focus are lying on the x-axis, so the hyperbola is a horizontal hyperbola and the standard equation of horizontal hyperbola when center is at origin: \frac{x^{2}}{a^{2}}  - \frac{y^{2}}{b^{2}}    = 1

The distance from center to focus is 'c' and here focus is at (-50,0)

So, c= 50

Now if the distance from center to the directrix line is 'd', then

d= \frac{a^{2}}{c}

Here the directrix line is given as : x= 2304/50

Thus, \frac{a^{2}}{c}  = \frac{2304}{50}

⇒ \frac{a^{2}}{50}  = \frac{2304}{50}

⇒ a² = 2304

⇒ a = √2304 = 48

For hyperbola, b² = c² - a²

⇒ b² = 50² - 48² (By plugging c=50 and a = 48)

⇒ b² = 2500 - 2304

⇒ b² = 196

⇒ b = √196 = 14

So, the equation of the hyperbola is : \frac{x^{2}}{48^2}  - \frac{y^{2}}{14^2}  = 1

5 0
3 years ago
Read 2 more answers
Divide 124 in the ratio of 3:1 plsssss
UNO [17]
The answer is 93 to 31 :)
6 0
4 years ago
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