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11Alexandr11 [23.1K]
3 years ago
9

Number 2 help me plz

Mathematics
2 answers:
Verizon [17]3 years ago
7 0

Answer:

option (b)...........

nadya68 [22]3 years ago
3 0

Answer:

D

Step-by-step explanation:

cuz 2.5% students voted for the hippo and not 25%

can I get brainliest if I got it right?

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Doris made a list of all the whole number from 1 to 100 how many times did she write the digit 2
STALIN [3.7K]
She wrote it 22 times
3 0
4 years ago
Given that the diameter of Circle A is 6 cm , and the radius of Circle B is 18 cm , what can be concluded about the two circles?
ELEN [110]

Answer:

The radius of circle B is 6 times greater than the radius of circle A

The area of circle B is 36 times greater than the area of circle A

Step-by-step explanation:

we have

<em>Circle A</em>

D=6\ cm

The radius of circle A is

r=6/2=3\ cm -----> the radius is half the diameter

<em>Circle B</em>

r=18\ cm

Compare the radius of both circles

3\ cm< 18\ cm

18=6(3)

The radius of circle B is six times greater than the radius of circle A

Remember that , if two figures are similar, then the ratio of its areas is equal to the scale factor squared

All circles are similar

In this problem the scale factor is 6

so

6^{2}=36

therefore

The area of circle B is 36 times greater than the area of circle A

8 0
3 years ago
5. The recursive algorithm given below can be used to compute gcd(a, b) where a and b are non-negative integer, not both zero.
s2008m [1.1K]

Implementating the given algorithm in python 3, the greatest common divisors of <em>(</em><em>124</em><em> </em><em>and</em><em> </em><em>244</em><em>)</em><em> </em>and <em>(</em><em>4424</em><em> </em><em>and</em><em> </em><em>2111</em><em>)</em><em> </em>are 4 and 1 respectively.

The program implementation is given below and the output of the sample run is attached.

def gcd(a, b):

<em>#initialize</em><em> </em><em>a</em><em> </em><em>function</em><em> </em><em>named</em><em> </em><em>gcd</em><em> </em><em>which</em><em> </em><em>takes</em><em> </em><em>in</em><em> </em><em>two</em><em> </em><em>parameters</em><em> </em>

if a>b:

<em>#checks</em><em> </em><em>if</em><em> </em><em>a</em><em> </em><em>is</em><em> </em><em>greater</em><em> </em><em>than</em><em> </em><em>b</em>

return gcd (b, a)

<em>#if</em><em> </em><em>true</em><em> </em><em>interchange</em><em> </em><em>the</em><em> </em><em>Parameters</em><em> </em><em>and</em><em> </em><em>Recall</em><em> </em><em>the</em><em> </em><em>function</em><em> </em>

elif a == 0:

return b

elif a == 1:

return 1

elif((a%2 == 0)and(b%2==0)):

<em>#even</em><em> </em><em>numbers</em><em> </em><em>leave</em><em> </em><em>no</em><em> </em><em>remainder</em><em> </em><em>when</em><em> </em><em>divided</em><em> </em><em>by</em><em> </em><em>2</em><em>,</em><em> </em><em>checks</em><em> </em><em>if</em><em> </em><em>a</em><em> </em><em>and</em><em> </em><em>b</em><em> </em><em>are</em><em> </em><em>even</em><em> </em>

return 2 * gcd(a/2, b/2)

elif((a%2 !=0) and (b%2==0)):

<em>#checks</em><em> </em><em>if</em><em> </em><em>a</em><em> </em><em>is</em><em> </em><em>odd</em><em> </em><em>and</em><em> </em><em>B</em><em> </em><em>is</em><em> </em><em>even</em><em> </em>

return gcd(a, b/2)

else :

return gcd(a, b-a)

<em>#since</em><em> </em><em>it's</em><em> </em><em>a</em><em> </em><em>recursive</em><em> </em><em>function</em><em>,</em><em> </em><em>it</em><em> </em><em>recalls</em><em> </em><em>the function</em><em> </em><em>with </em><em>new</em><em> </em><em>parameters</em><em> </em><em>until</em><em> </em><em>a</em><em> </em><em>certain</em><em> </em><em>condition</em><em> </em><em>is</em><em> </em><em>satisfied</em><em> </em>

print(gcd(124, 244))

print()

<em>#leaves</em><em> </em><em>a</em><em> </em><em>space</em><em> </em><em>after</em><em> </em><em>the</em><em> </em><em>first</em><em> </em><em>output</em><em> </em>

print(gcd(4424, 2111))

Learn more :brainly.com/question/25506437

6 0
3 years ago
How? Just how I'm horrible at math​
Oksanka [162]

Answer:

<h3>c)</h3><h3>\frac{x + 2}{x - 3} . \frac{4}{x + 2}</h3>

3 0
3 years ago
Joel wants to fence off a triangular portion of his yard for his chickens. The three pieces of fencing he haas to use are 8 feet
Sedaia [141]

No, it will not be able to make a right triangle.

<em><u>Solution:</u></em>

Given that,

Joel wants to fence off a triangular portion of his yard for his chickens.

The length of three peices of fencing are 8 feet , 15 feet and 20 feet

For making it a right triangle it must satisfy the "Pythagorus theorem"

Pythagorean theorem, states that the square of the length of the hypotenuse is equal to the sum of squares of the lengths of other two sides of the right-angled triangle.

For a right angled triangle with sides a, b, c

c^2 = a^2+b^2

Given sides are:

a = 8 feet

b = 15 feet

c = 20 feet

Substituting the values in above formula

20^2 = 8 ^2+15^2\\\\400 = 64 + 225\\\\400\neq 289

Since the pythogoras formula is not satisfied, the given sides cannot form a right angled triangle

Thus , No, it will not be able to make a right triangle.

7 0
3 years ago
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