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Natalka [10]
2 years ago
8

1. Based on the circular

Mathematics
1 answer:
kherson [118]2 years ago
7 0

Answer:

B. More than 90

Step-by-step explanation:

Hope this helps!

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Rectangle with a side length of 11" and a diagonal of 14" what is the perimeter
notka56 [123]

Answer:

10sqrt3+22

Step-by-step explanation:

Ok, let us imagine it as a sort of rectangle split upon its diagonal.

Using that, we can Pythag it out,

11^2+b^2=14^2

121+b^2=196

b^2=75

b=sqrt75

b=5sqrt3

Ok, using this info, we find the perimeter,

5sqrt3+5sqrt3+11+11

10sqrt3+22

The answer is 10sqrt3+22

7 0
3 years ago
Read 2 more answers
One company estimates same-day delivery as more than three less than half the total number of miles. Which
-BARSIC- [3]

Answer:

D

Step-by-step explanation:

I took the quiz and got it right.

Also, the equation is y > x/2 - 3

8 0
3 years ago
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PLEASE HELP I NEED THE VALUES OF M AND P!!!!!!!! WILL GIVE BRAINLIST!!!!
ki77a [65]

Answer:

m=5 and p=4

Step-by-step explanation:

8 0
3 years ago
PLS ANSWER CORRECTLY AND FAST I NEED TO SUBMIT PLS HELP PLS PLS PLS PLS......​
astra-53 [7]

Step-by-step explanation:

  • 1

vector AB(3-(-6); 5-7)

vector AB(9;-2)

AB= \sqrt{9^{2}+(-2)^{2}  } = \sqrt{85}

  • 2

M is the midpoint of AB

we have B(-5;10) and M(1;7)

let A(x;y)

(x-5)/2 = 1 ⇒ x-5 = 2⇒ x = 7

(10=y)/2 = 7⇒ 10+y = 14 ⇒y= 4

so : A(7;4)

  • 3

the center of the circle is the midponit of the line joining both ends of the diameter

let A(x;y) be the other end

(-2+x)/2 = 2 ⇒ -2+x = 4⇒ x= 6

(5=y) = -1 ⇒ 5+y = -2 ⇒ y= -7

so the coordinates of the other end are (6; -7)

  • 4

A,B and C are collinear such as AB=BC so b is the midpoint of AC

(-5+1)/2 = y ⇒ y = -4/2 ⇒ y = -2

((-3=x)/2 = 7 ⇒ -3+x = 14 ⇒ x = 17

so x= 17 and y = -2

7 0
3 years ago
Using the bijection rule to count binary strings with even parity.
AleksandrR [38]

Answer:

Lets denote c the concatenation of strings. For a binary string <em>a</em> in B9, we define the element f(a) in E10 this way:

  • f(a) = a c {1} if a has an odd number of 1's
  • f(a) = a c {0} if a has an even number of 1's

Step-by-step explanation:

To show that the function f defined above is a bijective function, we need to prove that f is well defined, injective and surjective.

f   is well defined:

To see this, we need to show that f sends elements fromo b9 to elements of E10. first note that f(a) has 1 more binary integer than a, thus, it has 10. if a has an even number of 1's, then f(a) also has an even number because a 0 was added. On the other hand, if a has an odd number of 1's, then f(a) has one more 1, as a consecuence it will have an even number of 1's. This shows that, independently of the case, f(a) is an element of E10. Thus, f is well defined.

f is injective (or one on one):

If a and b are 2 different binary strings, then f(a) and f(b) will also be different because the first 9 elements of f(a) form a and the first elements of f(b) form b, thus f(a) is different from f(b). This proves that f in injective.

f is surjective:

Let y be an element of E10, Let x be the first 9 elements of y, then f(x) = y:

  • If x has an even number of 1's, then the last digit of y has to be 0, and f(x) = x c {0} = y
  • If x has an odd number of 1's, then the last digit of y has to be a 1, otherwise it wont be an element of E10, and f(x) = x c {1} = y

This shows that f is well defined from B9 to E10, injective, and surjective, thus it is a bijection.

3 0
3 years ago
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