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Elanso [62]
3 years ago
6

What is the length of bc? AC = 8 AB = 17

Mathematics
1 answer:
padilas [110]3 years ago
5 0

Using Pythagoras theorem,

17^2=8^2+(BC)^2\implies BC=\sqrt{17^2-8^2}=15.

Hope this helps.

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slavikrds [6]
100x 0.18 as it equals 18 when you multiply it 
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3 • X = 5.25.<br> What is X?
Archy [21]
Your answer will be x=1.75
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Read 2 more answers
-3(2x+4) = 3(x-6)<br> i need steps too
Bezzdna [24]

If you are looking to find the value of x then you have to use distributive property.

1st: multiply -3 by 2x and 4                -3(2x+4)

                                                             -6x -12

2nd: do the same thing on the other side which is multiply 3 by x and -6

                             3(x-6)

                               3x-18

3rd: Then you have  -6x-12=3x-18

                                        +12     +12

add 12 to each side so then you have -6x=3x-6

then divide by -6 on each side and the answer is x= 2/3

6 0
3 years ago
(4+5)÷3×4= <br>can someone help me with this question please
Darina [25.2K]
PEMDAS
Parentheses
Exponents
Multiplication
Division
Addition
Subtraction
(4+5)÷3×4
9÷3•4
3•4
12
8 0
3 years ago
a) Suppose f"(z) exists on an interval I and f(s) has a zero at three distinct points a &lt; b&lt; c on I. Show there is a pbint
Fynjy0 [20]

Answer:

We can find a root in the average point (a+b+c)/3

Step-By-Step Explanation:

We can use the Rolle Theorem. Since f is two times deribable on both [a,b] and [b,c], then there exists points x in (a,b) and y in (b,c) such that f'(x) = f'(y) = 0. Now, again by using Rolle Theorem, since f' is derivable in [x,y] (because it is a closed interval in I), then there exists s in [x,y] such that f''(s) = 0. This proves a.

The cube (x-a)(x-b)(x-c) has 3 roots, a, b and c and it is two times derivable because it is a polynomial. Hence we can use (a) to ensure that there is a root on I. Nevertheless, we can try to find the root manually:

f'(x) = (x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)

f''(x) = (x-c)+(x-b)+(x-c)+(x-a)+(x-b)+(x-a) = 2(x-a)+2(x-b)+2(x-c) = 2( (x-a) + (x-b) + (x-c) ) = 6x - 2 (a+b+c).

We want x such that

6x - 2(a+b+c) = 0, or, equivalently,

3x - (a+b+c) = 0

Hence

x = (a+b+c)/3

Is a root of f'' in I.

3 0
3 years ago
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