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laiz [17]
3 years ago
8

_-16(3)^2 = 80(3) +0 =0

Mathematics
1 answer:
Varvara68 [4.7K]3 years ago
3 0

Answer:

55

Step-by-step explana

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What is the length (magnitude) of the vector (2,-1)?
KiRa [710]

Answer:

\[\sqrt{5}\]

Step-by-step explanation:

The given vector is represented by (2,-1).

This can be represented in general form as (x,y) where x=2 and y=-1.

Magnitude of the vector represented as (x,y) is given by [\sqrt{x^{2}+y^{2}}\]

Evaluating for the given values of x and y,

\[\sqrt{2^{2}+(-1)^{2}}\]\[=\sqrt{4+1}\]

\[=\sqrt{5}\]

Length of the vector is \[\sqrt{5}\]

5 0
2 years ago
4 1/3 ÷ 5 1/6 the respricol of a fraction must be found to solve the problem .what is reciprocal fraction that should be used ?
emmasim [6.3K]

The denominator would be 6.


4 0
3 years ago
What is the lcm 3 and 10
alukav5142 [94]

30, you could just multiply both numbers, but it's also better to list out the factor until both match

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3 years ago
Read 2 more answers
What is the area of the octagon?
Anna007 [38]

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A=2(1+2)a2

Step-by-step explanation:

8 0
3 years ago
Find all values of c in the open interval (a, b) such that f'(c)=(f(b)-f(a))/(b-a)
timama [110]
<h3>Answer:   c = 7/4</h3>

================================================

Work Shown:

Compute the function value at the endpoints

f(x) = \sqrt{4-x}\\\\f(-5) = \sqrt{4-(-5)} = 3\\\\f(4) = \sqrt{4-4} = 0\\\\

With a = -5 and b = 4, we have

f'(c) = \frac{f(b)-f(a)}{b-a}\\\\f'(c) = \frac{f(4)-f(-5)}{4-(-5)}\\\\f'(c) = \frac{0-3}{9}\\\\f'(c) = -\frac{1}{3}\\\\

So,

f(x) = \sqrt{4-x}\\\\f'(x) = -\frac{1}{2\sqrt{4-x}}\\\\f'(c) = -\frac{1}{3}\\\\-\frac{1}{2\sqrt{4-c}} = -\frac{1}{3}\\\\

Use algebra to solve for c

-\frac{1}{2\sqrt{4-c}} = -\frac{1}{3}\\\\\frac{1}{2\sqrt{4-c}} = \frac{1}{3}\\\\3 = 2\sqrt{4-c}\\\\2\sqrt{4-c} = 3\\\\\sqrt{4-c} = \frac{3}{2}\\\\4-c = \frac{9}{4}\\\\c = 4-\frac{9}{4}\\\\c = \frac{16-9}{4}\\\\c = \frac{7}{4}\\\\

6 0
3 years ago
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