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3 years ago

The circle given by

Mathematics

2 answers:

VikaD [51]3 years ago

6 0

Answer:

the value of k is 1 like the other person said

Step-by-step explanation:

mixer [17]3 years ago

5 0

Typically, the equation of a circle is given like this: (x-h)^2 + (y-k)^2 = r^2, where r is the radius of the circle and the point (h,k) is the center. To find the value k, you need to put the first equation you have into this form. Standard form requires that you only have the1 y variable, as seen in the parentheses, so you first need to complete the square.

x^2 + y^2 - 2y = 11
x^2 + y^2 - 2y +1 = 11 + 1
x^2 + (y - 1)^2 = 12

And that puts the equation in standard form. k = 1.

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A=72 and c=75 solve b

Andrei [34K]

Answer:

b = 3 or -3 i suppose

Step-by-step explanation:

3 0

3 years ago

Assume that x is a char variable has been declared and already given a value. Write an expression whose value is true if and onl

Lorico [155]

Answer:

bool b = isupper(x);

Step-by-step explanation:

I have written the expression for a char variable x.The isupper(x) will return true if the character x is upper case and false if the character x is lower case.

I have stored the returned value to a bool variable b .So the value of variable b will be true only when the x is in uppercase and false when b is lower case.

4 0

2 years ago

Calculus application

Ugo [173]

Answer:

49m/s

59.07 m/s

Step-by-step explanation:

Given that :

Distance (s) = 178 m

Acceleration due to gravity (a) = g(downward) = 9.8m/s²

Velocity (V) after 5 seconds ;

The initial velocity (u) = 0

Using the relation :

v = u + at

Where ; t = Time = 5 seconds ; a = 9.8m/s²

v = 0 + 9.8(5)

v = 0 + 49

V = 49 m/s

Hence, velocity after 5 seconds = 49m/s

b) How fast is the ball traveling when it hits the ground?

V² = u² + 2as

Where s = height = 178m

V² = 0 + 2(9.8)(178)

V² = 0 + 3488.8

V² = 3488.8

V = √3488.8

V = 59.07 m/s

8 0

2 years ago

Simplify square root of 2 over cube root of 2.

Fed [463]

Answer:

2 to the power of one sixth

Step-by-step explanation:

Assuming you don't already know this, any type of root can be expressed as an exponent. Generally speaking:

$\sqrt[n]{x} = {x}^{ \frac{1}{n} }$

So you can rewrite the given fraction as

$\frac{ {2}^{ \frac{1}{2} } }{ {2}^{ \frac{1}{3} } }$

and then reduce as you normally would. That is, if the bases of the numerator and denominator are the same, then you can subtract the denominator's exponent from the numerator's exponent like so:

$\frac{ {2}^{ \frac{1}{2} } }{ {2}^{ \frac{1}{3} } } = {2}^{ \frac{1}{2} - \frac{1}{3} }$