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prohojiy [21]
3 years ago
15

What does k equal  in 17 + k = -7

Mathematics
2 answers:
lesya692 [45]3 years ago
5 0

17 + k = -7

k = -7 - 17

k = -24

MakcuM [25]3 years ago
4 0
K equals -24 because to find k you subtract 17 from -7
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What is the length of the ladder? It is 6 ft. from the house at the bottom and touches the wall 14 ft. up at the top. Simplify y
Sauron [17]

Answer:

15.2

{c }^{2}  =  {a}^{2}    +  {b}^{2}  \:  \\  {c}^{2}  ={6}^{2}  +  {14}^{2}  \\  {c}^{2}  = 36 + 196  \\  {c}^{2}  = 232 \\  \sqrt{c }  =  \sqrt{232}  \\ c = 15.23154621

4 0
3 years ago
Determine the singular points of the given differential equation. Classify each singular point as regular or irregular. (Enter y
Zina [86]

Answer:

Step-by-step explanation:

The given differential equation is:

x^3y'' + 2x^2y' + 4y

the main task here is to determine the singular points of the given differential equation and Classify each singular point as regular or irregular.

So, for a regular singular point ;  x=x_o is  located at the first power in the denominator of P(x) likewise at the Q(x) in the second power of the denominator. If that is not the case, then it is termed as an irregular singular point.

Let first convert it to standard form by dividing through with x³

y'' + \dfrac{2x^2y'}{x^3} + \dfrac{4y}{x^3} =0

y'' + \dfrac{2y'}{x} + \dfrac{4y}{x^3} =0

The standard form of the differential equation is :

\dfrac{d^2y}{dy} + P(x) \dfrac{dy}{dx}+Q(x)y =0

Thus;

P(x) = \dfrac{2}{x}

Q(x) = \dfrac{4}{x^3}

The zeros of x,x^3  is 0

Therefore , the singular points of above given differential equation is 0

Classify each singular point as regular or irregular.

Let p(x) = xP(x)    and q(x) = x²Q(x)

p(x) = xP(x)

p(x) = x*\dfrac{2}{x}

p(x) = 2

q(x) = x²Q(x)

q(x) = x^2 * \dfrac{4}{x^3}

q(x) =\dfrac{4}{x}

The function (f) is analytic if at a given point a it is represented by power series in x-a either with a positive or infinite radius of convergence.

Thus ; from above; we can say that q(x) is not analytic  at x = 0

Q(x) = \dfrac{4}{x^3}  do not satisfy the condition,at most to the second power in the denominator of Q(x).

Thus, the point x =0 is an irregular singular point

6 0
3 years ago
Please help I don’t understand
charle [14.2K]

Answer:

\huge\boxed{\sf

Step-by-step explanation:

Since AB is parallel to CD,

∠h = 80°    (alternate angles)

\rule[225]{225}{2}

Hope this helped!

<h3>~AH1807</h3>
6 0
3 years ago
Read 2 more answers
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