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3 years ago

Plzzzz its due in 15mins1.

Mathematics

1 answer:

BigorU [14]3 years ago

5 0

P = -3
m = -22
p = 4
n = -9

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I want to know the answer of this question

Molodets [167]

$\bf \textit{circumference of a circle}\\\\
C=2\pi r\qquad 
\begin{cases}
r=radius\\
-----\\
r=5x^2
\end{cases}\implies C=2\pi 5x^2\implies C=10\pi x^2
\\\\\\
\textit{area of a circle}\\\\
A=\pi r^2\qquad 
\begin{cases}
r=radius\\
-----\\
r=5x^2
\end{cases}\implies A=\pi (5x^2)^2\implies A=\pi (5^2x^{2\cdot 2})
\\\\\\
A=\pi (25x^4)\implies A=25\pi x^4$

4 0

3 years ago

In the equation y = X - 5, the y-intercept is

Semenov [28]

-5 is the intercept since y=mx+b, b being the intercept.

8 0

3 years ago

Read 2 more answers

I can't figure it out

kap26 [50]

Answer:

the first one

Step-by-step explanation:

this picture has the workings

8 0

3 years ago

The function h is a quadratic function whose graph is a translation 3 units left and 4 units up of the parent

jeka57 [31]

Answer:

D ( if you add +4 to the (x + 3)^2)

Step-by-step explanation:

Parent function is f(x) = x^2

A translation 3 units left gives y = )x + 3)^2

- and 4 up gives y = (x + 3)^2 + 4 - vertex form.

Standard form :

y = x^2 + 6x + 9 + 4

= x^2 + 6x + 13.

4 0

3 years ago

Suppose n(x)= 22 and n(y) is 13 what is the maximum number of elements X ∪Y can have ?

Nikolay [14]

The inclusion/exclusion principle states that

$|X\cup Y|=|X|+|Y|-|X\cap Y|$

That is, the union has as many members as the sum of the number of members of the individual sets, minus the number of elements contained in both sets (to avoid double-counting).

Therefore, $|X\cup Y|$ will have the most elements when the sets $X$ and $Y$ are disjoint, i.e. $X\cap Y=\emptyset$ , which would mean the most we can can in this case would be

$|X\cup Y|=|X|+|Y|=22+13=35$

(Note that $n(X)=|X|$ denotes the cardinality of the set $X$ .)

3 0

4 years ago