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LuckyWell [14K]
4 years ago
10

Simplify. 5 2+20 5 30 45 72 (The 5 2 is an exponent.)

Mathematics
2 answers:
Masja [62]4 years ago
7 0

                         \Huge\boxed{Your\;answer\;is\;45.}

___________________________________________________________

                                                  \boxed{Explanation}

<em />

<em>So, we have the following equation...</em>

<em />

<em />5^2+20=(?)<em />

<em />

<em>Remember, 5² also means 5x5.</em>

<em />

<em>Therefore, </em>5x5=25\\25+20=45

<em>This means your final answer is 45.</em>

<em>___________________________________________________________</em>

<em />

Glad to help!

Could i possibly get brainliest?

\huge\boxed{Thanks,\;Plip.}

<em />

<em />

nekit [7.7K]4 years ago
4 0

Answer:C.45

Step-by-step explanation:

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erica [24]
What is the focus of the parabola given by the equation y = x2 − 2x − 3?
y = x2 − 2x − 3
y = x2 − 2x − 3 -1 +1y = (x - 1)^2 - 4 h = 1 and  k = - 4 and  a = 1

Vertex (a, k) so it is  (1,-4)

Now focus is
(1, -4 + 1/4) = (1,-3 3/4)
or
(1,-3.75)
4 0
3 years ago
Find the exact value of each trigonometric function for the given angle θ.
irina [24]

Answer:

[rad] 2.41

Step-by-step explanation:

Since it gave you the point and the angle to find, you simply just have to solve for the inverse of cot. Remember cot is the opposite of tan, so cot is cos/sin. In that case, we plug into the calc (in radians):

cot^-1(-√5/2)

And we should get 2.41 as our answer!

3 0
3 years ago
Y=10x+16 looks like what on a graph
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Here is my graph. Hope it helps

7 0
3 years ago
The indicated function y1(x) is a solution of the given differential equation. use reduction of order or formula (5) in section
Ber [7]
Given that y_1=e^{2x/3}, we can use reduction of order to find a solution y_2=v(x)y_1=ve^{2x/3}.

\implies {y_2}'=\dfrac23ve^{2x/3}+v'e^{2x/3}=\left(\dfrac23v+v'\right)e^{2x/3}
\implies{y_2}''=\dfrac23\left(\dfrac23v+v'\right)e^{2x/3}+v''e^{2x/3}=\left(\dfrac49v+v'+v''\right)e^{2x/3}

\implies9y''-12y'+4y=0
\implies 9\left(\dfrac49v+v'+v''\right)e^{2x/3}-12\left(\dfrac23v+v'\right)e^{2x/3}+4ve^{2x/3}=0
\implies9v''-3v'=0

Let u=v', so that

9u'-3u=0\implies 3u'-u=0\implies u'-\dfrac13u=0
e^{-x/3}u'-\dfrac13e^{-x/3}u=0
\left(e^{-x/3}u\right)'=0
e^{-x/3}u=C_1
u=C_1e^{x/3}

\implies v'=C_1e^{x/3}
\implies v=3C_1e^{x/3}+C_2

\implies y_2=\left(3C_1e^{x/3}+C_2\right)e^{2x/3}
\implies y_2=3C_1e^x+C_2e^{2x/3}

Since y_1 already accounts for the e^{2x/3} term, we end up with

y_2=e^x

as the remaining fundamental solution to the ODE.
7 0
3 years ago
A tire shop had to fill 2 1 ⁄2 tires with air. It took a small air compressor 2 1 ⁄2 seconds to fill them up. How long would it
Brums [2.3K]

Let 2_1/2 = 5/2.

Let t = time to fill 8 tires

(5/2)/8 = (5/2)/t

8(5/2) = (5/2)t

20 = (5/2)t

20 ÷ 5/2 = t

20 • 2/5 = t

8 seconds = t

Done.

6 0
3 years ago
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