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3 years ago

Can you help me please? ( I didn't mean to pick a answer)

Mathematics

1 answer:

SpyIntel [72]3 years ago

8 0

Answer:

#3:8n-4

Step-by-step explanation:

combine like terms

3n+n=4n

-7+5=-2

4n-2

multiply that by 2 to get the perimeter, which is 8n-4

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What’s 2percent of 300

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Step-by-step explanation:

2%=0.02

0.02*300=6

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3 years ago

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Step-by-step explanation:

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3 years ago

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3 years ago

Question 2(Multiple Choice Worth 4 points) (08.02)A pair of equations is shown below. x + y = 5 y = one halfx + 2 If the two equ

8_murik_8 [283]

Answer:

$(2,3)$

Step-by-step explanation:

The first equation is $x+y=5$

The second equation is $y=\frac{1}{2}x+2$

When we graph these two equations, <em>they will meet at a point which represent the solution of the two equations</em>.

We can solve the two equations simultaneously to determine their point of intersection.

Let us substitute the second equation into the first equation to get;

$x+\frac{1}{2}x+2=5$

Multiply through by 2 to get;

$2x+x+4=10$

Group similar terms to obtain;

$2x+x=10-4$

Simplify;

$3x=6$

Divide both sides by 3;

$\Rightarrow x=2$

Put $x=2$ into the second equation;

$y=\frac{1}{2}(2)+2$

$\Rightarrow y=1+2$

$\Rightarrow y=3$

Therefore the graphs of the two functions intersect at (2,3)

See graph in attachment.

3 0

3 years ago

Will someone help with this ! I was allowed a retry for it and I don’t know how to do it

pashok25 [27]

Answer:

$\frac{dN}{dt}=k(725-N)\\separating the variables and integrating\\\int \frac{dN}{725-N}=\int kdt+c\\-log (725-N)=kt+c\\log (725-N)=-kt-c\\(725-N)=e^{-kt-c} =e^{-kt} *e^{-c} =Ce^{-kt} \\when t=0,N=400\\725-400=Ce^{0} =C\\C=325\\when t=3,N=650\\725-650=325e^{-3k} \\\frac{75}{325}=(e^{-k})^3\\\\e^{-k} =(\frac{3}{13}) ^{\frac{1}{3} } \\when t=5\\725-N=325(\frac{3}{13}) ^{\frac{5}{3} } =325*\frac{3}{13}*(\frac{3}{13} )^{\frac{2}{3}}\\N=725-75*(\frac{3}{13} )^{\frac{2}{3} }=725-28=697$

Step-by-step explanation:

6 0

3 years ago

Can you help me please? ( I didn't mean to pick a answer)​

Can you help me please? ( I didn't mean to pick a answer)