Cedro collects state quarter. He has 42 out of 50 available quarters.What is 42 out of 50 as a percent

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Answer 1

Answer:

42/50*2/2=84/100=84%

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What is the lateral surface area of a cube with side length 9 cm?

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$a=9cm\n\nsurface\ area:A_S=6a^2\to A_S=6\cdot9^2=6\cdot81=486\ (cm^2)$

1. (k+7)² =2892. (2s-1)² =225
3. (x-4)² =169

please help me answer these :)

Answers

$1. \ \ \ (k+7)^2 =289\n\n\Leftrightarrow\ \ \ k+7= √(289) \ \ \ or\ \ \ \ k+7=- √(289)\n\nk=17-7=10\ \ \ \ \ \ \ or\ \ \ \ k=-17-7=-24 \n\n2.\ \ \ (2s-1)^2 =225\n\n\Leftrightarrow\ \ \ 2s-1= √(225) \ \ \ or\ \ \ \ 2s-1=- √(225)\n\n2s=15+1\ \ \ \ \ \ \ or\ \ \ \ 2s=-15+1\n2s=16\ \ \ \ \ \ \ \ \ \ \ \ or\ \ \ \ 2s=-14\ns=8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ or\ \ \ \ s=-7 \n\n$

$3.\ \ \ (x-4)^2 =169\n\n\Leftrightarrow\ \ \ x-4= √(169) \ \ \ or\ \ \ \ x-4=- √(169)\n\nx=13+4=17\ \ \ \ \ \ \ or\ \ \ \ x=-13+4=-9$

$(k+7)^2 =289\n |k+7|=17\n k+7=17\vee k+7=-17\n k=10 \vee k=-24$

$(2s-1)^2 =225\n |2s-1|=15\n 2s-1=15\vee 2s-1=-15\n 2s=16 \vee 2s=-14\n s=8 \vee s=-7$

$(x-4)^2 =169\n |x-4|=13\n x-4=13 \vee x-4=-13\n x=17 \vee x=-9$

The temperature of a chemical reaction ranges between 30 degrees Celsius and 70 degrees Celsius. The temperature is at its lowest point when t = 0, and the reaction completes 1 cycle during a 10-hour period. What is a cosine function that models this reaction?

Answers

Let us consider function as T= Acos(teta)+B, max and min values will occur when cos(teta) = +1 and -1;
so 30= -A+B;
70= A+B;
solving both B=50, A=20;
also at t=0 , cos(teta)= -1 and finishes one cycle after t=10hour;
so(teta) = pi+pi*t/10;
so, T= 20cos(pi+pi*t/10) + 50;

I need help pls asap