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goldenfox [79]
4 years ago
10

The temperature of a chemical compound, in degrees Celsius, changes when the compound is exposed to UV light. The table below sh

ows the temperature of the compound, in degrees Celsius, x minutes after the compound is exposed to UV light.
x
0
2
4
6
8
f(x)
15
15.5
17
19.5
23

Write the quadratic function that corresponds to the given table of values in standard form, f(x) = ax2 + bx + c, where a, b and c are constants.

Mathematics
2 answers:
Vsevolod [243]4 years ago
6 0
The temperature of the compound will follow this function: 1/8x^2+15 ( I could be wrong).

But how did I come up with this function ??
The standard of a quadratic equation is given...the x is given...the f(x) is given. Now plug these in correspondence to find the constants. The rest is algebra. If you are not getting this answer, let me see your work
ratelena [41]4 years ago
3 0

Hope I was able to help you

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All I need is the answer Please and thank you
Alex73 [517]

Should be 40ft. If not, I dearly apologize.

5 0
3 years ago
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I WILL GIVE BRAINIEST IF YOU ANSWER ASAP AND CORRECT!!!
Aleks04 [339]

Answer:

a-y=2x+9

b:y=2x-9

c:4x+9=y

d:y=3x+9

6 0
3 years ago
Which expression is equivalent to 162x^4-144x^2+32?
uysha [10]
I didn't really get the which part because we shape it to serveral factors, this is one of them

6 0
3 years ago
Otieno is k years old.He is 5 years older than kidole what will be their age fours to come​
EleoNora [17]

Answer:

Otieno's age in 4 years: k+4

Kidole's age in 4 years: k-1

Step-by-step explanation:

k = Otieno's age

k-5 = Kidole's age

Their ages will increase by 4 years when 4 years pass.

k+4 = Otieno's age in 4 years

k-5+4 = Kidole's age in 4 years

k-1 = Kidole's age in 4 years (after simplifying the expression)

I hope this helps!

6 0
3 years ago
Evaluate the integral Integral ∫ from (1,2,3 ) to (5, 7,-2 ) y dx + x dy + 4 dz by finding parametric equations for the line seg
n200080 [17]

\vec F(x,y,z)=y\,\vec\imath+x\,\vec\jmath+3\,\vec k

is conservative if there is a scalar function f(x,y,z) such that \nabla f=\vec F. This would require

\dfrac{\partial f}{\partial x}=y

\dfrac{\partial f}{\partial y}=x

\dfrac{\partial f}{\partial z}=3

(or perhaps the last partial derivative should be 4 to match up with the integral?)

From these equations we find

f(x,y,z)=xy+g(y,z)

\dfrac{\partial f}{\partial y}=x=x+\dfrac{\partial g}{\partial y}\implies\dfrac{\partial g}{\partial y}=0\implies g(y,z)=h(z)

f(x,y,z)=xy+h(z)

\dfrac{\partial f}{\partial z}=3=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=3z+C

f(x,y,z)=xy+3z+C

so \vec F is indeed conservative, and the gradient theorem (a.k.a. fundamental theorem of calculus for line integrals) applies. The value of the line integral depends only the endpoints:

\displaystyle\int_{(1,2,3)}^{(5,7,-2)}y\,\mathrm dx+x\,\mathrm dy+3\,\mathrm dz=\int_{(1,2,3)}^{(5,7,-2)}\nabla f(x,y,z)\cdot\mathrm d\vec r

=f(5,7,-2)-f(1,2,3)=\boxed{18}

8 0
3 years ago
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