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Galina-37 [17]
3 years ago
11

Hey can you please help me posted picture of question :)

Mathematics
2 answers:
otez555 [7]3 years ago
4 0
Answer:
The solutions are -3 and -6

Explanation:
First we would need to put the equation in standard form which is:
ax² + bx + c = 0
This can be done as follows:
x² + 18 = -9x
x² + 9x + 18 = 0
By comparison, we would find that:
a = 1
b = 9
c = 18

Now, to get the roots, we would use the quadratic formula shown in the attached image.

By substitution, we would find that:
either x = \frac{-9+ \sqrt{(9)^2-4(1)(18)} }{2(1)} = -3

or x = \frac{-9- \sqrt{(9)^2-4(1)(18)} }{2(1)} = -6

Hope this helps :)

soldi70 [24.7K]3 years ago
4 0
The solutioms for the equation are a and b or -3 & -6
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It cost $4 to enter a fair. Each ride costs $2.50. You have 21.50. How many rides can you go on
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3 years ago
For the cost function c equals 0.1 q squared plus 2.1 q plus 8​, how fast does c change with respect to q when q equals 11​? Det
Soloha48 [4]

Answer:

Rate of change of c with respect to q is 4.3

Percentage rate of change c with respect to q is  9.95%

Step-by-step explanation:

Cost function is given as,  c=0.1\:q^{2}+2.1\:q+8

Given that c changes with respect to q that is, \dfrac{dc}{dq}. So differentiating given function,  

\dfrac{dc}{dq}=\dfrac{d}{dq}\left (0.1\:q^{2}+2.1\:q+8 \right)

Applying sum rule of derivative,

\dfrac{dc}{dq}=\dfrac{d}{dq}\left(0.1\:q^{2}\right)+\dfrac{d}{dq}\left(2.1\:q\right)+\dfrac{d}{dq}\left(8\right)

Applying power rule and constant rule of derivative,

\dfrac{dc}{dt}=0.1\left(2\:q^{2-1}\right)+2.1\left(1\right)+0

\dfrac{dc}{dt}=0.1\left(2\:q\right)+2.1

\dfrac{dc}{dt}=0.2\left(q\right)+2.1

Substituting the value of q=11,

\dfrac{dc}{dt}=0.2\left(11\right)+21.

\dfrac{dc}{dt}=2.2+2.1

\dfrac{dc}{dt}=4.3

Rate of change of c with respect to q is 4.3

Formula for percentage rate of change is given as,  

Percentage\:rate\:of\:change=\dfrac{Q'\left(x\right)}{Q\left(x\right)}\times 100

Rewriting in terms of cost C,

Percentage\:rate\:of\:change=\dfrac{C'\left(q\right)}{C\left(q\right)}\times 100

Calculating value of C\left(q \right)

C\left(q\right)=0.1\:q^{2}+2.1\:q+8

Substituting the value of q=11,

C\left(q\right)=0.1\left(11\right)^{2}+2.1\left(11\right)+8

C\left(q\right)=0.1\left(121\right)+23.1+8

C\left(q\right)=12.1+23.1+8

C\left(q\right)=43.2

Now using the formula for percentage,  

Percentage\:rate\:of\:change=\dfrac{4.3}{43.2}\times 100

Percentage\:rate\:of\:change=0.0995\times 100

Percentage\:rate\:of\:change=9.95%

Percentage rate of change of c with respect to q is 9.95%

7 0
3 years ago
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