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3 years ago

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Twice a year, Billings and his team Drive 1,000 miles from Colorado to California to deliver statuettes and pick up grammium. Th

at's 2,000 less than 1/2 the amount of Grammium they use each year in pounds. Write and solve an equation to find out how much Grammium they use to each year to make the awards, using the variable G

1 answer:

8090 [49]3 years ago

4 0

Answer:

Step-by-step explanation:

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Find all critical points of the given plane autonomous system. (Enter your answers as a comma-separated list.) x′ = x 12 − x − 1

Andru [333]

Answer:

the critical points are (0,0) , (0, 20), (12, 0) , (4,16)

Step-by-step explanation:

To consider the autonomous system

$x' =x (12 -x - \dfrac{1}{2})$

$y' = y( 20 -y - x)$

The critical points of the above system can be derived by replacing x' = o and y' = 0.

i.e.

$x' =x (12 -x - \dfrac{y}{2}) = 0$

$\dfrac{x}{2} (24 -2x - y) = 0$

x = 0 or 24 - 2x - y = 0 ----- (1)

Also

$y' = y( 20-y-x) = 0$

y( 20 -y - x) = 0

y = 0 or 20 - y - x = 0 ----- (2)

By solving (1) and (2);

we get x = 4 and y = 16

Suppose x = 0 from (2)

y = 20

Also;

if y = 0 from (1)

x = 12

Thus, the critical points are (0,0) , (0, 20), (12, 0) , (4,16)

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2 years ago

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3 years ago

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Ipatiy [6.2K]

The answer is No, they are not proportional. This is because 8/42 is not equal to 20/105.

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3 years ago

Simplify 7.021 x 0.15 25pts

Helen [10]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

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